© 2010 Pearson Prentice Hall. All rights reserved Chapter Inferences on Two Samples 11

Chapter

Inferences on Two Samples

Section

Inference about Two Means: Dependent Samples

Objectives

1. Distinguish between independent and dependent sampling

2. Test hypotheses regarding matched-pairs data

3. Construct and interpret confidence intervals about the population mean difference of matched-pairs data

Objective 1

• Distinguish between Independent and Dependent Sampling

A sampling method is independent when the individuals selected for one sample do not dictate which individuals are to be in a second sample. A sampling method is dependent when the individuals selected to be in one sample are used to determine the individuals to be in the second sample.

Dependent samples are often referred to as matched-pairs samples.

For each of the following, determine whether the sampling method is independent or dependent.

a) A researcher wants to know whether the price of a one night stay at a Holiday Inn Express is less than the price of a one night stay at a Red Roof Inn. She randomly selects 8 towns where the location of the hotels is close to each other and determines the price of a one night stay.

b) A researcher wants to know whether the “state” quarters (introduced in 1999) have a mean weight that is different from “traditional” quarters. He randomly selects 18 “state” quarters and 16 “traditional” quarters and compares their weights.

Parallel Example 1: Distinguish between Independent and Dependent Sampling

a) The sampling method is dependent since the 8 Holiday Inn Express hotels can be matched with one of the 8 Red Roof Inn hotels by town.

b) The sampling method is independent since the “state” quarters which were sampled had no bearing on which “traditional” quarters were sampled.

Solution

Objective 2

• Test Hypotheses Regarding Matched-Pairs Data

“In Other Words”

Statistical inference methods on matched-pairs data use the same methods as inference on a single population mean with unknown, except that the differences are analyzed.

To test hypotheses regarding the mean difference of matched-pairs data, the following must be satisfied:

1. the sample is obtained using simple random sampling

2. the sample data are matched pairs,

3. the differences are normally distributed with no outliers or the sample size, n, is large (n ≥ 30).

Testing Hypotheses Regarding the Differenceof Two Means Using a Matched-Pairs Design

Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways, where d is the population mean difference of the matched-pairs data.

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

Step 3: Compute the test statistic

which approximately follows Student’s t-distribution with n-1 degrees of freedom. The values of and sd are the mean and standard deviation of the differenced data.

t0 d sd

Step 4: Use Table VI to determine the critical value using n-1 degrees of freedom.

Classical Approach

(critical value)

Two-Tailed

Classical Approach

(critical value)

Left-Tailed

Classical Approach

(critical value)

Right-Tailed

Step 5: Compare the critical value with the test statistic:

Classical Approach

Step 4: Use Table VI to determine the P-value using n-1 degrees of freedom.

P-Value Approach

Two-Tailed

P-Value Approach

Left-Tailed

P-Value Approach

Right-Tailed

Step 5: If P-value < , reject the null hypothesis.

P-Value Approach

Step 6: State the conclusion.

These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.

The following data represent the cost of a one night stay in Hampton Inn Hotels and La Quinta Inn Hotels for a random sample of 10 cities. Test the claim that Hampton Inn Hotels are priced differently than La Quinta Hotels at the =0.05 level of significance.

Parallel Example 2: Testing a Claim Regarding Matched-Pairs Data

City Hampton Inn La Quinta

Dallas 129 105

Tampa Bay 149 96

St. Louis 149 49

Seattle 189 149

San Diego 109 119

Chicago 160 89

New Orleans 149 72

Phoenix 129 59

Atlanta 129 90

Orlando 119 69

This is a matched-pairs design since the hotel prices come from the same ten cities. To test the hypothesis, we first compute the differences and then verify that the differences come from a population that is approximately normally distributed with no outliers because the sample size is small.

The differences (Hampton - La Quinta) are:

24 53 100 40 -10 71 77 70 39 50

with = 51.4 and sd = 30.8336.

Solution

No violation of normality assumption.

Solution

Step 1: We want to determine if the prices differ:

H0: d = 0 versus H1: d 0

Step 2: The level of significance is =0.05.

