Sfs4e ppt 10

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.

Chapter

Hypothesis Tests Regarding a Parameter

10


Section

The Language of Hypothesis Testing

10.1

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.10-3

1. Determine the null and alternative hypotheses

2. Explain Type I and Type II errors

3. State conclusions to hypothesis tests

Objectives


Objective 1

• Determine the Null and Alternative Hypotheses


A hypothesis is a statement regarding a characteristic of one or more populations.

In this chapter, we look at hypotheses regarding a single population parameter.


Examples of Claims Regarding a Characteristic of a Single Population

• In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

Source: ReadersDigest.com poll created on 2008/05/02



• According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.




• According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

• Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.



CAUTION!

We test these types of statements using sample data because it is usually impossible or impractical to gain access to the entire population. If population data are available, there is no need for inferential statistics.


Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations.


1. Make a statement regarding the nature of the population.

2. Collect evidence (sample data) to test the statement.

3. Analyze the data to assess the plausibility of the statement.

Steps in Hypothesis Testing


The null hypothesis, denoted H0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference and is assumed true until evidence indicates otherwise.


The alternative hypothesis, denoted H1, is a statement that we are trying to find evidence to support.


In this chapter, there are three ways to set up the null and alternative hypotheses:

1.Equal versus not equal hypothesis (two-tailed test)

H0: parameter = some value

H1: parameter ≠ some value



1. Equal versus not equal hypothesis (two-tailed test)



2. Equal versus less than (left-tailed test)


H1: parameter < some value



1. Equal versus not equal hypothesis (two-tailed test)



2. Equal versus less than (left-tailed test)


H1: parameter < some value

3. Equal versus greater than (right-tailed test)


H1: parameter > some value


“In Other Words”

The null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. We seek evidence that supports the statement in the alternative hypothesis.


For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed.

a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

b)According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

c)Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

Parallel Example 2: Forming Hypotheses


a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H0: p=0.62.

Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H1: p≠0.62.

Solution


b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

The hypothesis deals with a population mean, μ. If the mean call length on a cellular phone is no different than in 2006, it will be 3.25 minutes so the null hypothesis isH0: μ = 3.25.

Since the researcher believes that the mean call length has increased, the alternative hypothesis is:H1: μ > 3.25, a right-tailed test.

Solution


c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

The hypothesis deals with a population standard deviation, σ. If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H0: σ = 0.23.

Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H1: σ < 0.23, a left-tailed test.

Solution


Objective 2

• Explain Type I and Type II Errors


1. Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct.

Four Outcomes from Hypothesis Testing



2. Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct.





3. Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error.





3. Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error.

4. Do not reject the null hypothesis when the alternative hypothesis is true. This decision would be incorrect. This type of error is called a Type II error.



For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error?

a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

b)According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

Parallel Example 3: Type I and Type II Errors


a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

A Type I error is made if the researcher concludes thatp ≠ 0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%.

A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%.

Solution


b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

A Type I error occurs if the sample evidence leads the researcher to conclude that μ > 3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes.

A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes.

Solution


α = P(Type I Error)

= P(rejecting H0 when H0 is true)

β = P(Type II Error)

= P(not rejecting H0 when H1 is true)


The probability of making a Type I error, α, is chosen by the researcher before the sample data is collected.

The level of significance, α, is the probability of making a Type I error.


“In Other Words” As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa.


Objective 3

• State Conclusions to Hypothesis Tests


CAUTION!

We never “accept” the null hypothesis, because, without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis. This is just like the court system. We never declare a defendant “innocent”, but rather say the defendant is “not guilty”.


According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

a)Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.

b)Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion.

Parallel Example 4: Stating the Conclusion


a) Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.

The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null hypothesis (μ = 3.25) is rejected, there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes.

Solution


b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion.

Since the null hypothesis (μ = 3.25) is not rejected, there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes.

Solution


Section

Hypothesis Tests for a Population Proportion

10.2


Objectives

1. Explain the logic of hypothesis testing

2. Test the hypotheses about a population proportion

3. Test hypotheses about a population proportion using the binomial probability distribution.


Objective 1

• Explain the Logic of Hypothesis Testing


A researcher obtains a random sample of 1000 people and finds that 534 are in favor of the banning cell phone use while driving, so = 534/1000. Does this suggest that more than 50% of people favor the policy? Or is it possible that the true proportion of registered voters who favor the policy is some proportion less than 0.5 and we just happened to survey a majority in favor of the policy? In other words, would it be unusual to obtain a sample proportion of 0.534 or higher from a population whose proportion is 0.5? What is convincing, or statistically significant, evidence?


When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant. When results are found to be statistically significant, we reject the null hypothesis.


