Math 227

Solutions of Tech Demo

Homework Problems

Department of Mathematics

Los Angeles Mission College

August 2019

Stat Tables > Contingency > With Summary Select Column(s): Frequency Row Labels: Response Display: Percent of total Compute!

Click Open in StatCrunch

Input the results with the proper rounding in the relative frequency column.

Use the relative frequency table obtained from part (a). Input the New York relative frequency.

Graph Bar Plot > With Summary Categories in: Response Counts in: Frequency Type: Frequency Order by: Worksheet Display: Check Value above bar Compute!

From the frequency bar plot, click Options Edit Scroll down to Type: and select Relative Frequency Compute!

Close the relative frequency bar plot. Graph Pie Chart > With Summary Categories in: Response Counts in: Frequency Display: Highlight Count and Percent of Total Order by: Worksheet Compute!

To construct a frequency distribution, we will construct a frequency histogram of the data first which is the question for part (c). From the histogram, we will fill in the table. Graph Histogram Select Columns: var1 Type: Frequency Bins: Start at: 0; Width: 0.5 Display options: Check Value above bar: Compute!

From the histogram, each increment difference is 0.5. Each increment represents the lower class limit of each class. The upper class limit of the first class is the maximum number with 2 decimal places (based on the given raw data) that is smaller than 0.5. The upper limit of the second class is the maximum number with 2 decimal places that is smaller than 1. The upper limit of the second class is 0.99. Use the similar logic to complete the frequency distribution in a table format.

From the histogram output, click Options Edit Scroll down to Type: change it to Relative Frequency Compute!

The class limits for each class for part (a) are the same as for part (b). Pair each class with the appropriate relative frequency.

We have done this for part (a). Look at the histogram from part (a) or redo the steps for obtaining a frequency histogram, then select the appropriate answer.

We have done this for part (b). Look at the relative frequency histogram from part (b) or redo the steps for obtaining a relative frequency histogram, then select the appropriate answer.

Use part (a) as a reference. The only change is Bins: Width: 1 The 1st class limits are from 0 to 0.99; the 2nd class limits are from 1 to 1.99; etc.

Use part (b) as a reference. From the histogram of part (a), click Options Type: Relative Frequency The class limits for each class is the same as part (f). Fill in the relative frequency.

We have done this for part (f). Look at the histogram from part (f) or redo the steps for obtaining a frequency histogram, then select the appropriate answer.

We have done this for part (g). Look at the relative frequency histogram from part (g) or redo the steps for obtaining a relative frequency histogram, then select the appropriate answer.

Open worksheet in StatCrunch Graph Stem and Leaf Select column(s): var1 Leaf unit: 1 Compute!

StatCrunch created a double Stem and Leaf plot. In order to select the appropriate answer, we need to redo it as a single Stem and Leaf by joining the leaves from the double lines into a single line using the same stem.

0 89 1 9 2 127 3 2688 4 1223449 5 1335556

Open the worksheet in StatCrunch. From the choices of the answer, we need to graph a histogram. However, we need to figure out what is the starting point and the width for the Bins. There are 4 rectangles from 0 to 80 so the width is (80-0)/4 = 20. The starting point is 0. Graph Histogram Select column(s): var1 Bins: Start at: 0; Width: 20 Display options: Check Value above bar:

Based on the choices of your answer, the start point of the histogram is 0 and the width is 5. Open the worksheet in StatCrunch Graph Histogram Input the information as shown below.

Close the display for Histogram Click Stat Summary Stats Columns Click var1 for Select column(s): Take the default for Statistics: Compute!

Mean = 17.467 hours

Median = 17 hours

Since the distribution shape is symmetric, the mean is used for the central tendency of hours worked by college students outside the home.

Open worksheet in StatCrunch Stat Summary Stats Columns Click var1 for Select column(s): Take the default for Statistics: Compute!

Std. dev. = 0.065619808 = 0.07 grams

According to the Empirical Rule, 68% of the data lie within mean +/- one Std. dev. 95% of the data lie within mean +/- two Std. dev. 99.7% of the data lie within mean +/- three Std. dev. We need to find out 0.7 is how many Std. dev. from the mean (0.8396 grams). 0.84 - 0.07 - 0.07 = 0.84 - 2(0.07) = 0.7 0.7 is two Std. dev. from the mean (0.84 grams). We need to check if 0.98 is two Std. dev. above the mean. 0.84 + 2(0.07) = 0.98 The answer for (c) is 95% according to the Empirical Rule.

The answer obtained from part (c) is the theoretical percentage. To obtain the actual percentage, we will sort the raw data from the worksheet, then count how many candies actually weigh between 0.7 and 0.98 gram, inclusive. Convert the answer to a percentage. Close the summary statistics dialog box From the worksheet, click Data Sort Click var1 for Select Columns: Compute!

Look at the Sort(var1) column; count how many candies weigh in between 0.7 gram and 0.98 gram; divide the number of candies by the total (50) and convert it to a percentage. Based on this Sort(Var1) on the left side, there is no candy weighs less than 0.7 gram and no candy weighs more than 0.98 gram. This means all candies weigh between 0.7 gram and 0.98 gram. The answer for (d) is 100%.

How many Std dev. is 0.91 away from the mean (0.84)? 0.84 + 0.07 = 0.91 0.91 is 1 Std dev. above the mean (0.84). According to the Empirical rule, 68% of the data lie within mean +/- one Std. dev. This means 32% of the data lie outside mean +/- one Std. dev 16% of the data lie below mean minus one Std. dev. and 16% of the data lie above mean plus one Std. dev. The question only asks for the percentage of the data lie above 0.91 gram which is above mean plus one Std. dev.. The answer is 16%.

From the Sort(var1) for part (d), there are 8 candies weigh above 0.91 gram from the raw data. Percentage = 8/50*100% = 16% The actual percentage is 16% which happened to be the same as the theoretical percentage obtained from the Empirical rule for part (e).

Open the worksheet in StatCrunch Stat Summary Stats Columns Drag both Car1 and Car2 for Select column(s): Take the default for Statistics Compute! Use the results to answer all questions.

Open the worksheet in StatCrunch Stat Summary Stats Columns Click var1 for Select column(s): Take the default for Statistics Compute!

The five-number summary is Min, Q1, Median, Q3, Max.

Graph Boxplot Click var1 for Select column(s): Check Draw boxes horizontally Compute!

Graph Boxplot Drag both Name Brand and Store Brand for Select column(s): Check Draw boxes horizontally Compute!

Note: The multiple choice selections have the Name Brand's boxplot on top and the Store Brand at the bottom. Adjust the above accordingly.

The median of the name brand has a higher value than the median of the store brand.

The IQR (Q3 - Q1) of the name brand is smaller than the IQR (Q3-Q1) of the store brand.

For the accompanying data set, (a) draw a scatter diagram of the data, (b) compute the correlation coefficient, and (c) determine whether there is a linear relation between x and y. x y 7 3 6 2 6 6 7 9 9 5

Critical value Table n 3 0.997 4 0.950 5 0.878 6 0.811 7 0.754

(a) Draw a scatter diagram of the data. Choose the correct graph below.

Open StatCrunch through the data icon Select Graph Scatter Plot elect X column for x variable and Y column for y variable Compute! Be sure to notice the StatCrunch scatter plot begins at (6,2). The scatter plots to select your answer from begin at (0,0)

(b) Compute the correlation coefficient. Open StatCrunch Stat Correlation Select column(s) X & Y Compute! The correlation coefficient is r = 0.1490712 ≈ 0.149 (c) Determine whether there is a linear relation between x and y.

Because the coefficient is positive and the absolute value of the correlation coefficient 0.149 is not greater than the critical value for the data set, 0.878, no linear relation exists between x and y. (Round to three decimal places as needed.)

The data in the table to the right are based on the results of a survey comparing the commute time of adults to their score on a well-being test. Complete parts (a) through (d) below. Open StatCrunch through the data icon. We will use it shortly. Commute_Time Score Critical values for

3

69.4

Correlation coefficient

17 68.4 n 24 67.4 3 0.997 36 67.3 4 0.950 50 66.7 5 0.878 72 65.7 6 0.811 96 63.1 7 0.754

(a) Which variable is likely the explanatory variable and which is the response variable?

A. The explanatory variable is commute time and the response variable is the well-being score because commute time affects well-being score.

B. The explanatory variable is the well-being score and the response variable is commute time because commute time affects well-being score.

C. The explanatory variable is the well-being score and the response variable is commute time because the well-being score affects the commute time.

D. The explanatory variable is commute time and the response variable is the well-being score because the well-being score affects the commute time score.

(b) Draw a scatter diagram of the data. In StatCrunch, Select Graph Scatter Plot Commute Time for X variable and Score for Y

variable Compute!

Which of the following represents the data?

Be sure to notice the StatCrunch scatter plot begins at (0,63). The scatter plots to select your answer from begin at (0,60). (c) Determine the linear correlation coefficient between commute time and well-being score. Open StatCrunch Stat Summary Stats Correlation Select column(s): Select both commute time and score Compute! Correlation between Commute_Time and Score is: -0.97739224 r = -0.97739224 ≈ -0.977 (Round to three decimal places as needed.) (d) Does a linear relation exist between the commute time and well-being index score? We compare r obtained from part (c) with the critical value r from the Critical Value Table. Open the Critical Value Table and identify the r critical.

Critical values for Correlation Coefficient

n 3 0.997 4 0.950 5 0.878 6 0.811 7 0.754

A. Yes, there appears to be a positive linear association because r is positive and is greater than the critical value.

B. Yes, there appears to be a negative linear association because r is negative and is less than the negative of the critical value.

C. Yes, there appears to be a positive linear association because r is positive and is less than the critical value.

D. No, there is no linear association since r is positive and is less than the critical value.

A pediatrician wants to determine the relation that exists between a child's height, x, and head circumference, y. She randomly selects 11 children from her practice, measures their heights and head circumferences and obtains the accompanying data. Complete parts (a) through (g).

Height, x (inches)

Head circumference, y (inches)

27.75 17.5 24.5 17.1 25.5 17.1 25.5 17.1 25 16.9

27.75 17.6 27 17.3

27.75 17.5 26.75 17.3 26.75 17.5 27.5 17.5

Answers are on following pages. See below. (a) Find the least-squares regression line treating height as the explanatory variable and head circumference as the response variable. Open SatCrunch through the data icon Stat Regression Simple Linear Height for X variable and head circumference for Y variable Compute! Simple linear regression results: Dependent Variable: Head circumference, y (inches) Independent Variable: Height, x (inches) Head circumference, y (inches) = 12.539091 + 0.18 Height, x (inches) Sample size: 11 R (correlation coefficient) = 0.9176112 R-sq = 0.84201031 Estimate of error standard deviation: 0.096373642 Parameter estimates: Parameter Estimate Std. Err. Alternative DF T-Stat P-value Intercept 12.539091 0.68934872 ≠ 0 9 18.189765 <0.0001 Slope 0.18 0.025990038 ≠ 0 9 6.9257304 <0.0001

Analysis of variance table for regression model: Source DF SS MS F-stat P-value Model 1 0.4455 0.4455 47.965742 <0.0001 Error 9 0.083590909 0.0092878788 Total 10 0.52909091 (a) Find the least-squares regression line treating height as the explanatory variable and head circumference as the response variable. The least-squares regression line is 𝑦𝑦� = 0.1800x + 12.5391. (Round to four decimal places as needed.) (b) Interpret the slope and y-intercept, if appropriate. Interpret the slope. If height increase by 1 inch, head circumference increases by about 0.1800 inches on average. Interpret the y-intercept. Choose the correct answer below.