Step 3: The test statistic is

Solution

t0 51.4

30.833610

5.2716.

Step 4: This is a two-tailed test so the critical values at the =0.05 level of significance with n-1=10-1=9 degrees of freedom

are -t0.025=-2.262 and t0.025=2.262.

Solution: Classical Approach

Step 5: Since the test statistic, t0=5.27 is greater than the critical value t.025=2.262, we reject the null hypothesis.

Step 4: Because this is a two-tailed test, the P-value is two times the area under the t-distribution with n-1=10-1=9 degrees of freedom to the right of the test statistic t0=5.27. That is, P-value = 2P(t > 5.27) ≈ 2(0.00026)=0.00052 (using technology). Approximately 5 samples in 10,000 will yield results as extreme as we obtained if the null hypothesis is true.

Solution: P-Value Approach

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

Step 6: There is sufficient evidence to conclude that Hampton Inn hotels and La Quinta hotels are priced differently at the =0.05 level of significance.

Solution

Objective 3

• Construct and Interpret Confidence Intervals for the Population Mean Difference of Matched-Pairs Data

A (1-)100% confidence interval for d is given by

Lower bound:

Upper bound:

The critical value t/2 is determined using n-1 degrees of freedom.

Confidence Interval for Matched-Pairs Data

Note: The interval is exact when the population is normally distributed and approximately correct for nonnormal populations, provided that n is large.

Confidence Interval for Matched-Pairs Data

Construct a 90% confidence interval for the mean difference in price of Hampton Inn versus La Quinta hotel rooms.

Parallel Example 4: Constructing a Confidence Interval for Matched-Pairs Data

• We have already verified that the differenced data come from a population that is approximately normal with no outliers.

• Recall = 51.4 and sd = 30.8336.

• From Table VI with = 0.10 and 9 degrees of freedom, we find t/2 = 1.833.

Solution

• Lower bound =

• Upper bound =

We are 90% confident that the mean difference in hotel room price for Ramada Inn versus La Quinta Inn is between $33.53 and $69.27.

Solution

51.4 1.83330.8336

Section

Inference about Two Means: Independent Samples

Objectives

1. Test hypotheses regarding the difference of two independent means

2. Construct and interpret confidence intervals regarding the difference of two independent means

Suppose that a simple random sample of size n1 is taken from a population with unknown mean 1 and unknown standard deviation 1. In addition, a simple random sample of size n2 is taken from a population with unknown mean 2 and unknown standard deviation 2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30, n2 ≥ 30) , then

approximately follows Student’s t-distribution with the smaller of n1-1 or n2-1 degrees of freedom where is the sample mean and si

is the sample standard deviation from population i.

Sampling Distribution of the Difference of Two Means: Independent Samples with Population

Standard Deviations Unknown (Welch’s t)

t x 1 x 2 1 2

Objective 1

• Test Hypotheses Regarding the Difference of Two Independent Means

To test hypotheses regarding two population means, 1 and 2, with unknown population standard deviations, we can use the following steps, provided that:

1. the samples are obtained using simple random sampling;

2. the samples are independent;3. the populations from which the samples are drawn

are normally distributed or the sample sizes are large (n1 ≥ 30, n2 ≥ 30).

Testing Hypotheses Regarding the Difference of Two Means

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

which approximately follows Student’s t- distribution.

t0 x 1 x 2 1 2

Step 4: Use Table VI to determine the criticalvalue using the smaller of n1 -1 or n2 -1 degrees of freedom.

Classical Approach

(critical value)

Two-Tailed

Classical Approach

(critical value)

Left-Tailed

Classical Approach

(critical value)

Right-Tailed

Classical Approach

Step 4: Use Table VI to determine the P-value using the smaller of n1 -1 or n2 -1 degrees of freedom.

P-Value Approach

Two-Tailed

P-Value Approach

Left-Tailed

P-Value Approach

Right-Tailed

P-Value Approach

These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.

A researcher wanted to know whether “state” quarters had a weight that is more than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the following data.