To determine if a sample proportion of 0.534 is statistically significant, we build a probability model.

Since np(1 – p) = 100(0.5)(1 – 0.5) = 250 ≥ 10 and the sample size (n = 1000) is sufficiently smaller than the population size, we can use the normal model to describe the variability in . The mean of the distribution of is and the standard

deviation is

öp0.5

öp

0.5 1 0.5

10000.016.


Sampling distribution of the sample proportion


We may consider the sample evidence to be statistically significant (or sufficient) if the sample proportion is too many standard deviations, say 2, above the assumed population proportion of 0.5.

The Logic of the Classical Approach


Recall that our simple random sample yielded a sample proportion of 0.534, so

standard deviations above the hypothesized proportion of 0.5.

Therefore, using our criterion, we would reject the null hypothesis.

z öp öp

öp

0.534 0.5

0.5 1 0.5 1000

2.15


Why does it make sense to reject the null hypothesis if the sample proportion is more than 2 standard deviations away from the hypothesized proportion? The area under the standard normal curve to the right ofz = 2 is 0.0228.


If the null hypothesis were true (population proportion is 0.5), then 1 – 0.0228 = 0.9772 = 97.72% of all sample proportions will be less than

.5 + 2(0.016) = 0.532and only 2.28% of the sample proportions will be more than 0.532.


If the sample proportion is too many standard deviations from the proportion stated in the null hypothesis, we reject the null hypothesis.

Hypothesis Testing Usingthe Classical Approach


A second criterion we may use for testing hypotheses is to determine how likely it is to obtain a sample proportion of 0.534 or higher from a population whose proportion is 0.5. If a sample proportion of 0.534 or higher is unlikely (or unusual), we have evidence against the statement in the null hypothesis. Otherwise, we do not have sufficient evidence against the statement in the null hypothesis.

The Logic of the P-Value Approach


We can compute the probability of obtaining a sample proportion of 0.534 or higher from a population whose proportion is 0.5 using the normal model.


Recall

So, we compute

The value 0.0158 is called the P-value, which means about 2 samples in 100 will give a sample proportion as high or higher than the one we obtained if the population proportion really is 0.5. Because these results are unusual, we take this as evidence against the statement in the null hypothesis.

z 0.534 0.5

0.5 1 0.5 1000

2.15

P öp 0.534 P(z 2.15) 0.0158


If the probability of getting a sample proportion as extreme or more extreme than the one obtained is small under the assumption the statement in the null hypothesis is true, reject the null hypothesis.

Hypothesis TestingUsing the P-value Approach


Objective 2

• Test hypotheses about a population proportion.


Recall:

• The best point estimate of p, the proportion of the population with a certain characteristic, is given by

where x is the number of individuals in the sample with the specified characteristic and n is the sample size.

ˆ p x

n


Recall:

• The sampling distribution of is approximately normal, with mean and standard deviation

provided that the following requirements are satisfied:

1. The sample is a simple random sample.2. np(1-p) ≥ 10.3. The sampled values are independent of each

other.

ˆ p

ˆ p p

ˆ p p(1 p)

n


Testing Hypotheses Regarding a Population Proportion, p

To test hypotheses regarding the population proportion, we can use the steps that follow, provided that:•The sample is obtained by simple random sampling.• np0(1 – p0) ≥ 10.•The sampled values are independent of each other.


Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:


Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.


Step 3: Compute the test statistic

Note: We use p0 in computing the standard error rather than . This is because, when we test a hypothesis, the null hypothesis is always assumed true.

z0 ˆ p p0

p0(1 p0)

n

ˆ p

Classical Approach


Classical Approach

Two-Tailed

(critical value)

Use Table V to determine the critical value.


Classical Approach

Left-Tailed

(critical value)


Classical Approach

Right-Tailed

(critical value)


Step 4: Compare the critical value with the test statistic:

Classical Approach


By Hand Step 3: Compute the test statistic.

P-Value Approach

z0 ˆ p p0

p0(1 p0)n


P-Value Approach

Two-Tailed

Use Table V to estimate the P-value.


P-Value Approach

Left-Tailed


P-Value Approach

Right-Tailed


Technology Step 3: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value.The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, MINITAB, Excel, and StatCrunch are in the Technology Step-by-Step in the text.

P-Value Approach


Step 4: If the P-value < α, reject the null hypothesis.

P-Value Approach


Step 5: State the conclusion.


Parallel Example 1: Testing a Hypothesis about a Population Proportion: Large Sample Size

In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the α = 0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997?