A. It is not appropriate to interpret the y-intercept. It is outside the scope of the model. B. The y-intercept is the child's head circumference when the child's height is 0 inches. C. If height increases by 1 inch, head circumference increases by about 0.1800 inch, on

average.

(c) Use the regression equation to predict the head circumference of a child who is 25 inches tall. You can use StatCrunch to perform this calculation. If your window is still open, Select the Options on the upper left of the results window Edit Scroll down to Prediction of Y:, input 25 for the V value Compute! Predicted values: X value Pred. Y 25 17.039091 I deleted the extra information. The predicted value of the head circumference of a child who is 25 inches tall is 17.04 inches. (Round to two decimal places as needed.) (d) Compute the residual based on the observed head circumference of the 25-inch-tall child in the table. Is the head circumference of this child above average or below average? This, too, can be found with SatCrunch. You can go back to the Options and edit; scroll down in the window to Save: and click on Residuals then Compute! You will now have the residuals for each y value saved in your data table view. You can also calculate it very easily Take the observed value (16.9) and subtract the predicted value of 17.04. 16.9-17.04 = −0.14 inches. (Round to two decimal places as needed.) (e) Draw the least-squares regression line on the scatter diagram of the data and label the residual from part (d). Choose the correct graph below.

If your window is still open, click on the > symbol on the lower right corner for the graph.

Is the head circumference of this child above average or below average? Below average Above average The residual is a negative and therefore below average. It is also below the line, indicating that it is below average. (f) Notice that two children are 26.75 inches tall. One has a head circumference of 17.3 inches; the other has a head circumference of 17.5 inches. How can this be?

A. It is not appropriate to use this model to predict the value of the head circumference for children who are 26.75 inches tall because 26.75 inches is outside the scope of the model.

B. For children who are 26.75 inches tall, head circumference varies C. It is a mistake among the measurements.

(g) Would it be reasonable to use the least-squares regression line to predict the head circumference of a child who was 32 inches tall? Yes No

Open StatCrunch Stat Calculators Binomial Input n:7 and p: 0.45 Input P(x<4)

Compute!

Use n = 9, and p = 0.4 to complete parts (a) through (c) below. (a) Find the probabilities and construct a binomial probability distribution with the given parameters. (b) Compute the mean and standard deviation of the random variable. (c) Draw the probability histogram, comment on its shape, and label the mean on the histogram. (a) To construct a binomial probability distribution, complete the table. (Round to four decimal places as needed.) If we refer to PowerPoint slide 25 of 34, it tells us to Open StatCrunch: and input the possible values of the random variable. Stat Calculators Binomial Input n: 9 and p: 0.4 Input each value for x, i.e. x = 0, x = 1, x = 2, x = 3, x = 4, x = 5, x = 6, x = 7, x = 8, x = 9,Compute! after each value of x and then record the results.

x P(x) 0 0.0101 1 0.0605 2 0.1612 3 0.2508 4 0.2508 5 0.1672 6 0.0743 4 0.0212 8 0.0035 9 0.0003

(b) Compute the mean and standard deviation of the random variable. Slide 14 tells us to find the mean and standard deviation of the discrete random variable in part (a) to Open StatCrunch Enter the data: X and P (X) Stat Calculators Custom Chose X for “Values in,” P (X) for “Weights in” Compute!

𝜇𝜇𝑥𝑥 = 3.60 (Round to two decimal places as needed.) 𝜎𝜎𝑥𝑥 = 1.5 (Round to one decimal place as needed.) Method II : (Short Cut Formulas for a Binomial Distribution Only) Since this is a Binomial distribution, we can use the short cut formulas of 𝜇𝜇𝑥𝑥= n * p and �𝑛𝑛 ∗ 𝑝𝑝 ∗ (1 − 𝑝𝑝) We will use a scientific calculator to obtain the answers. 𝜇𝜇𝑥𝑥 = 9 * 0.4 = 3.6 𝜎𝜎𝑥𝑥 = �9 ∗ 0.4 ∗ (1 − 0.4) = √9 ∗ .04 ∗ .06 ≈ 1.5 If the distribution is a binomial distribution, Method II is the preferred way of finding the mean and standard deviation of the discrete random variable. (c) Draw the probability histogram, comment on its shape, and label the mean on the histogram. Choose the correct answer below. From StatCrunch output, the shape of the distribution is approximately a bell-shape curve.

However, based on the available choices provided for part (c), the results most resembles the shape is answer "B".

According to a study done by a university student, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.62. Suppose you sit on a bench in a mall and observe people's habits as they sneeze. (a) What is the probability that among 50 randomly observed individuals exactly 39 do not cover their mouth when sneezing? (b) What is the probability that among 50 randomly observed individuals fewer than 20 do not cover their mouth when sneezing? (c) Would you be surprised if, after observing 50 individuals, fewer than half covered their mouth when sneezing? Why? (a) The probability that exactly 39 do not cover their mouth when sneezing is.0071. Just like in the previous problem, we Open StatCrunch Stat Calculators Binomial Input n: 50 and p: 0.62 InputP(X = 39) =___ Compute! (Round to four decimal places as needed.)

(b) The probability that fewer than 20 do not cover their mouth when sneezing is 0.0005. Just like in the previous problem, we Open StatCrunch Stat Calculators Binomial Input n: 50 and p: 0.62 InputP(X <20) =___ Compute! (Round to four decimal places as needed.)

(c) Would you be surprised if, after observing 50 individuals, fewer than half covered their mouth when sneezing? Fewer than half of 50, means fewer than 25. Just like in the previous problem, we Open StatCrunch Stat Calculators Binomial Input n: 50 and p: 0.62 InputP(X <25) =___ Compute! (Round to four decimal places as needed.)

We see that P(X <25)= 0.0305. Since this is less than 0.05, this will be unusual and we would be surprise.

One graph in the figure represents a normal distribution with mean 𝜇𝜇 = 16 and standard deviation σ = 2. The other graph represents a normal distribution with mean 𝜇𝜇 = 11 and standard deviation σ = 2. Determine which graph is which and explain how you know.

Choose the correct answer below.

A. Graph A has a mean of μ = 11 and graph B has a mean of μ =16 because a larger mean shifts the graph to the right.

B. Graph A has a mean of μ = 16 and graph B has a mean of μ = 11 because a larger mean shifts the graph to the left.

C. Graph A has a mean of μ = 11 and graph B has a mean of μ = 16 because a larger mean shifts the graph to the left.

D. Graph A has a mean of 𝜇𝜇 = 11 and graph B has a mean of μ = 11 because a larger mean shifts the graph to the right. Fundamentals of Statistics Informed Decisions Using Data, Fourth Edition Page 331

Pg 332

The graph of a normal curve is given below. Use the graph to identify the values of 𝜇𝜇 and 𝜎𝜎. −14 −12 −10 −8 −6 −4 −2 𝜇𝜇 = −8 midpoint of the curve 𝜎𝜎 = 2 difference between 2 consecutive standard deviation point (−14) – (−12) = 2

X

Suppose the monthly charge for cell phone plans is normally distributed with mean µ = $59 and a standard deviation 𝜎𝜎 = $16. 59-16 = 43, 59+16 = 75 (a) Draw a normal curve with the parameters labeled. Open StatCrunch Stat Calculators Normal Between, input Mean: 59, Std. Dev.: 16, P (43 < x < 75) = ______ Compute! You do not get the exact number you searched shown on the display numerically, but it is shown by the shaded area.

43 75 59

(b) Shade the region that represents the proportion of plans that charge more than $75. This time we only want the area greater than 75. Go back to your curve created in part (a) and click on Standard, Input the mean = 59, std. dev. = 16 and P( x > 75) ____ Compute!

(c) Choose the correct interpretation below.

A. 15.87% of cell phone plans cost more than $75 per month. B. 15.87% of cell phone plans cost less than $75 per month. C. The probability is 0.1587 that a randomly selected cell phone plan is less than $75 per

month. D. 84.13% of cell phone plans cost more than $75 per month. Look back at the curve you produced in part (b) and it tells you that the area corresponding to P ( x > 75 ) ≈ 0.1587 or 15.87%

Michael went to the driving range with his range finder and hit 75 golf balls with his pitching wedge and measured the distance each ball traveled (in yards). The accompanying table shows his data. Complete parts a and b below. Open in StatCrunch by clicking here. Click the icon to view the table of drive distances.

(a). Use technology to construct a relative frequency histogram. Comment on the shape of the distribution. Draw a normal density function on the relative frequency histogram. Open in StatCrunch Graph Histogram Select Column(s): var2, Type: Relative Frequency Compute!

Comment on the shape of the distribution.

A. The shape of the distribution is roughly normal. B. The shape of the distribution is roughly uniform. C. The shape of the distribution seems to have no pattern.

Go back to your histogram, select options, edit Under Display options: from the Display overlay distrib.: click the drop down arrow and select Normal Compute!

(b). Do you think the normal density function describes the distance Michael hits with a pitching wedge? Why?

A. Yes, because the histogram shape resembles a normal curve, and the area of each bar is roughly equal to the area under the normal curve for the same region.

B. No, because the bars in the histogram are sometimes higher than the normal curve and sometimes lower.

C. Yes, because the distance Michael hits the ball is a random variable, so it is normally distributed.

D. No, because the normal distribution is continuous, but the distance Michael hits the ball is a discrete variable.

We are going to use StatCrunch to determine the area under the standard normal curve that lies to the right of a specific Z value. Recall: Z is a random variable for the standard normal distribution with mean = 0 and std. dev. = 1. (a) Determine the area under the standard normal curve that lies to the right of Z = -1.65 (Round to four decimal places as needed.) Open StatCrunch Stat Calculators Normal take the default setup of Mean: 0 and Std. Dev.: 1 change ≤ to ≥ and input the Z-score into P (X ≥ -1.65) =__ Compute! record the result. The area to the right of Z = -1.65 is 0.9505

(b) Determine the area under the standard normal curve that lies to the right of Z = -1.23 (Round to four decimal places as needed.) Continue to use the setup from part (a) but change the Z = -1.23 into P(X ≥ -1.23) Compute! The area to the right of Z = -1.23 is 0.8907.

(c) Determine the area under the standard normal curve that lies to the right of Z = 0.82 (Round to four decimal places as needed.) Change the Z = 0.82 into P(X ≥ 0.82) Compute! The area to the right of Z = 0.82 is 0.2061.

(d) Determine the area under the standard normal curve that lies to the right of Z = -0.82 (Round to four decimal places as needed.) Change the Z = - 0.82 into P(X ≥ - 0.82) ◊ Compute! The area to the right of Z = -0.82 is 0.7939.