Parallel Example 1: Testing Hypotheses Regarding Two Means

Test the claim that “state” quarters have a mean weight that is more than “traditional” quarters at the =0.05 level of significance.

NOTE: A normal probability plot of “state” quarters indicates the population could be normal. A normal probability plot of “traditional” quarters indicates the population could be normal

No outliers.

Step 1: We want to determine whether state quarters weigh more than traditional quarters:

H0: 1 = 2 versus H1: 1 > 2

Step 2: The level of significance is =0.05.

Step 3: The test statistic is

Solution

t0 5.7022 5.6494

0.04972

18 0.06892

Step 4: This is a right-tailed test with =0.05. Since n1-1=17 and n2-1=15, we will use 15 degrees of freedom. The corresponding critical value is t0.05=1.753.

Step 5: Since the test statistic, t0=2.53 is greater than the critical value t.05=1.753, we reject the null hypothesis.

Step 4: Because this is a right-tailed test, the P-value is the area under the t-distribution to the right of the test statistic t0=2.53. That is, P-value = P(t > 2.53) ≈ 0.01.

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

Step 6: There is sufficient evidence at the =0.05 level to conclude that the state quarters weigh more than the traditional quarters.

Solution

NOTE: The degrees of freedom used to determine the critical value in the last example are conservative. Results that are more accurate can be obtained by using the following degrees of freedom:

Objective 3

• Construct and Interpret Confidence Intervals Regarding the Difference of Two Independent Means

A simple random sample of size n1 is taken from a population with unknown mean 1 and unknown standard deviation 1. Also, a simple random sample of size n2 is taken from a population with unknown mean 2 and unknown standard deviation 2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1≥30 and n2≥30), a (1-)100% confidence interval about 1 - 2 is given by

Lower bound:

Upper bound:

Constructing a (1-) 100% ConfidenceInterval for the Difference of Two Means

x 1 x 2 t2

Construct a 95% confidence interval about the difference between the population mean weight of a “state” quarter versus the population mean weight of a “traditional” quarter.

Parallel Example 3: Constructing a Confidence Interval for the Difference of Two Means

• We have already verified that the populations are approximately normal and that there are no outliers.

• Recall = 5.702, s1 = 0.0497, =5.6494 and s2 = 0.0689.

• From Table VI with = 0.05 and 15 degrees of freedom, we find t/2 = 2.131.

Solution

• Lower bound =

• Upper bound =

Solution

5.702 5.649 2.1310.04972

0.06892

160.0086

5.702 5.649 2.1310.04972

0.06892

160.0974

We are 95% confident that the mean weight of the “state” quarters is between 0.0086 and 0.0974 ounces more than the mean weight of the “traditional” quarters. Since the confidence interval does not contain 0, we conclude that the “state” quarters weigh more than the “traditional” quarters.

Solution

When the population variances are assumed to be equal, the pooled t-statistic can be used to test for a difference in means for two independent samples. The pooled t-statistic is computed by finding a weighted average of the sample variances and using this average in the computation of the test statistic.

• The advantage to this test statistic is that it exactly follows Student’s t-distribution with n1+n2-2 degrees of freedom.

• The disadvantage to this test statistic is that it requires that the population variances be equal.

Section

Inference about Two Population Proportions

Objectives

1. Test hypotheses regarding two population proportions

2. Construct and interpret confidence intervals for the difference between two population proportions

3. Use McNemar’s Test to compare two proportions from matched-pairs data

4. Determine the sample size necessary for estimating the difference between two population proportions within a specified margin of error.

Suppose that a simple random sample of size n1 is taken from a

population where x1 of the individuals have a specified characteristic,

and a simple random sample of size n2 is independently taken from a

different population where x2 of the individuals have a specified

characteristic. The sampling distribution of , where

and , is approximately normal, with mean

and standard deviation

provided that and .

Sampling Distribution of the Difference between Two Proportions

ˆ p 1 ˆ p 2

ˆ p 1 x1 n1

ˆ p 1 ˆ p 2p1 p2

ˆ p 2 x2 n2

ˆ p 1 ˆ p 2

p1 1 p1 n1

p2 1 p2

n1ˆ p 1 1 ˆ p 1 10

n2ˆ p 2 1 ˆ p 2 10

The standardized version of is then written as

which has an approximate standard normal distribution.