Source: Gallup Poll


Solution

We want to know if p > 0.46. First, we must verify the requirements to perform the hypothesis test:

1. This is a simple random sample.

2. np0(1 – p0) = 1010(0.46)(1 – 0.46) = 250.8 > 10

3. Since the sample size is less than 5% of the population size, the assumption of independence is met.


Solution

Step 1: H0: p = 0.46 versus H1: p > 0.46

Step 2: The level of significance is α = 0.05.

Step 3: The sample proportion is .

The test statistic is then

ˆ p 525

10100.52

z0 0.52 0.46

0.46(1 0.46)

1010

3.83


Solution: Classical Approach

Step 4: Since this is a right-tailed test, we determine the critical value at the α = 0.05 level of significance to be

z0.05 = 1.645.

Step 5: Since the test statistic, z0 = 3.83, is

greater than the critical value 1.645, we reject the null hypothesis.


Solution: P-Value Approach

Step 4: Since this is a right-tailed test, the P- value is the area under the standard normal distribution to the right of the test statistic z0=3.83. That is, P-value =

P(Z > 3.83) ≈ 0.

Step 5: Since the P-value is less than the level of significance, we reject the null

hypothesis.


Solution

Step 6: There is sufficient evidence at the α = 0.05 level of significance to

conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.


Objective 3

• Test hypotheses about a population proportion using the binomial probability distribution.


For the sampling distribution of to be approximately normal, we require np(1– p) be at least 10. What if this requirement is not met?

We stated that an event was unusual if the probability of observing the event was less than 0.05. This criterion is based on the P-value approach to testing hypotheses; the probability that we computed was the P-value. We use this same approach to test hypotheses regarding a population proportion for small samples.

ˆ p


Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size

In 2006, 10.5% of all live births in the United States were to mothers under 20 years of age. A sociologist claims that births to mothers under 20 years of age is decreasing. She conducts a simple random sample of 34 births and finds that 3 of them were to mothers under 20 years of age. Test the sociologist’s claim at the α = 0.01 level of significance.



Approach:

Step 1: Determine the null and alternative hypotheses

Step 2: Check whether np0(1–p0) is greater than or equal to 10, where p0 is the proportion stated in the null hypothesis. If it is, then the sampling distribution of is approximately normal and we can use the steps for a large sample size. Otherwise we use the following Steps 3 and 4.

öp



Approach:

Step 3: Compute the P-value. For right-tailed tests, the P-value is the probability of obtaining x

or more successes. For left-tailed tests, the P- value is the probability of obtaining x or fewer successes. The P-value is always computed with the proportion given in the null hypothesis.

Step 4: If the P-value is less than the level of significance, α, we reject the null hypothesis.


Solution

Step 1: H0: p = 0.105 versus H1: p < 0.105

Step 2: From the null hypothesis, we have p0 = 0.105. There were 34 mothers sampled, so np0(1– p0)=3.57 < 10.

Thus, the sampling distribution of is not approximately normal.

ˆ p


Solution

Step 3: Let X represent the number of live births in the United States to mothers under 20

years of age. We have x = 3 successes in n = 34 trials so = 3/34= 0.088. We want to

determine whether this result is unusual if the population mean is truly 0.105. Thus,

P-value = P(X ≤ 3 assuming p=0.105 )

= P(X = 0) + P(X =1)

+ P(X =2) + P(X = 3)

= 0.51

ˆ p


Solution

Step 4: The P-value = 0.51 is greater than the level of significance so we do not

reject H0. There is insufficient evidence to conclude that the percentage of live births in the United States to mothers under the age of 20 has decreased below the 2006 level of 10.5%.


Section

Hypothesis Tests for a Population Mean

10.3


Objectives

1. Test hypotheses about a mean 2. Understand the difference between statistical

significance and practical significance.


Objective 1

• Test Hypotheses about a Mean


To test hypotheses regarding the population mean assuming the population standard deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace σ with s,

follows Student’s t-distribution with n –1 degrees of freedom.

x s

n


1. The t-distribution is different for different degrees of freedom.

Properties of the t-Distribution



2. The t-distribution is centered at 0 and is symmetric about 0.




2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area under the curve is 1. Because of the symmetry, the area under the curve to the right of 0 equals the area under the curve to the left of 0 equals 1/2.



4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0.



4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0.

5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution because using s as an estimate of σ introduces more variability to the t-statistic.



6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get closer to the values of σ by the Law of Large Numbers.



Testing Hypotheses Regarding aPopulation Mean

To test hypotheses regarding the population mean, we use the following steps, provided that:•The sample is obtained using simple random sampling.•The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).•The sampled values are independent of each other.


Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:


Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.


Classical Approach

Step 3: Compute the test statistic

which follows the Student’s t-distribution with n – 1 degrees of freedom.

t0

x 0

s

n


Classical Approach

Use Table VI to determine the critical value.