Determine the area under the standard normal curve that lies between (a) Z = −0.71 and Z = 0.71, (b) Z = −1.79 and Z = 0, and (c) Z = 0.73 and Z = 2.26. (a) The area that lies between Z = −0.71 and Z = 0.71 is 0.5223. (Round to four decimal places as needed.) Open StatCrunch Stat Calculators Normal Select Between on the Standard|Between box take the default setup of Mean:0 and Std. Dev.:1 Input the Z values into P (−0.71 ≤ X ≤ 0.71) = ___ Compute!

(b) The area that lies between Z = −1.79 and Z = 0 is 0.4633. (Round to four decimal places as needed.) Change the Z values to P (−1.79 ≤ X ≤ 0) ◊ Compute!

(c) The area that lies between Z = 0.73 and Z = 2.26 is 0.2208. (Round to four decimal places as needed.) Change the Z values to P (0.73 ≤ X ≤ 2.26) ◊ Compute!

Determine the total area under the standard normal curve in parts (a) through (c) below. Click the icon to view a table of areas under the normal curve. Again, we are using StatCrunch. (a) Find the area under the normal curve to the left of Z = −2 plus the area under the normal curve to the right of Z = 2. Open StatCrunch Stat Calculators Normal Between take the default setup of Mean : 0 and St. Dev.: 1Click the inequality of ≤ and input the Z-value for P (x ≤ -2) Compute! record the result. Change the inequality to ≥ and input the Z- value for P (x ≥ 2) Compute! record the result.

The combined area is 0.02275013 + 0.02275013 = 0.04550026 ≈ 0.455. Alternative way of obtaining the answer: Knowing that the area under the normal distribution is 1, we will find the area between the given values and subtract the results from 1 to obtain the answer. Open StatCrunch Stat Calculators Normal Between take the default setup of Mean:0 and St. Dev.: 1 Input the Z- values in P(−2 ≤ x ≤ 2) = ___ Compute! Subtract your result from 1. P(−2 ≤ x ≤ 2) = 0.95449974

The area to the left of Z = -2 plus the area to the right of Z = 2 is (1 - 0.95449974) ≈ 0.0455.

(Round to four decimal places as needed.)

(b) Find the area under the normal curve to the left of Z = −1.53 plus the area under the normal curve to the right of Z = 2.53. Change the Z- values to P(−1.53 ≤ x ≤ 2.53) = ___ ◊ Compute! Subtract your result from 1.

The combined area is 1 – 0.9313 = 0.0687. (Round to four decimal places as needed.) (c) Find the area under the normal curve to the left of Z = −0.22 plus the area under the normal curve to the right of Z = 1.20. Change the Z- values to P(−0.22 ≤ x ≤ 1.20) = ___ Compute! Subtract your result from 1.

The combined area is 1 – 0.4720 = 0.5280. (Round to four decimal places as needed.)

Find the Z-score such that the area under the standard normal curve to the left is 0.26. Instead of finding what the area is equal to for specific Z- values, we are now going to find the Z-scores for a specific area. Open StatCrunch Stat Calculators Standard (default setting) Input the value of the area we are looking for P (x ≤ ___) = 0.26. Be sure to select the appropriate inequality symbol. Compute! −0.64 is the Z-score such that the area under the curve to the left is 0.26. (Round to two decimal places as needed.)

Find the Z- score such that the area under the standard normal curve to the right is 0.26. Similar to the previous questions, Open StatCrunch Stat Calculators Normal Standard (Default Setting) Input the value of the area we are looking for P (x ≥ ___) = 0.26. Be sure to select the appropriate inequality symbol. Compute! The approximate Z-score that corresponds to a right tail area of 0.26 is 0.64 (Round to two decimal places as needed.)

Find the z-scores that separate the middle 85% of the distribution from the area in the tails of the standard normal distribution. Open StatCrunch Stat Calculators Normal Select Between Standard (Default Setting) P (__≤ x ≤ __) = 0.85 Compute!

The Z-scores are −1.44, 1.44. (Use a comma to separate answers as needed. Round to two decimal places as needed.)

Assume that the random variable X is normally distributed, with mean μ=51 and standard deviation σ=8. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. P(x ≤ 47) Which of the following shaded regions corresponds to P (X ≤ 47)?

Now we will construct this curve with StatCrunch. Stat Calculators Normal Input the mean: 51, Std. Dev.: 8, and P (x ≤ 47) Compute!

Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ =7. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. P (35 < x < 59) Which of the following normal curves corresponds to P (35 < x < 59)?

Now we will use StatCrunch to calculate P (35 ≤ x ≤ 59). StatCrunch. Stat Calculators Normal Between, input mean: 50 Std. Dev.: 7 P (35 ≤ x ≤ 59) = ___ Compute!

P (35 < x < 59) = 0.8847 (Round to four decimal places as needed.)

A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 63.6 seconds. Assuming drive-through times are normally distributed with a standard deviation of 2.9 seconds, complete parts (a) through (d) below. So we have μ=63.6 and σ=2.9 (a) What is the probability that a randomly selected car will get through the drive-through in less than 59.9 seconds? Open StatCrunch Stat Calculators Normal Input Mean: 63.6, Std. Dev.: 2.9 and P (x ≤ 59.9) Compute!

The probability is 0.1010. (Round to four decimal places as needed.) (b) What is the probability that a randomly selected car will spend more than 70.1seconds in the drive-through? Open StatCrunch Stat Calculators Normal Input Mean: 63.6, Std. Dev.: 2.9 and P (x ≥ 70.1) Compute!

The probability is 0.0125. (Round to four decimal places as needed.)

(c) What proportion of cars spend between 59.9 and 70.1 seconds in the drive-through? Open StatCrunch Stat Calculators Normal Input Mean: 63.6, Std. Dev.: 2.9 and P (59.9 ≤ x ≤ 70.1) = ___ Compute!

0.8865 (Round to four decimal places as needed.) (d) Would it be unusual for a car to spend more than 70.1 seconds in the drive-through?

A. Yes, because the probability of randomly choosing a car that waits more than 70.1 seconds in the drive-through is greater than 5%.

B. Yes, because the probability of randomly choosing a car that waits more than 70.1 seconds in the drive-through is less than 5%.

C. No, because the probability of randomly choosing a car that waits more than 70.1 seconds in the drive-through is greater than 5%.

D. No, because the probability of randomly choosing a car that waits more than 70.1 seconds in the drive-through is less than 5%.

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1263 chips and a standard deviation of 118 chips. (a) Determine the 26th percentile for the number of chocolate chips in a bag. Open StatCrunch Stat Calculators Normal Input mean: 1263, Std. Dev.: 118 and P (<__) = 0.26 Compute!

(a) The 26th percentile for the number of chocolate chips in a bag of chocolate chip cookies is 1187 chocolate chips. (Round to the nearest whole number as needed.) (b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags. Open StatCrunch Stat Calculators Normal Between, Input mean: 1263, Std. Dev.: 118 and P (__≤ x ≤ __) = 0.95 Compute!

(b) The number of chocolate chips in a bag that make up the middle 95% of bags is 1032 to 1494 chocolate chips. (Round to the nearest whole number as needed. Use ascending order.)

Suppose a sample of O-rings was obtained and the wall thickness (in inches) of each was recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed.

Graph QQ Plot Compute!

Could the population be normally distributed?

A. Yes, since the relationship between the expected z-values and the observed is not linear. B. No, since the relationship between the expected z-values and the observed is not linear. C. Yes, since the relationship between the expected z-values and the observed is

approximately linear. D. No, since the relationship between the expected z-values and the observed is

approximately linear.

Open in StatCrunch

A random sample of 16 undergraduate students receiving student loans wasobtained, and the amount of their loans for the school year was recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed.

Open StatCrunch Input the Data manually (You can try it through the shortcut above, mine gave some really weird data values) Graph QQ Plot Select column(s): var1 Compute!

Choose the correct answer below.

A. No, since the relationship between the expected Z-scores and the observed values is not linear.

B. No, since the relationship between the expected Z-scores and the observed values is approximately linear.

C. Yes, since the relationship between the expected Z-scores and the observed values is approximately linear.

D. Yes, since the relationship between the expected Z-scores and the observed values is not linear

Do not Open in StatCrunch

A simple random sample of size n = 34 is obtained from a population with μ = 90 and σ = 3. Does the population need to be normally distributed for the sampling distribution of �̅�𝑥 to be approximately normally distributed? Why? What is the sampling distribution of �̅�𝑥 ? Does the population need to be normally distributed for the sampling distribution of �̅�𝑥 to be approximately normally distributed? Why? Recall: If n ≥ 30, 𝑥𝑥 is approximately normally distributed. If n < 30, 𝑥𝑥 is approximately normally distributed if the sampling data came for a population that is normally distributed.

A. **No because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of �̅�𝑥 becomes approximately normal as the sample size, n, increases.

B. Yes because the Central Limit Theorem states that only for underlying populations that are normal is the shape of the sampling distribution of �̅�𝑥 normal, regardless of the sample size, n.

C. No because the Central Limit Theorem states that only if the shape of the underlying population is

normal or uniform does the sampling distribution of �̅�𝑥 become approximately normal as the sample size, n, increases.

D. Yes because the Central Limit Theorem states that the sampling variability of nonnormal populations

will increase as the sample size increases. What is the sampling distribution of �̅�𝑥? Select the correct choice below and fill in the answer boxes within your choice. (Type integers or decimals rounded to three decimal places as needed.) Note: Since n = 34, the sampling distribution of 𝑥𝑥 is approximately normally distributed.

A. The sampling distribution of �̅�𝑥 is skewed left with 𝜇𝜇�̅�𝑥 = ___ and 𝜎𝜎�̅�𝑥 = ___. B. *The sampling distribution of �̅�𝑥 is approximately normal with 𝜇𝜇�̅�𝑥 = 90 and 𝜎𝜎�̅�𝑥 = 0.514. C. The sampling distribution of �̅�𝑥 is uniform with 𝜇𝜇�̅�𝑥 = ___ and 𝜎𝜎�̅�𝑥 = ___. D. The sampling distribution of �̅�𝑥 follows Student's t-distribution with 𝜇𝜇�̅�𝑥 = ___ and 𝜎𝜎�̅�𝑥 = ___.

*B. 3 ÷ √34 = 0.514

Summary taken from e-book pg. 375: SUMMARY: SHAPE, CENTER, AND SPREAD OF THE DISTRIBUTION OF 𝒙𝒙 �

Distribution of the Sample Mean Shape, Center, and Spread of the

Population Shape Center Spread

Population is normal with mean μ and standard deviation σ

Regardless of the sample size n, the shape of the distribution of the sample mean is normal μ �̅�𝑥 =μ σ 𝑥𝑥 � = σ/√ n

SUMMARY: SHAPE, CENTER, AND SPREAD OF THE DISTRIBUTION OF 𝒙𝒙 � Distribution of the Sample Mean

Shape, Center, and Spread of the Population Shape Center Spread

Population is not normal with mean μ and standard deviation σ

As the sample size n increases, the distribution of the sample mean becomes approximately normal μ �̅�𝑥 = μ σ 𝑥𝑥 �= σ /√n

A simple random sample of size n =40 is obtained from a population with μ = 67 and 𝜎𝜎= 19 (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of �̅�𝑥. (b) Assuming the normal model can be used, determine P(�̅�𝑥 < 71.4). (c) Assuming the normal model can be used, determine P(�̅�𝑥 ≥ 69.4). (a) What must be true regarding the distribution of the population?