Sampling Distribution of the Difference between Two Proportions

ˆ p 1 ˆ p 2

Z ˆ p 1 ˆ p 2 p1 p2

p1 1 p1 n1

p2 1 p2

Objective 1

• Test Hypotheses Regarding Two Population Proportions

The best point estimate of p is called the pooled estimate of p, denoted , where

Test statistic for Comparing Two Population Proportions

ˆ p x1 x2

z0 ˆ p 1 ˆ p 2 ˆ p 1 ˆ p 2

ˆ p 1 ˆ p 2

ˆ p 1 ˆ p 1

To test hypotheses regarding two population proportions, p1 and p2, we can use the steps that follow, provided that:

1. the samples are independently obtained using simple random sampling,

2. and and

3. n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment.

Hypothesis Test Regarding the Differencebetween Two Population Proportions

n1ˆ p 1 1 ˆ p 1 10

n2ˆ p 2 1 ˆ p 2 10,

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

z0 ˆ p 1 ˆ p 2

ˆ p 1 ˆ p 1

ˆ p x1 x2

Step 4: Use Table V to determine the criticalvalue.

Classical Approach

(critical value)

Two-Tailed

Classical Approach

(critical value)

Left-Tailed

Classical Approach

(critical value)

Right-Tailed

Classical Approach

Step 4: Use Table V to estimate the P-value..

P-Value Approach

Two-Tailed

P-Value Approach

Left-Tailed

P-Value Approach

Right-Tailed

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P-Value Approach

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An economist believes that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. He obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Test the economist’s claim at the =0.05 level of significance.

Parallel Example 1: Testing Hypotheses Regarding Two Population Proportions

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We must first verify that the requirements are satisfied:

– The samples are simple random samples that were obtained independently.

– x1=338, n1=800, x2=292 and n2=750, so

3. The sample sizes are less than 5% of the population size.

Solution

ˆ p 1 338

8000.4225 and ˆ p 2

7500.3893. Thus,

n1ˆ p 1 1 ˆ p 1 800(0.4225)(1 0.4225) 195.195 10

n2ˆ p 2 1 ˆ p 2 750(0.3893)(1 0.3893) 178.309 10

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Step 1: We want to determine whether the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. So,

H0: p1 = p2 versus H1: p1 > p2

or, equivalently,

H0: p1 - p2=0 versus H1: p1 - p2 > 0

Step 2: The level of significance is = 0.05.

Solution

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Step 3: The pooled estimate of is:

The test statistic is:

Solution

z0 0.4225 0.3893

0.4065 1 0.4065 1

ˆ p x1 x2

338 292

800 7500.4065.

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Step 4: This is a right-tailed test with =0.05. The critical value is z0.05=1.645.

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Step 5: Since the test statistic, z0=1.33 is less than the critical value z.05=1.645, we fail to reject the null hypothesis.

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Step 4: Because this is a right-tailed test, the P-value is the area under the normal to the right of the test statistic z0=1.33. That is, P-value = P(Z > 1.33) ≈ 0.09.

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

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Step 6: There is insufficient evidence at the =0.05 level to conclude that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access.

Solution

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Objective 2

• Construct and Interpret Confidence Intervals for the Difference between Two Population Proportions

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To construct a (1-)100% confidence interval for the difference between two population proportions, the following requirements must be satisfied:

1. the samples are obtained independently using simple random sampling,

2. , and

3. n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment.

Constructing a (1-) 100% Confidence Interval for the Difference between Two

Population Proportions

n1ˆ p 1 1 ˆ p 1 10

n2ˆ p 2 1 ˆ p 2 10

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Provided that these requirements are met, a (1-)100% confidence interval for p1-p2 is given by

Lower bound:

Upper bound:

ˆ p 1 ˆ p 2 z2

ˆ p 1 1 ˆ p 1

ˆ p 2 1 ˆ p 2

ˆ p 1 ˆ p 2 z2

ˆ p 1 1 ˆ p 1

ˆ p 2 1 ˆ p 2

Constructing a (1-) 100% Confidence Interval for the Difference between Two

Population Proportions

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An economist obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Find a 99% confidence interval for the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access.