Classical Approach

Two-Tailed


Classical Approach

Left-Tailed


Classical Approach

Right-Tailed


Classical Approach

Step 4: Compare the critical value with the test statistic:


P-Value Approach

By Hand Step 3: Compute the test statistic

which follows the Student’s t-distribution with n – 1 degrees of freedom.

t0

x 0

s

n


P-Value Approach

Use Table VI to approximate the P-value.


P-Value ApproachTwo-Tailed


P-Value ApproachLeft-Tailed


P-Value ApproachRight-Tailed


P-Value Approach

Technology Step 3: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, MINITAB, Excel, and StatCrunch, are in the Technology Step-by-Step in the text.


P-Value Approach

Step 4: If the P-value < α, reject the null hypothesis.


Step 5: State the conclusion.


The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test. However, for small samples, if the data have outliers, the procedure should not be used.


Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample

Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be 5708.07 with a standard deviation of 992.05.

At the α = 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?


Solution

We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day.

Since the sample size is large, we follow the steps for testing hypotheses about a population mean for large samples.


Solution

Step 1: H0: μ = 5710 versus H1: μ ≠ 5710


Step 3: The sample mean is = 5708.07 and the sample standard deviation is s = 992.05. The test statistic is

t0 5708.07 5710

992.05 45 0.013

x



Step 4: Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n –1 = 45 – 1 = 44 degrees of freedom to be approximately –t0.025 = –2.021

and t0.025 = 2.021.

Step 5: Since the test statistic, t0 = –0.013, is between

the critical values, we fail to reject the null hypothesis.



Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n –1 = 45 – 1

= 44 degrees of freedom to the left of –t0.025= –0.013

and to the right of t0.025 = 0.013. That is, P-

value = P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013). 0.50 < P-value.

Step 5: Since the P-value is greater than the level of significance (0.05<0.5), we fail to reject the null hypothesis.


Solution

Step 6: There is insufficient evidence at the α = 0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day.


Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample

According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data.

5.70 5.67 5.73 5.61 5.70 5.67

5.65 5.62 5.73 5.65 5.79 5.73

5.77 5.71 5.70 5.76 5.73 5.72

At the α = 0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams?


Solution

We assume that the weight of the state quarters is 5.67 grams. This is a two-tailed test since we are interested in determining whether the weight differs from 5.67 grams.

Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers before proceeding to Steps 1-6.


Assumption of normality appears reasonable.


No outliers.


Solution

Step 1: H0: μ = 5.67 versus H1: μ ≠ 5.67


Step 3: From the data, the sample mean is calculated to be 5.7022 and the sample standard deviation is s = 0.0497. The test statistic is

t0 5.7022 5.67

.0497 182.75



Step 4: Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n – 1 = 18 – 1 = 17 degrees of freedom to be –t0.025 = –2.11 and t0.025 = 2.11.

Step 5: Since the test statistic, t0 = 2.75, is greater than

the critical value 2.11, we reject the null hypothesis.



Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n – 1 = 18 – 1

= 17 degrees of freedom to the left of –t0.025 = –2.75

and to the right of t0.025 = 2.75. That is, P-value

= P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75). 0.01 < P-value < 0.02.

Step 5: Since the P-value is less than the level of significance (0.02<0.05), we reject the null hypothesis.


Solution

Step 6: There is sufficient evidence at the α = 0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams.


Objective 2

• Understand the Difference between Statistical Significance and Practical Significance


When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance.


Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important.


Parallel Example 7: Statistical versus Practical Significance

In 2003, the average age of a mother at the time of her first childbirth was 25.2. To determine if the average age has increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5 with a standard deviation of 4.8, determine whether the mean age has increased using a significance level of α = 0.05.


Solution

Step 1: To determine whether the mean age has increased, this is a right-tailed test with

H0: μ = 25.2 versus H1: μ > 25.2.


Step 3: Recall that the sample mean is 25.5. The test statistic is then

t0

25.5 25.2

4.8 12002.17


Solution

Step 4: Since this is a right-tailed test, the critical value with α = 0.05 and 1200 – 1 = 1199 degrees of freedom is

P-value = P(t > 2.17).

From Table VI: .001 < P-value < .005

Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis.


Solution

Step 6: There is sufficient evidence at the 0.05 significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2.

Although we found the difference in age to be significant, there is really no practical significance in the age difference (25.2 versus 25.5).


Large Sample Sizes

Large sample sizes can lead to statistically significant results while the difference between the statistic and parameter is not enough to be considered practically significant.


Section

Putting It Together: Which Method Do I Use?

10.4


Objective

1. Determine the appropriate hypothesis test to perform


Objective 1

• Determine the Appropriate Hypothesis Test to Perform