A. The population must be normally distributed. B. The sampling distribution must be assumed to be normal. C. The population must be normally distributed and the sample size must be large. D. Since the sample size is large enough, the population distribution does not need to be normal.

Assuming the normal model can be used, describe the sampling distribution �̅�𝑥 Choose the correct answer below.

A. Approximately normal, with 𝜇𝜇�̅�𝑥 = 67 and 𝜎𝜎�̅�𝑥 = 40√19

B. Approximately normal, with 𝜇𝜇�̅�𝑥 = 67 and 𝜎𝜎�̅�𝑥 = 19√40

C. Approximately normal, with 𝜇𝜇�̅�𝑥 = 67 and 𝜎𝜎�̅�𝑥 = 19 Open StatCrunch stat Calculators Normal input mean: 67, Std. dev.: 3.004 (calculate from data given) and p(x< 71.4) = __ Compute! (b) Assuming the normal model can be used, determine P(𝑥𝑥 < 71.4). Note: Mean is 𝜇𝜇𝑥𝑥 = µ = 67 and Std. Dev. is 𝜎𝜎𝑥𝑥 = 𝜎𝜎

√𝑛𝑛 = 19

√40 ≈ 3.004 (see results obtained from part a ).

Open StatCrunch Stat Calculators Normal input mean: 67, Std. Dev.: 3.004 (calculate from part b) and p(x < 71.4) = __ Compute! Note: x above meant 𝑥𝑥 .

P(𝑥𝑥 �< 71.4) = 0.9285. (Round to four decimal places as needed.)

(c) Assuming the normal model can be used, determine P(𝑥𝑥 ≥ 69.4). Since 𝜇𝜇𝑥𝑥 and 𝜎𝜎𝑥𝑥 are the same as part b, we just need to change the inequality sign to ≥ and input x as 69.4. (Again, x meant 𝑥𝑥)

P(�̅�𝑥 ≥ 69.4)= 0.2122 (Round to four decimal places as needed.)

The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is unknown. However, records indicate that the mean time is 11.2 minutes, and the standard deviation is 3.3 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? Note: Since the problem did not indicate the population distribution is normally distributed, a large sample size is necessary to guarantee that 𝑥𝑥 is normally distributed.

A. The sample size needs to be less than or equal to 30. B. Any sample size could be used. C. The normal model cannot be used if the shape of the distribution is

unknown. D. The sample needs to be greater than or equal to 30.

(b) What is the probability that a random sample of n = 35 oil changes results in a sample mean time less than 10 minutes? The question asks for P ( 𝑥𝑥 <10 ). 𝑥𝑥 is normally distributed with Mean is µ𝑥𝑥 = µ =11.2 and Std. Dev. is 𝜎𝜎𝑥𝑥 = 𝜎𝜎

√𝑛𝑛 = 3.3

√35 ≈ .0.5578.

StatCrunch Stat Calculators Normal input the mean: 11.2, Std. Dev.: 0.5578 and P (x< 10) Compute! Note: x above meant 𝑥𝑥

The probability of the sample mean ( 𝑥𝑥 ) time less than 10 minutes is approximately 0.0157. (Round to four decimal places as needed.) (c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager. The question asks for the cut off sample mean time such that P( 𝑥𝑥 ≤ ?? ) = 0.10. StatCrunch Stat Calculators Normal input the mean: 11.2, Std. Dev.: 0.5578 and P (x ≤ __) = 0.10 Compute! Note: x above meant 𝑥𝑥

There is a 10% chance of being at or below a mean oil-change time of 10.5 minutes. (Round to one decimal place as needed.)

Without doing any computation, decide which has a higher probability, assuming each sample is from a population that is normally distributed with μ = 100 and σ = 15. Explain your reasoning. (a) P(90 ≤ �̅�𝑥 ≤ 110) for a random sample of size n = 10 (b) P(90 ≤ �̅�𝑥 ≤ 110) for a random sample of size n = 20 Explanation: Since the population is normally distributed, x is normally distributed for any size of n with Mean: µ𝑥𝑥 = µ =100 and Std. Dev. is σ𝑥𝑥 = 𝜎𝜎

√𝑛𝑛 .

As n increases, the standard deviation of the sample mean, σ𝑥𝑥 , decreases. For a normal distribution, Z = 𝑥𝑥−𝑀𝑀𝑀𝑀𝑀𝑀𝑛𝑛

𝑆𝑆𝑆𝑆𝑆𝑆.𝐷𝐷𝑀𝑀𝐷𝐷.. As n increases, σ𝑥𝑥 decreases, thus Z increases which

means the in between area becomes bigger. Try to picture the in between area. Choose the correct answer below.

A. P(90 ≤ �̅�𝑥 ≤ 110) for a random sample of size n = 20 has a higher probability. As n increases, the standard deviation increases.

B. P(90 ≤ �̅�𝑥 ≤ 110) for a random sample of size n = 20 has a higher probability. As

n increases, the standard deviation decreases.

C. P(90 ≤ �̅�𝑥 ≤ 110) for a random sample of size n = 10 has a higher probability. As n increases, the standard deviation increases.

D. P(90 ≤ �̅�𝑥 ≤ 110) for a random sample of size n = 10 has a higher probability. As

n increases, the standard deviation decreases.

According to a study conducted by a statistical organization, the proportion of Americans who are satisfied with the way things are going in their lives is 0.82. Suppose that a random sample of 100 Americans is obtained. Complete parts (a) through (c). (a) Describe the sampling distribution of p�. Choose the correct answer below.

A. The distribution of p� can be approximated by a normal distribution with 𝜇𝜇𝑝𝑝�= 0.82 and 𝜎𝜎𝑝𝑝� = 0.076.

B. The distribution of p� is uniform. C. The distribution of p� can be approximated by a normal distribution with 𝜇𝜇𝑝𝑝�= 0.82 and

𝜎𝜎𝑝𝑝� = 0.038. D. The distribution of p� cannot be approximated by the normal distribution. Instead, a

binomial distribution with n = 100 and p = 0.82 should be used.

(b) Using the distribution from part (a), what is the probability that at least 82 Americans in the sample are satisfied with their lives?

Recall: �̂�𝑝 = 𝑥𝑥

𝑛𝑛 = 82

100 = 0.82

The question asks for P( 𝑝𝑝 � ≥ 0.82 ). Open StatCrunch Stat Calculators Normal Standard, input mean: 0.82, Std. Dev.: 0.38418745, P (x ≥ 0.82) Compute!

The probability that at least 82 Americans in the sample are satisfied is 0.5000. (Round to four decimal places as needed.) (c) Using the distribution from part (a), what is the probability that 77 or fewer Americans in the sample are satisfied with their lives? Recall: �̂�𝑝 = 𝑥𝑥

𝑛𝑛 = 77

100 = 0.77

The question asks for P( 𝑝𝑝 � ≤ 0.77 ).

Sice the Mean (𝜇𝜇𝑝𝑝�= p) and Std. Dev. (𝜎𝜎𝑝𝑝� = �𝑝𝑝(1−𝑝𝑝)𝑛𝑛

) are the same as part b, we just need to

change the inequality to ≤ and the x-value (�̂�𝑝 value) to 0.77 Compute!

(Round to four decimal places as needed.) The probability that 77 or fewer Americans in the sample are satisfied is 0.0965.

Exit polling is a popular technique used to determine the outcome of an election prior to results being tallied. Suppose a referendum to increase funding for education is on the ballot in a large town (voting population over 100,000). An exit poll of 500 voters finds that 260 voted for the referendum. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.49? Based on your result, comment on the dangers of using exit polling to call elections. Setup calculations: p� = 260/500 = 0.52

μp� = p = 0.49 σp� = �0.49(0.51)500

= 0.022356

How likely are the results of your sample (�̂�𝑝 = 𝑥𝑥

𝑛𝑛 = 260

500 = 0.52) if the population proportion of

voters in the town in favor of the referendum is 0.49 (p = 0.49)? Open StatCrunch Stat Calculators Normal Standard, Input Mean: 0.49, Std. Dev.: 0.022356, P(x ≥ 0.52) = ___ Compute!

The probability that more than 260 people voted for the referendum is 0.0898. (Round to four decimal places as needed.) Comment on the dangers of using exit polling to call elections. Choose the correct answer below. A. The result is not unusual because the probability that p� is equal to or more extreme

than the sample proportion is greater than 5%. Thus, it is not unusual for a wrong call to be made in an election if exit polling alone is considered

B. The result is unusual because the probability that p� is equal to or more extreme than the sample proportion is greater than 5%. Thus, it is not unusual for a wrong call to be made in an election if exit polling alone is considered.

C. The result is not unusual because the probability that p� is equal to or more extreme than the sample proportion is less than 5%. Thus, it is unusual for a wrong call to be made in an election if exit polling alone is considered.

A simple random sample of size n = 450 individuals who are currently employed is asked if they work at home at least once per week. Of the 450 employed individuals surveyed, 33 responded that they did work at home at least once per week. Construct a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week. Summary: We are given n = 450 (This is our observed.) Out of the 450, 33 is the number of successes We are looking for a 99% Confidence Interval for p. We need the lower bound and the upper bound. To find these values, we Open StatCrunch Stat Proportion Stats One sample With summary (because we are not given a group of data to sort; we were given the summary info. listed above) Input the # of successes: 33, the # of observations: 450 and select the confidence interval for p, level: 0.99 Compute!

One sample proportion confidence interval: p : Proportion of successes Method: Standard-Wald 99% confidence interval results: Proportion Count Total Sample Prop. Std. Err. L. Limit U. Limit

p 33 450 0.073333333 0.012288708 0.041679719 0.10498695

The Lower bound is 0.042. (Round to three decimal places as needed.) The Upper bound is 0.105. (Round to three decimal places as needed.)

A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be �̅�𝑥 = 121.3 and the sample standard deviation is found to be s = 12.6. Construct a 99% confidence interval for the population mean. Again, let’s look at what we are given, n = 40 �̅�𝑥 = 121.3 s = 12.6 We are asked to construct a 99% Confidence Interval for 𝜇𝜇. Open StatCrunch Stat T – Stats One Sample with summary Input the mean: 121.3, Sample std. Dev.: 12.6, sample size: 40 and Perform: Confidence interval for 𝝁𝝁 Level 0.99 Compute!

One sample T confidence interval: μ : Mean of population 99% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit μ 121.3 1.9922349 39 115.9052 126.6948

The lower bound is 115.91. (Round to two decimal places as needed.) The upper bound is 126.69. (Round to two decimal places as needed.)

In a random sample of 100 audited estate tax returns, it was determined that the mean amount of additional tax owed was $3451with a standard deviation of $2572. Construct and interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. Summarize: n = 100 x� = $3451 S = $2572 Construct 90% Confidence interval for 𝜇𝜇 Open StatCrunch Stat T-Stats One Sample With Summary Input the mean: 3451, Sample Std. Dev.: 2572, Sample size: 100 and Confidence level for 𝝁𝝁, Level: 0.90 Compute!