Parallel Example 3: Constructing a Confidence Interval for the Difference between Two Population Proportions

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We have already verified the requirements for constructing a confidence interval for the difference between two population proportions in the previous example.

Recall

Solution

ˆ p 1 338

8000.4225 and ˆ p 2

7500.3893.

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Lower bound =

Upper bound =

Solution

0.4225 0.3893

2.5750.4225(1 0.4225)

0.3893(1 0.3893)

750 0.0310

0.4225 0.3893

2.5750.4225(1 0.4225)

0.3893(1 0.3893)

7500.0974

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We are 99% confident that the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access is between -0.03 and 0.10. Since the confidence interval contains 0, we are unable to conclude that the proportion of urban households with Internet access is greater than the proportion of rural households with Internet access.

Solution

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Objective 3

• Use McNemar’s Test to Compare Two Proportions from Matched-Pairs Data

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McNemar’s Test is a test that can be used to compare two proportions with matched-pairs data (i.e., dependent samples)

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Testing a Hypothesis Regarding the Difference of Two Population

Proportions: Dependent Samples

To test hypotheses regarding two population proportionsp1 and p2, where the samples are dependent, arrange thedata in a contingency table as follows:

Treatment A

Treatment B

Success Failure

Success f11 f12

Failure f21 f22

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Testing a Hypothesis Regarding the Difference of Two Population

Proportions: Dependent Samples

We can use the steps that follow provided that:1. the samples are dependent and are obtained

randomly and2. the total number of observations where the

outcomes differ must be greater than or equal to 10.That is, f12 + f21 ≥ 10.

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Step 1: Determine the null and alternative hypotheses.

H0: the proportions between the two populations are equal (p1 =

H1: the proportions between the two populations differ (p1 ≠ p2)

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z0 f12 f21 1

f12 f21

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Step 4: Use Table V to determine the criticalvalue. This is a two tailed test. However, z0 is always positive, so we only need to find the right critical value z/2.

Classical Approach

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Classical Approach

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Step 5: If z0 > z/2, reject the null hypothesis.

Classical Approach

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Step 4: Use Table V to determine the P-value..Because z0 is always positive, we find the area to the right of z0 and then double this area (since this is a two-tailed test).

P-Value Approach

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P-Value Approach

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P-Value Approach

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A recent General Social Survey asked the following two questions of a random sample of 1483 adult Americans under the hypothetical scenario that the government suspected that a terrorist act was about to happen:

• Do you believe the authorities should have the right to tap people’s telephone conversations?

• Do you believe the authorities should have the right to detain people for as long as they want without putting them on trial?

Parallel Example 4: Analyzing the Difference of Two Proportions from Matched-Pairs Data

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The results of the survey are shown below:

Do the proportions who agree with each scenario differ significantly? Use the =0.05 level of significance.

Parallel Example 4: Analyzing the Difference of Two Proportions from Matched-Pairs Data

Detain

Agree Disagree

Agree 572 237

Disagree 224 450

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The sample proportion of individuals who believe that the

authorities should be able to tap phones is

The sample proportion of individuals who believe that

the authorities should have the right to detain people is

. We want to

determine whether the difference in sample

proportions is due to sampling error or to the fact that

the population proportions differ.

Solution

ˆ p T 572 237

14830.5455

ˆ p D 572 224

14830.5367

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The samples are dependent and were obtained randomly. The total number of individuals who agree with one scenario, but disagree with the other is 237+224=461, which is greater than 10. We can proceed with McNemar’s Test.

Step 1: The hypotheses are as follows

H0: the proportions between the two populations are equal (pT = pD)

H1: the proportions between the two populations differ (pT ≠ pD)

Solution

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Step 3: The test statistic is:

Solution

z0 224 237 1

237 2240.56

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Step 4: The critical value with an =0.05 level of significance is z0.025=1.96.