One sample T confidence interval: μ : Mean of population 90% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit μ 3451 257.2 99 3023.9474 3878.0526 The lower bound is $3023.95. (Round to the nearest cent as needed.) The upper bound is $3878.05. (Round to the nearest cent as needed.)

Interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. Choose the correct answer below.

A. One can be 90% confident that the mean additional tax owed is between the lower and upper bounds. B. One can be 90% confident that the mean additional tax owed is less than the lower bound. C. One can be 90% confident that the mean additional tax owed is greater than the upper bound.

In a survey of 1051adults, a polling agency asked, "When you retire, do you think you will have enough money to live comfortably or not?" Of the 1051surveyed, 564 stated that they were worried about having enough money for retirement. Construct a 90% confidence interval for the proportion of adults who are worried about having enough money for retirement. Summarize: n = 1051 # of successes: 564 Construct 90% Confidence Interval for p. Open StatCrunch Stat Proportion Stats One Sample With Summary Input # of successes: 564, # of observations: 1051, Confidence Level for p, Level: 0.90 Compute!

One sample proportion confidence interval: p : Proportion of successes Method: Standard-Wald 90% confidence interval results: Proportion Count Total Sample Prop. Std. Err. L. Limit U. Limit p 564 1051 0.53663178 0.015381545 0.51133139 0.56193217 A 90% confidence interval for the proportion of adults who are worried about having enough money for retirement is (0.5113, 0.5619). (Use ascending order. Round to four decimal places as needed.)

The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers.

52 30 44 35 47 37 56 26 30 39 26 52

Mean = 39.5 St. Dev. = 10.59 (a) Because the sample size is small, we must verify that the data come from a population that is normally distributed and that the sample size does not contain any outliers. Are the conditions for constructing a confidence interval about the mean satisfied?

A. No, the population is not normally distributed. B. No, the sample contains an outlier. C. Yes, the population is normally distributed and the sample does not contain any outliers.

(b) Construct a 95% confidence interval for the mean age at which a baby first crawls. Select the correct choice below and fill in any answer boxes in your choice. Open StatCrunch Stat T-Stat One Sample with Summary Input Mean: 39.5, Std. Dev.: 10.59, Sample size: 12 Perform: Confidence level for 𝜇𝜇 Level: 95% Compute!

One sample T confidence interval: μ : Mean of population 95% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit μ 39.5 3.0570697 11 32.771435 46.228565

A. (32.8, 46.2) (Use ascending order. Round to one decimal place as needed.) B. A 95% confidence interval cannot be constructed.

(c) What could be done to increase the accuracy of the interval without changing the level of confidence?

A. Decrease the sample size. B. Increase the sample size C. Nothing can be done D. Either increase or decrease the sample size

Find the critical values 𝜒𝜒21−𝛼𝛼 2⁄ and 𝜒𝜒2 α/2 for a 99% confidence level and a sample size of n =15. Open StatCrunch Stat Calculators Chi-Square DF 14 Between P(___≤ x ≤ ____) = 0.99 Compute! and record Note: DF = n - 1

𝜒𝜒2 1−𝛼𝛼 2⁄ = 4.075 (Round to three decimal places as needed.) 𝜒𝜒2 α/2 = 31.319 (Round to three decimal places as needed.)

A simple random sample of size 20 is drawn from a population that is known to be normally distributed. The sample variance, 𝑠𝑠2, is determined to be 12.4. Construct a 90% confidence interval for 𝜎𝜎2. Open Statcrunch Stat Variance Stats One Sample With Summary Sample Variance: 12.4, Sample Size: 20, Confidence Level for 𝝈𝝈𝟐𝟐 Level: 0.90 Compute!

One sample variance confidence interval: σ2 : Variance of population 90% confidence interval results: Variance Sample Var. DF L. Limit U. Limit σ2 12.4 19 7.81594 23.287506 The lower bound is 7.82. (Round to two decimal places as needed.) The upper bound is 23.29. (Round to two decimal places as needed.)

A jar of peanuts is supposed to have 16 ounces of peanuts. The filling machine inevitably experiences fluctuations in filling, so a quality-control manager randomly samples 12 jars of peanuts from the storage facility and measures their contents. The data are in the table. Complete parts (a) through (d).

(a) Are the given data normally distributed? (Check by constructing a normal probability plot.) Open StatCrunch through icon by the data Graph QQ Plot Select column(s): var1 Compute!

(a) Are the given data normally distributed? Yes

No (b) Determine the sample standard deviation. Stat Summary Stats Select column(s): var1 Highlight Std Dev. Compute! Summary statistics: Column Std. dev. var1 0.33671767 The sample standard deviation is 0.337 oz. (Round to three decimal places as needed.) (c) Construct a 90% confidence interval for the population standard deviation of the number of ounces of peanuts. OpenStatCrunch Stat Input Data Variance Stats One Sample with data Columns Var 1 Perform: Confidence interval for 𝝈𝝈𝟐𝟐 Level: 0.90 Compute and record the results. One sample variance confidence interval: σ2 : Variance of variable 90% confidence interval results: Variable Sample Var. DF L. Limit U. Limit var1 0.11337879 11 0.063387952 0.27261587 (Note: The results obtained are the limits in variance. To convert variances to standard deviations, you will need to manually calculate the square root of each limit.) √0.063387952 = 0.25176964074327945070647846610393 ≈0.252 √0.27261587 = 0.52212629698187008669252231346955 ≈ 0.522 The lower bound is 0.252 oz. (Round to three decimal places as needed.) The upper bound is 0.523 oz. (Round to three decimal places as needed.) (d) The quality control manager wants the machine to have a population standard deviation below 0.21 ounce. Does the confidence interval validate this desire by indicating that the population standard deviation is below 0.21? Yes No

(Note: 0.21 is not between 0.252 and 0.523.)

In a clinical trial, 531 out of 1040 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 50% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 50% of this drug's users experience flulike symptoms as a side effect at the alpha equals 0.05 level of significance?

n = 1040, 0 0.50p = , 0 0(1 ) (1040)(.50)(1 .50) 260 10np p− = − = >

What are the null and alternative hypotheses? The null hypothesis is p = 0.50 and the alternative is p > 0.50. 𝐻𝐻0: p = 0.50 versus 𝐻𝐻1: p > 0.50 Use technology to find the P-value. Open StatCrunch Stat Proportion Stats One sample With summary Input # successes and # of observations: Perform: select Hypothesis test for p 𝐻𝐻0: p = 0.50, 𝐻𝐻𝐴𝐴: p > 0.5 Compute!.

One sample proportion hypothesis test: p : Proportion of successes H0 : p = 0.5 HA : p > 0.5

Hypothesis test results: Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 531 1040 0.51057692 0.015504342 0.68219104 0.2476 Test statistic is 0.68

Hypothesis test results: Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 531 1040 0.51057692 0.015504342 0.68219104 0.2476 P-value = 0.248. (Round to three decimal places as needed.) Choose the correct answer below.

A. Since P-value < α, reject the null hypothesis reject the null hypothesis and conclude that there is sufficient evidence that a majority of adults in the United States believe they will not have enough money in retirement.

B. Since P-value < α, do not reject the null hypothesis and conclude that there is sufficient evidence that

a majority of adults in the United States believe they will not have enough money in retirement.

C. Since P-value > α, do not reject the null hypothesis do not reject the null hypothesis and conclude that there is not sufficient evidence that a majority of adults in the United States believe they will not have enough money in retirement.

D. Since P-value > α, reject the null hypothesis and conclude that there is not sufficient evidence that a

majority of adults in the United States believe they will not have enough money in retirement.

In a previous year, 65% of females aged 15 years of age and older lived alone. A sociologist tests whether this percentage is different today by conducting a random sample of 700 females aged 15 years of age and older and finds that 427 are living alone. Is there sufficient evidence at the α = 0.1 level of significance to conclude the proportion has changed? Population: p = 0.65 Sample: �̂�𝑝 = 𝑥𝑥

𝑛𝑛 = 427

700 = 0.61

Identify the null and alternative hypotheses for this test.

*Find the test statistic for this hypothesis test. **Determine the p-value for this hypothesis test. Open StatCrunch Stat Proportion Stats One Sample, with summary Input # of Successes, and # of Observations: Perform: select Hypothesis test value for p 𝐻𝐻𝑜𝑜= 0.65, 𝐻𝐻𝐴𝐴: p ≠ 0.65 Compute!

One sample proportion hypothesis test: p : Proportion of successes H0 : p = 0.65 HA : p ≠ 0.65 Hypothesis test results: Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 427 700 0.61 0.018027756 -2.2188008 0.0265 *Find the test statistic for this hypothesis test. z = - 2.22. (Round to two decimal places as needed.) **Determne the p-value for this hypothesis test.

P-value = 0.027 (Round to two decimal places as needed.) State the conclusion for this hypothesis test.

A. Do not reject H0. There is not sufficient evidence at the 𝛼𝛼 = 0.1 level of significance to conclude that the proportion of females who are living alone has changed.

B. Reject H0. Thee is sufficient evidence at the 𝛼𝛼 = 0.1 level of significance to conclude that the proportion of females who are living alone has changed.

C. Do not reject H0. Thee is sufficient evidence at the 𝛼𝛼 = 0.1 level of significance to conclude that the proportion of females who are living alone has changed.

A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 705. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 50 high-income individuals and found the sample mean credit score to be 715 with a standard deviation of 85. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α = 0.05 level of significance. Choose the correct hypotheses. H0: 𝜇𝜇 = 705 H1: 𝜇𝜇 > 705 Find the test statistic. Open StatCrunch Stat Calculator T –Stats, one sample, with summary Sample Mean: 705, Std. Dev.: 85, Sample size: 50, Hypothesis test for µ: 𝐻𝐻0: = 705, 𝐻𝐻𝐴𝐴: > 705 change the direction of the inequality to ≥ for the right-tailed test Compute and record results

One sample T hypothesis test: μ : Mean of population H0 : μ = 705 HA : μ > 705 Hypothesis test results:

t0 = 0 (Round to two decimal places as needed.) Find the P- value. The P-value is 0.5. Since 0.5 is not less than 0.05, we do not reject the null hypothesis.

What conclusion can be drawn?

A. Do not reject H0. There is not sufficient evidence to conclude that male teenagers consume more than the recommended daily amount of calcium.

B. Reject H0. There is sufficient evidence to conclude that male teenagers consume less than the recommended daily amount of calcium.

C. Reject H0. There is not sufficient evidence to conclude that male teenagers consume less

than the recommended daily amount of calcium.

D. Do not reject H0. There is sufficient evidence to conclude that male teenagers consume more than the recommended daily amount of calcium.

A golf association requires that golf balls have a diameter that is 1.68 inches. To determine if golf balls conform to the standards, a random sample of golf balls was selected. Their diameters are shown in the table. Complete parts (a) and (b). Click the icon to view the data table. (a) Because the sample size is small, the engineer must verify that the diameter is normally distributed and the sample does not contain any outliers. The normal probability plot and boxplot are shown below. View the given graphs.

There are three conditions for testing hypotheses regarding a population mean with σ unknown.

• First, the sample must be obtained using simple random sampling. • Second, the sample must have no outliers. • Third, the population from which the sample is drawn must be normally distributed or the sample

size, (n, ≥ 30). If the normal probability plot is roughly linear and all the data lie within the bounds, there is reason to believe the data come from a population that is approximately normal. If a data value is less than the lower fence or greater than the upper fence, it is considered an outlier. From the boxplot, determine if there are outliers in the sample. Since the data appear to be normally distributed, and there are no outliers, the conditions for testing the hypothesis are satisfied. Are the conditions for testing the hypothesis satisfied?

Yes No (b) Do the golf balls conform to the standards? Conduct a hypothesis test using the P-value approach and a level of significance of α = 0.01. First determine the appropriate hypotheses. Since the standards require the population mean to be exactly 1.68, this is a two-tailed test. Therefore, the null and alternative hypotheses are:

H0: µ = 1.68 H1: µ ≠ 1.68

Open StatCrunch with icon provided. --> Stat T-Stat, one sample, with data Select column: Diameter_(in.), Hypothesis test value Under Optional graphs an tables: select P-value plot Compute! One sample T hypothesis test: μ : Mean of variable H0 : μ = 1.68 HA : μ ≠ 1.68 Hypothesis test results:

Variable Sample Mean Std. Err. DF T-Stat P-value Diameter_(in.) 1.681 0.001331438 11 0.72663608 0.4826 Find the test statistic. t0 = ± 0.75 (Round to two decimal places as needed. Since the test is a two-tailed test, find the area under the t-distribution for n−1 = 11 degrees of freedom to the left of −𝑡𝑡 0 = -0.75 and to the right of 𝑡𝑡0 = 0.75. Use technology to find the P-value. (This was done in the previous step.) Select the > in the lower right corner and you will access the graph below,

Find the P-value. The P-value is 0.483. (Round to three decimal places as needed.)

A P-value is the probability of observing a sample statistic as extreme as or more extreme than the one observed under the assumption that the null hypothesis is true. Put another way, the P-value is the likelihood or probability that a sample will result in a sample mean such as the one obtained if the null hypothesis is true. Use the α = 0.05level of significance. What can be concluded from the hypothesis test? The null hypothesis is rejected if the P-value is < α Note that for this two-tailed test, the alternative hypothesis claims the golf balls do not conform to the standards, but it makes no claim if they tend to be under the weight standard or if they tend to be over it.

A. There is sufficient evidence to conclude that the golf balls weigh more than the association's standards.

B. There is not sufficient evidence to conclude that the golf balls do not conform to the association's standards.

C. There is sufficient evidence to conclude that the golf balls weigh less than the association's standards D. There is sufficient evidence to conclude that the golf balls do not conform to the association's

standards.

A simple random sample of size n = 200 drivers were asked if they drive a car manufactured in a certain country. Of the 200 drivers surveyed, 119 responded that they did. Determine if more than half of all drivers drive a car made in this country at the α= 0.05 level of significance. Complete parts (a) through (d). Summarize: n= 200 x = 119 p = 0.5 𝛼𝛼 = 0.05

(a) Determine the null and alternative hypotheses. H0: p = 0.5 H1: p > 0.5 b) Calculate the P-value. �̂�𝑝 = 119

200 = 0.595

=−

−=

npp

ppz)1(

ˆ

00

00

0.595−0.5

�0.5∗(1−0.5)200

= 2.6870057685088805927232085759984 ≈ 2.687

Find the P-value associated with 𝑧𝑧0 = 2.687, rounding to three decimal places. Open StatCrunch Stat Calculators Normal P(> 2.687 Compute!

P – value = 0.004. (Rounding to three decimal places.) (c) State the conclusion for the test. If the P-value is less than the α , reject H0. If the P-value is greater than the α, do not reject H0. Compare the P-value to the level of significance. Choose the correct answer below.

A. Reject H0 because the P-value is less than the α = 0.05 level of significance. B. Reject H0 because the P-value is greater than the α = 0.05 level of significance. C. Do not reject H0 because the P-value is greater than the α = 0.05 level of significance. D. Do not reject H0 because the P-value is less than the α = 0.05 level of significance.

(d) State the conclusion in context of the problem. There is sufficient evidence at the α = 0.05 level of significance to conclude that more than half of all drivers drive a car made in this country.

The question did not provide StatCrunch and we are not going to look up the critical value form

the Chi-Square Distribution. Please either open StatCrunch from the left navigation button or use

the back arrow then open StatCrunch from the previous question.

Stat Calculators Chi-Square Standard DF: 16 P(x ≥ ___ ) = 0.01 Compute!

Change the DF to 30 (DF = n - 1) P(x ≤ ___ ) = 0.1 Compute!

The critical value for this left-tailed test is 20.600 or 20.6 . (Round to three decimal places)

Change to "Between" DF: 25 (DF = n - 1) P( __ ≤ x ≤ __ ) = 0.05 Compute!

The critical value for this two-tailed test are 23.909 and 24.776. (Round to three decimal places)

To test 𝐻𝐻0: σ = 1.2 versus 𝐻𝐻1: σ < 1.2, a random sample of size n = 17 is obtained from a population that is known to be normally distributed. (a) If the sample standard deviation is determined to be s = 2.4, compute the test statistic. 𝜒𝜒02= (𝑛𝑛−1)𝑠𝑠2

𝜎𝜎2 = (17−1)2.42

1.22 = 64.000

𝜒𝜒02 = 64.000 (Round to three decimal places as needed.) (b) If the researcher decides to test this hypothesis at the α = 0.10 level of significance, determine the critical value. Open StatCrunch Stat Calculators Chi-Square Input DF: 16, P (x< __) = 0.10 – . Compute!

The critical value is 9.312. (Round to three decimal places as needed.)

(d) Will the researcher reject the null hypothesis? Why? Choose the correct answer below.

To test H0: σ=4.2 versus H1: σ≠4.2,a random sample of size n = 11 is obtained from a population that is known to be normally distributed. (a) If the sample standard deviation is determined to be s= 5.2, compute the test statistic. (b) If the researcher decides to test this hypothesis at the α = 0.10 level of significance, use technology to determine the P-value. (c) Will the researcher reject the null hypothesis? (a) To test a hypothesis about a population standard deviation, the test statistic is 𝜒𝜒02 = (𝑛𝑛−1)𝑠𝑠2

𝜎𝜎02 = (11−1)5.22

4.202 =

270.417.64

= 15.32879818594104308390022675737 ≈15.33 The test statistic is 𝜒𝜒02 = 15.33 (Round to two decimal places as needed.) (b) The P –value is: Open StatCrunch Variance Stats One Sample, with summary Sample variance: (5.2)2 = 27.04, sample size = 11 Select H0, sigma squared = (4.2)2 = 17.64 Compute! One sample variance hypothesis test: σ2 : Variance of population H0 : σ2 = 17.64 HA : σ2 ≠ 17.64 Hypothesis test results: Variance Sample Var. DF Chi-square Stat P-value σ2 27.04 10 15.328798 0.2411 (b) The P-value is 0.241. (Round to three decimal places as needed.) (c) Since the P-value is greater than the level of significance, α = 0.10, the researcher will not reject the null

hypothesis H0: σ = 4.2.

Conduct the following test at the α = 0.10 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume that the samples were obtained independently using simple random sampling. Test whether p1 ≠ p2. Sample data are x1 = 28, n1 = 255, x2 = 36, and n2 = 301. (a) Determine the null and alternative hypotheses. Choose the correct answer below.

A. 𝐻𝐻0: P1 = P2 versus H1: P1 > P2

B. 𝐻𝐻0: P1 = P2 versus H1: P1 ≠ P2

C. 𝐻𝐻0: P1 = 0 versus H1: P1 = 0

D. 𝐻𝐻0: P1 = P2 versus H1: P1 < P2 For (b and c) Open StatCrunch Stat Proportion Stats Two sample, with summary input # of Successes and # of trials for each sample check the hypothesis information Compute!

Two sample proportion hypothesis test: p1 : proportion of successes for population 1 p2 : proportion of successes for population 2 p1 - p2 : Difference in proportions H0 : p1 - p2 = 0 HA : p1 - p2 ≠ 0 Hypothesis test results:

Difference Count1 Total1 Count2 Total2 Sample Diff. Std. Err. Z-Stat P-value p1 - p2 28 255 36 301 -0.0097974073 0.027163255 -0.36068606 0.7183 (b) The test statistic z0 is −0.36. (Round to two decimal places as needed.) (c) The P-value is 0.718. (Round to three decimal places as needed.) Test the null hypothesis. Choose the correct conclusion below.

A. Reject the null hypothesis because there is not sufficient evidence to conclude that P1 < P2.

B. Do not reject the null hypothesis because there is sufficient evidence to conclude that P1 > P2.

C. Reject the null hypothesis because there is sufficient evidence to conclude that P1 ≠ P2

D. Do not reject the null hypothesis because there is not sufficient evidence to conclude that P1 ≠ P2

Construct a confidence interval for p1−p2 at the given level of confidence. X1 = 353, n1 = 541, x2 = 407, n2 = 584, 90% confidence. Open StatCrunch Stat Proportions, select two sample, with summary Input # of successes and # of trials for each sample, check confidence interval and input desired level of confidence Compute!

Two sample proportion confidence interval: p1 : proportion of successes for population 1 p2 : proportion of successes for population 2 p1 - p2 : Difference in proportions 90% confidence interval results: Differenc

e Count

1 Total

1 Count

2 Total

2 Sample Diff. Std. Err. L. Limit U. Limit

p1 - p2 353 541 407 584 -0.044422429

0.027942909

-0.090384424

0.0015395657

The 90% confidence interval for p1 – p2 is (-0.090, 0.002). (Use ascending order. Round to three decimal places as needed.)

In a survey, respondents were asked, "Would you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction?" Treat the respondents as a simple random sample of adults. Complete parts (a) and (b). Click the icon to view the survey results. (a) What proportion of the males who took the survey is willing to pay higher taxes to reduce the deficit? What proportion of the females who took the survey is willing to pay higher taxes to reduce the deficit? Open in StatCrunch through the icon Stat Tables Contingency With Data Row variable: Gender, Column variable: Response: Display: Row percent Compute)

Contingency table results: Rows: Gender Columns: Response

Cell format Count (Row percent)

No Yes Total Female 60

(73.17%) 22

(26.83%) 82

(100%) Male 87

(73.73%) 31

(26.27%) 118

(100%) Total 147

(73.5%) 53

(26.5%) 200

(100%)

The proportions of the males and females who took the survey who are willing to pay higher taxes to reduce the deficit are 0.263 and 0.268 respectively. (Round to three decimal places as needed.) (b) Is there significant evidence to suggest the proportions of males and females who are willing to pay higher taxes to reduce the deficit differ at the α = 0.10 level of significance? Identify the null and alternative hypotheses for this test. Let p1 represent the population proportion of adult males who are willing to pay higher taxes to reduce the deficit and p2 represent the population proportion of adult females who are willing to pay higher taxes to reduce the deficit.