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Step 5: Since the test statistic, z0=0.56 is less than the critical value z.025=1.96, we fail to reject the null hypothesis.

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Step 4: The P-value is two times the area under the normal curve to the right of the test statistic z0=0.56. That is, P-value = 2*P(Z > 0.56) ≈ 0.5754.

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

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Step 6: There is insufficient evidence at the =0.05 level to conclude that there is a difference in the proportion of adult Americans who believe it is okay to phone tap versus detaining people for as long as they want without putting them on trial in the event that the government believed a terrorist plot was about to happen.

Solution

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Objective 4

• Determine the Sample Size Necessary for Estimating the Difference between Two Population Proportions within a Specified Margin of Error

11-143

Sample Size for Estimating p1-p2

The sample size required to obtain a (1-)100% confidence interval with a margin of error, E, is givenby

rounded up to the next integer, if prior estimates of p1

and p2, , are available. If prior estimates ofp1 and p2 are unavailable, the sample size is

rounded up to the next integer.

n n1 n2 ˆ p 1 1 ˆ p 1 ˆ p 2 1 ˆ p 2 z 2

ˆ p 1 and ˆ p 2

n n1 n2 0.5z 2

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A doctor wants to estimate the difference in the proportion of 15-19 year old mothers that received prenatal care and the proportion of 30-34 year old mothers that received prenatal care. What sample size should be obtained if she wished the estimate to be within 2 percentage points with 95% confidence assuming:

a) she uses the results of the National Vital Statistics Report results in which 98% of the 15-19 year old mothers received prenatal care and 99.2% of 30-34 year old mothers received prenatal care.

b) she does not use any prior estimates.

Parallel Example 5: Determining Sample Size

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We have E=0.02 and z/2=z0.025=1.96.

a) Letting ,

The doctor must sample 265 randomly selected 15-19 year old mothers and 265 randomly selected 30-34 year old mothers.

Solution

ˆ p 1 0.98 and ˆ p 2 0.992

n1 n2 0.98(1 0.98) 0.992(1 0.992) 1.96

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b) Without prior estimates of p1 and p2, the sample size is

The doctor must sample 4802 randomly selected 15-19 year old mothers and 4802 randomly selected 30-34 year old mothers. Note that having prior estimates of p1 and p2 reduces the number of mothers that need to be surveyed.

Solution

n1 n2 0.51.96

Section

Inference about Two Population Standard Deviations

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Objectives

1. Find critical values of the F-distribution

2. Test hypotheses regarding two population standard deviations

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Objective 1

• Find Critical Values of the F-distribution

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Requirements for Testing Claims Regarding Two Population Standard

Deviations

1. The samples are independent simple random samples.

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Requirements for Testing Claims Regarding Two Population Standard

Deviations

1. The samples are independent simple random samples.2. The populations from which the samples are drawn are normally distributed.

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CAUTION!

If the populations from which the samples are drawnare not normal, do not use the inferential procedures

discussed in this section.

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Notation Used When Comparing Two Population Standard Deviations

: Variance for population 1

: Variance for population 2

: Sample variance for population 1

: Sample variance for population 2

n1 : Sample size for population 1

n2 : Sample size for population 2

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Fisher's F-distribution

If and and are sample variances fromindependent simple random samples of size n1 andn2, respectively, drawn from normal populations, then

follows the F-distribution with n1-1 degrees offreedom in the numerator and n2-1 degrees of freedom in the denominator.

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Characteristics of the F-distribution

1. It is not symmetric. The F-distribution is skewed right.

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1. It is not symmetric. The F-distribution is skewed right.2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the 2 distribution and Student’s t-distribution, whose shapes depend upon their degrees of freedom.

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1. It is not symmetric. The F-distribution is skewed right.2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.3. The total area under the curve is 1.

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1. It is not symmetric. The F-distribution is skewed right.2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.3. The total area under the curve is 1.4. The values of F are always greater than or equal to zero.

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is the critical F with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator and an area of to the right of the critical F.