In an experiment, 19 babies were asked to watch a climber attempt to ascend a hill. On twooccasions, the baby witnesses the climber fail to make the climb. Then, the baby witnesses either a helper toy push the climber up the hill, or a hinderer toy preventing the climber from making the ascent. The toys were shown to each baby in a random fashion. A second part of this experiment showed the climber approach the helper toy, which is not a surprising action, and then the climber approached the hinderer toy, which is a surprising action. The amount of time the baby watched the event was recorded. The mean difference in time spent watching the climber approach the hinderer toy versus watching the climber approach the helper toy was 1.25 seconds with a standard deviation of 1.55 seconds. Complete parts a through c. (a) State the null and alternative hypotheses to determine if babies tend to look at the hinderer toy longer than the helper toy. Let µhinderer − 𝜇𝜇helper, where 𝜇𝜇hinderer is the population mean time babies spend watching the climber approach the hinderer toy and 𝜇𝜇helper is the population mean time babies spend watching the climber approach the helper toy.

(b) Assuming the differences are normally distributed with no outliers, test if the difference in the amount of time the baby will watch the hinderer toy versus the helper toy is greater than 0 at the 0.01 level of significance. n = 19 sd = 1.55 �̅�𝑑 = 1.25 Find the test statistic for this hypothesis test. 3.52 (Round to two decimal places as needed.) Determine the P-value for this hypothesis test. 0.001 (Round to two decimal places as needed.)

State the conclusion for this hypothesis test.

A. Do not reject H0. There is sufficient evidence at the α= 0.01level of significance to conclude that the difference is greater than 0.

B. Do not reject H0. There is not sufficient evidence at the α= 0.01level of significance to conclude that the difference is greater than 0.

C. Reject H0. There is not sufficient evidence at the α= 0.01level of significance to conclude that the difference is greater than 0.

D. Reject H0. There is sufficient evidence at the α= 0.01level of significance to conclude that the difference is greater than 0.

c) What do you think the results of this experiment imply about babies' ability to assess surprising behavior?

A. There is sufficient evidence that babies have the ability to assess surprising behavior. B. There is not sufficient evidence that babies have the ability to assess surprising behavior. C. There is sufficient evidence that babies do not have the ability to assess surprising

behavior. D. The experiment does not imply anything about babies' ability to assess surprising

behavior. E. There is not sufficient evidence that babies do not have the ability to assess surprising

behavior.

A researcher takes measurements of water clarity at the same location in a lake on the same dates during the course of a year and repeats the measurements on the same dates 5 years later. The researcher immerses a weighted disk painted black and white and measures the depth(in inches) at which it is no longer visible. The collected data is given in the accompanying table. Let X represent the initial depth and let Y represent the depth 5 years later. Complete parts(a) through(c) below. Click the icon to view the water clarity data

Observation 1 2 3 4 5 6 Date 1/25 3/19 5/30 7/3 9/13 11/7

Initial depth, X1 40 62 69 69 51 40 Depth 5 years later, Y1 52 68 76 69 46 49

(a) Why is it important to take the measurements on the same date?

A. Using the same dates makes it easier to remember to take samples B. Using the same dates makes the second sample dependent on the first. C. Using the same dates maximizes the difference in water clarity. D. Those are the same dates that all biologists use to take water clarity samples.

(b) Does the evidence suggest that the clarity of the lake is improving at the α = 0.05 level ofsignificance? Note that the normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Choose the correct conclusion below. Let d1 = Yi−Xi. What are the correct null and alternative hypotheses?

What is the P-value? Open the spreadsheet in StatCrunch Stat T-Stats Paired Sample 1: Y1 Sample 2: X1 Check the box for Save Differences Change the alternative inequality to > (greater than) Compute!

Paired T hypothesis test: μD = μ1 - μ2 : Mean of the difference between After_five_years and Initial H0 : μD = 0 HA : μD > 0 Hypothesis test results:

Difference Mean Std. Err. DF T-Stat P-value After_five_years - Initial 4.8333333 2.54842 5 1.8966 0.0582 Differences stored in column, Differences. P-value = 0.058 (Round to three decimal places as needed.) Choose the correct conclusion below.

A. Reject H0. There is not sufficient evidence to conclude that there has been an improvement in the clarity of the water in the 5-year period.

B. Do not reject H0. There is not sufficient evidence to conclude that there has been an improvement in

the clarity of the water in the 5-year period.

C. Reject H0. There is sufficient evidence to conclude that there has been an improvement in the clarity of the water in the 5-year period.

D. Do not reject H0. There is sufficient evidence to conclude that there has been an improvement in the

clarity of the water in the 5-year period. (c) Draw a boxplot of the differences data. Choose the correct boxplot below. Graph Boxplot Column: Differences, Check Other Options: Draw boxes horizontally Compute!

Our Boxplot is like Boxplot D.

Does this visual evidence support the results obtained in part (b)? Yes No If the differences are mostly positive, then it supports the rejection of the null hypothesis. If the differences are not mostly positive, then it does not support the rejection of the null hypothesis. The differences are mostly positive. Notice that 0 is far to the left of the distribution. Therefore, the boxplot does support the results obtained in (b).

The response variable is the time to graduate. The explanatory variable is the use of community college or not.

Since the hypothesis we are seeking evidence for is whether it takes longer time for community college transfer students, this becomes the alternate hypothesis.

To find the test statstic and P-value, we will run a test in StatCrunch StatCrunch Stat T-Stats Two sample, with summary Input mean, std dev. and sample size for Sample 1 and Sample 2, uncheck pooled variances, select Hypothesis test, change alternative inequality to > Compute!

Here are the results of the test:

c) T-statistic = 12.50(Round to three decimal places as needed.) P-value =0(Round to three decimal places as needed.)

Since the P-value is very small, smaller than 0.10, we will reject the null hypothesis and found evidence for the alternative hypothesis .

Select the options tab and edit, select confidence interval for μ1 - μ2, Level: 0.99 Compute!

Here are the results:

(0.753, 1.145) (Round to three decimal places as needed.)

Yes! In part c) we found evidence that it takes longer time for community college transfers and part d) shows that the difference of the two means is a positive number, which means the first mean (mean time for community college transfers) is more than no transfers.

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 22 with a standard deviation of 4, while the 200 students in group 2 had a mean score of 17.6 with a standard deviation of 2.7. Group 1 n1 = 200, �̅�𝑥 R1 = 22, s 1= 4 Group 2 n2 = 200, �̅�𝑥2 = 17.6, s2= 2.7

Complete parts (a) and (b) below. (a) Determine the 90% confidence interval for the difference in scores, μ1−μ2. Interpret the interval. Open StatCrunch Stat T-Stats Two Sample, with summary Input: Sample 1 mean, std. dev. and size, repeat for Sample 2;, select confidence interval; Level 0.90 Compute! Two sample T confidence interval: μ1 : Mean of Population 1 μ2 : Mean of Population 2 μ1 - μ2 : Difference between two means (without pooled variances) 90% confidence interval results: Difference Sample Diff. Std. Err. DF L. Limit U. Limit μ1 - μ2 4.4 0.34124771 349.16531 3.8372043 4.9627957 (3.837, 4.963) (Round to three decimal places as needed.) Interpret the interval. Choose the correct answer below.

A. The researchers are 90% confident that the difference of the means is in the interval. B. There is a 90% probability that the difference of the means is in the interval. C. The researchers are 90% confident that the difference between randomly selected individuals will be in the

interval. D. There is a 90% probability that the difference between randomly selected individuals will be in the interval.

(b) What does this say about priming? If a confidence interval for the difference between two population means contains zero, then it is possible that the two populations have the same mean. If the confidence interval does not contain zero, then it is improbable that the two populations have the same mean.

A. Since the 90% confidence interval does not contain zero, the results suggest that priming does not have an effect on scores.

B. Since the 90% confidence interval contains zero, the results suggest that priming does not have an effect on scores.

C. Since the 90% confidence interval does not contain zero, the results suggest that priming does have an effect on scores.

D. Since the 90% confidence interval contains zero, the results suggest that priming does have an effect on scores.

Test the hypothesis that σ1 < σ2 at the =0.05 level of significance for the given sample data. Assume that the populations are normally distributed. Use the P-value approach.

Population 1 Population 2 n 51 25 s 7.5 10.8

State the null and alternative hypotheses for this test. 𝐻𝐻0: 𝜎𝜎1 = 𝜎𝜎2 𝐻𝐻1: 𝜎𝜎1 < 𝜎𝜎2 Use technology to find the P-value for this test. Open StatCrunch Stat Variance Stats Two sample, with summary Input the Variance and size for each sample (Remember the summary you were given includes the std. dev. for each sample, therefore you must square it to get the variance.), Hypothesis: alternative inequality < Compute! Two sample variance hypothesis test: σ1

2 : Variance of population 1 σ2

2 : Variance of population 2 σ1

2/σ22 : Ratio of two variances

H0 : σ12/σ2

2 = 1 HA : σ1

2/σ22 < 1

Hypothesis test results: Ratio Num. DF Den. DF Sample Ratio F-Stat P-value

σ12/σ2

2 50 24 0.48225309 0.48225309 0.0149 The P-value is 0.015. (Round to three decimal places as needed.) Make a statement regarding the null hypothesis and draw a conclusion for this test. Compare the obtained P-value to the given level of significance α. If P-value < α, reject the null hypothesis.Otherwise, do not reject the null hypothesis. If the null hypothesis is rejected, then there is sufficient evidence to support the alternative hypothesis at the given level of significance. If the null hypothesis is not rejected, then there is not sufficient evidence to support the alternative hypothesis at the given level of significance. Use this information to make a statement regarding the null hypothesis and draw a conclusion for this test.

Since the P-value is less than the α = 0.05 level of significance, reject the null hypothesis. There is sufficient evidence to conclude that the standard deviation of population 1 is significantly less than the standard deviation of Population 2 at the α = 0.05 level of significance.

In a randomized, double-blind experiment, 100 babies were randomly divided into a treatment group left parenthesis n 1 equals 50 right parenthesis(n1=50) and a control group left parenthesis n 2 equals 50 right parenthesis(n2=50). After the study, the treatment group had a mean serum retinol concentration of 49.14 micrograms per deciliter left parenthesis mu g divided by dL right parenthesis(μg/dL) with a standard deviation of 16.83 μg/dL, while the control group had a mean serum retinol concentration of 10.24 μg/dL with a standard deviation of 5.89 μg/dL. Does the treatment group have a higher standard deviation for serum retinol concentration than the control group at the α = 0.01 level of significance? It is known that serum retinol concentration is normally distributed. Use the P-value approach to perform the test. Summary: (This is not required, I just like to have the numbers organized.) nt = 50, �̅�𝑥𝑡𝑡 = 49.14 mg/dl, st = 16.83g/dl nc = 50, �̅�𝑥𝑐𝑐 = 10.24 mg/dl, sc = 5.89mg/dl Let 𝜎𝜎1 represent the population standard deviation for the treatment group and σ2 represent the population standard deviation for the control group. State the null and alternative hypotheses for this test. 𝐻𝐻0: 𝜎𝜎1 = 𝜎𝜎2 𝐻𝐻1: 𝜎𝜎1 > 𝜎𝜎2 Use technology to find the P-value for this test. Open StatCrunch Variance Stats Two sample, with summary Input Sample 1 and 2 information (remembering to square the std. dev. for the variance values), check direction of the inequality for the alternative Compute!

Two sample variance hypothesis test: σ12 : Variance of population 1 σ22 : Variance of population 2 σ12/σ22 : Ratio of two variances

H0 : σ12/σ22 = 1 HA : σ12/σ22 > 1 Hypothesis test results: Ratio Num. DF Den. DF Sample Ratio F-Stat P-value σ1

2/σ22 49 49 8.1646513 8.1646513 <0.0001

The P-value is 0.000. (Round to three decimal places as needed.) Make a statement regarding the null hypothesis and draw a conclusion for this test. Since the P-value is less than the α = 0.01 level of significance, reject the null hypothesis. There is sufficient evidence to conclude that the treatment group has a higher standard deviation for serum retinol concentration than the control group at the α = 0.01 level of significance.

The table to the right contains observed values and expected values in parentheses for two categorical variables, X and Y, where variable X has three categories and variable Y has two categories. Use the table to complete parts(a) and(b) below.

X1 X2 X3 Y1 30

(33.27) 41

(43.18) 55

(49.55) Y2 17

(13.73) 20

(17.82) 15

(20.45)

Open in StatCrunch Stat Tables Contingency With summary Select column(s): select all three X1, X2 and X3, Row labels: -- Compute!

Contingency table results: Rows: _ Columns: None

X1 X2 X3 Total Y1 30 42 53 125 Y2 16 22 16 54 Total 46 64 69 179 Chi-Square test:

Statistic DF Value P-value Chi-square 2 2.5982143 0.2728

(a) Compute the value of the chi-square test statistic. 𝜒𝜒02 = 2.598 (Round to three decimal places as needed.)

(b) Test the hypothesis that X and Y are independent at the α = 0.1 level of significance.

What is the P-value?

P-value = 0.273 (Round to three decimal places as needed.) Should the null hypothesis be rejected?

A. No, Do not reject H0. There is not sufficient evidence at the α (0.1) level of significance to conclude that X and Y are dependent because the P-value (0.273) > α.

B. Yes, reject H0. There is not sufficient evidence at the α = 0.1 level of significance to conclude that X and Y are dependent because the P-value < α.

C. No, Do not reject H0. There is sufficient evidence at the α = 0.1 level of significance to conclude that X and Y are dependent because the P-value < α.

D. Yes, reject H0. There is not sufficient evidence at the α = 0.1 level of significance to conclude that X and Y are dependent because the P-value > α.

A highway safety institution conducts experiments in which cars are crashed into a fixed barrier at 40 mph. In the institute's 40-mph offset test, 40% of the total width of each vehicle strikes a barrier on the driver's side. The barrier's deformable face is made of aluminum honeycomb, which makes the forces in the test similar to those involved in a frontal offset crash between two vehicles of the same weight, each going just less than 40 mph. You are in the market to buy a family car and you want to know if the mean head injury resulting from this offset crash is the same for large family cars, passenger vans, and midsize utility vehicles (SUVs). The data in the accompanying table were collected from the institute's study. Complete parts (a) through (d) below. Click the icon to view the data table (a) State the null and alternative hypotheses.

(b) Normal probability plots indicate that the sample data come from normal populations. Are the requirements to use the one-way ANOVA procedure satisfied? A one-way ANOVA procedure can be used provided the following requirements are met. 1. There are k simple random samples from k populations. 2. The k samples are independent of each other; that is, the subjects in one group cannot be

related in any way to subjects in a second group. 3. The populations are normally distributed. 4. The populations have equal variances. This condition is met if the largest sample standard

deviation is no more than two times larger than the smallest sample standard deviation.

Summary statistics: Column Std. dev. Unadj. variance Unadj. std. dev.

Large_Family_Cars 200.58818 34487.673 185.70857 Passenger_Vans 185.56349 29514.694 171.79841 Midsize_Utility_Vehicles 126.87639 13797.959 117.46471 A. No, because the largest sample standard deviation is more than twice the smallest sample

standard deviation. B. Yes, all the requirements for use of a one-way ANOVA procedure are satisfied. C. No, because the populations are not normally distributed.

D. No, because the samples are not independent. (c) Test the hypothesis that the mean head injury for each vehicle type is the same at the α = 0.01 level of significance. Use technology to find the F-test statistic for this data set. Open StatCrunch Stat Anova One Way Selected Columns: select all three Compute! Analysis of Variance results: Data stored in separate columns. Column statistics

Column n Mean Std. Dev. Std. Error Large_Family_Cars 7 327.42857 200.58818 75.815207 Passenger_Vans 7 393.14286 185.56349 70.136407 Midsize_Utility_Vehicles 7 318.42857 126.87639 47.954769 ANOVA table Source DF SS MS F-Stat P-value

Columns 2 23290.381 11645.19 0.38489267 0.686 Error 18 544602.29 30255.683

Total 20 567892.67

F0 = 0.385 (Round to three decimal places as needed.) Determine the P-value and state the appropriate conclusion below. Since the P-value is 0.6860 there is insufficient evidence to reject the null hypothesis. Thus, we cannot conclude that the means are different at the α = 0.01 level of significance. (Round to four decimal places as needed.) (d) Draw boxplots of the three vehicle types to support the results obtained in part (c). Graph Boxplot Select all three , Draw horizontally Compute!

Given the following ANOVA output, complete parts (a) through (c).

Source DF SS MS F P Factor A 1 12.2 12.2 0.27 0.610 Factor B 2 28.0 14.0 0.31 0.737 Interaction 2 17.2 8.6 0.19 0.829 Error 18 814.9 45.3

(a) Is there evidence of an interaction effect? Hypothesis Testing for Interaction: H0: There is no interaction between factors A and B H1: There is an interaction between factors A and B If the P-value is small, reject the null hypothesis. Since 0.829 is a large P-value which is not less than α (0.05), do not reject Ho. There is no evidence of an interaction effect. No Yes (b) Based on the P-value, is there evidence of a difference in the means from factor A? Hypothesis Testing for Factor A effect: H0: The means do not differ from factor A H1: The means differ from factor A Since 0.610 is a large P-value, do not reject Ho. There is no evidence of a difference in the means from factor A. Yes No (b) Based on the P-value, is there evidence of a difference in the means from factor B? Hypothesis Testing for Factor B effect: H0: The means do not differ from factor B H1: The means differ from factor B Since 0.737 is a large P-value, do not reject Ho. There is no evidence of a difference in the means from factor B.

Yes No (c) What is the mean square error? A mean square is an average (mean) of squared values. The denominator of the F-test statistic is called the mean square due to error and denoted MSE. Find MSE in the ANOVA output. MSE = 45.3 The mean square error is 45.3.

Assume that the accompanying data come from populations that are normally distributed with the same variance. Complete parts (a) through (d). Click the icon to view the data table – View in StatCrunch

(a) Determine whether or not there is significant interaction between factor A and factor B. View Data in StatCrunch Stat ANOVA Two Way Responses in: Response, Row factor in: Factor A, Column factor in: Factor B Options: Check Plot interactions and Compute Tuckey HSD (check Level: value) Compute!

Two Way Analysis of Variance results: Responses: Response Row factor: Factor A Column factor: Factor B ANOVA table

Source DF SS MS F-Stat P-value Factor A 1 35.770417 35.770417 0.38276177 0.5439 Factor B 2 886.96333 443.48167 4.7454809 0.0221 Interaction 2 5.6633333 2.8316667 0.030300283 0.9702 Error 18 1682.1625 93.453472

Total 23 2610.5596

Tukey HSD results (95% level) for Factor A: Level 1 subtracted from

Difference Lower Upper P-value Level 2 -2.4416667 -10.733146 5.8498125 0.5439 Tukey HSD results (95% level) for Factor B: Level 1 subtracted from

Difference Lower Upper P-value Level 2 -6.2 -18.536051 6.1360511 0.4226 Level 3 8.625 -3.7110511 20.961051 0.2029 Level 2 subtracted from

Difference Lower Upper P-value Level 3 14.825 2.4889489 27.161051 0.0174 Tukey HSD results (95% level) for Factor A*Factor B: Level 1,Level 1 subtracted from

Difference Lower Upper P-value

Level 1,Level 2 -7.375 -29.09908 14.34908 0.8834 Level 1,Level 3 7.875 -13.84908 29.59908 0.8531 Level 2,Level 1 -3.725 -25.44908 17.99908 0.9933 Level 2,Level 2 -8.75 -30.47408 12.97408 0.7917 Level 2,Level 3 5.65 -16.07408 27.37408 0.9586 Level 1,Level 2 subtracted from

Difference Lower Upper P-value Level 1,Level 3 15.25 -6.4740797 36.97408 0.2719 Level 2,Level 1 3.65 -18.07408 25.37408 0.9939 Level 2,Level 2 -1.375 -23.09908 20.34908 0.9999 Level 2,Level 3 13.025 -8.6990797 34.74908 0.4302 Level 1,Level 3 subtracted from

Difference Lower Upper P-value Level 2,Level 1 -11.6 -33.32408 10.12408 0.5506 Level 2,Level 2 -16.625 -38.34908 5.0990797 0.1971 Level 2,Level 3 -2.225 -23.94908 19.49908 0.9994 Level 2,Level 1 subtracted from

Difference Lower Upper P-value Level 2,Level 2 -5.025 -26.74908 16.69908 0.9747 Level 2,Level 3 9.375 -12.34908 31.09908 0.7426 Level 2,Level 2 subtracted from

Difference Lower Upper P-value Level 2,Level 3 14.4 -7.3240797 36.12408 0.3271 If the interaction P-value is large, there is no significant interaction. The P-value is large if it is more than 0.05. There is no significant interaction between factor A and factor B because 0.970 is a large P-value. There is significant interaction. There is no significant interaction. (b) If there is no significant interaction, determine if there is a significant difference in the means for the two levels of factor A.

Review the two-way ANOVA output to determine the P-value for factor A. If it is large, there is no significant difference in the means for the two levels of factor A. There is no significant difference in the means for the two levels of factor A because 0.970 is a large P-value. Review the two-way ANOVA output to determine the P-value for factor B. If it is large, there is no significant difference in the means for the two levels of factor B.

There is a significant difference in the means for the two levels of factor B because 0.221 is a small P-value. If there is no significant interaction, determine if there is a significant difference in the means for the three levels of factor B.

(c) Draw an interaction plot of the data to support the results of parts (a) and (b). This was created in the beginning of the StatCrunch calculations. It can be accessed through the > symbol in the lower right corner of the results.

The lines in the interaction plot are close to parallel, supporting the conclusion that there is no interaction effect. (d) If there is a significant difference in the means for the two levels of factor A, use Tukey's test to determine which pairwise means differ using a familywise error rate of α = 0.05. If there is a significant difference in the means for the three levels of factor B, use Tukey's test to determine which pairwise means differ using a familywise error rate of α = 0.05. Obtained in part (b), there is no a significant difference in the means for the two levels of factor A. Thus, it is not necessary to use Tukey's test to determine which pairwise means of factor A differ. If the interval does not include 0, reject the null hypothesis that the means do not differ.

If there is a significant difference in the means for the three levels of factor B, use Tukey's test to determine which pairwise means differ using a familywise error rate of α = 0.05. Choose the correct answer below.