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To find the critical F with an area of α to the left, use the following:

F1 ,n1 1,n2 1 1

F,n2 1,n1 1

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Find the critical F-value:

a) for a right-tailed test with =0.1, degrees of freedom in the numerator = 8 and degrees of freedom in the denominator = 4.

b) for a two-tailed test with =0.05, degrees of freedom in the numerator = 20 and degrees of freedom in the denominator = 15.

Parallel Example 1: Finding Critical Values for the F-distribution

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a) F0.1,8,4 = 3.95

b) F.025,20,15 = 2.76; = 0.39

Solution

F.975,20,15 1

F.025,15,20

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If the number of degrees of freedom is not found in the table, we follow the practice of choosing the degrees of freedom closest to that desired. If the degrees of freedom is exactly between two values, find the mean of the values.

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Objective 2

• Test Hypotheses Regarding Two Population Standard Deviations

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To test hypotheses regarding two population standard deviations, 1 and 2, we can use the following steps, provided that

1. the samples are obtained using simple random sampling,

2. the sample data are independent, and

3. the populations from which the samples are drawn are normally distributed.

Test Hypotheses Regarding Two Population Standard Deviations

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

Two-Tailed Left-Tailed Right-Tailed

H0: 1 = 2 H0: 1 = 2 H0: 1 = 2

H1: 1 ≠ 2 H1: 1 < 2 H1: 1 > 2

Note: 1 is the population standard deviation for population 1 and 2 is the population standard deviation for population 2.

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which follows Fisher’s F-distribution with n1-1 degrees of freedom in the numerator and n2-1 degrees of freedom in the denominator.

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Step 4: Use Table VIII to determine the criticalvalue(s) using n1-1 degrees of freedom in the numerator and n2-1 degrees of freedom in the denominator.

Classical Approach

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Classical Approach

(critical value)

Two-Tailed

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Classical Approach

(critical value)

Left-Tailed

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Classical Approach

(critical value)

Right-Tailed

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Classical Approach

Two-Tailed Left-Tailed Right-Tailed

If or , reject the null hypothesis.

If , reject the null hypothesis.

F0 F1 2,n1 1,n2 1

F0 F 2,n1 1,n2 1

F0 F1 ,n1 1,n2 1

F0 F,n1 1,n2 1

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Step 4: Use technology to determine the P-value.

P-Value Approach

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P-Value Approach

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CAUTION!

The procedures just presented are not robust,minor departures from normality will adversely affect the results of the test. Therefore, the test should be used only when the requirement of normality has been verified.

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A researcher wanted to know whether “state” quarters had a standard deviation weight that is less than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the data on the next slide. A normal probability plot indicates that the sample data could come from a population that is normal. Test the researcher’s claim at the = 0.05 level of significance.

Parallel Example 2: Testing Hypotheses Regarding Two Population Standard Deviations

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Step 1: The researcher wants to know if “state” quarters have a standard deviation weight that is less than “traditional” quarters. Thus

H0: 1= 2 versus H1: 1< 2

This is a left-tailed test.

Solution

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Step 3: The standard deviation of “state” quarters was found to be 0.0497 and the standard deviation of “traditional” quarters was found to be 0.0689. The test statistic is then

Solution

F0 0.04972

0.06892 0.52

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Step 4: Since this is a left-tailed test, we determine the critical value at the 1- = 1-0.05 = 0.95 level of significance with n1-1=18-1=17 degrees of freedom in the numerator and n2-1=16-1=15 degrees of freedom in the denominator. Thus,

Note: we used the table value F0.05,15,15 for the above calculation since this is the closest to the required degrees of freedom available from Table VIII.

F.95,17,15 1

F.05,15,17

2.400.42.

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Step 5: Since the test statistic F0= 0.52 is greater than the critical value F0.95,17,15=0.42, we fail to reject the null hypothesis.

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Step 4: Using technology, we find that the P-value is 0.097. If the statement in the null hypothesis were true, we would expect to get the results obtained about 10 out of 100 times. This is not very unusual.

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

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Step 6: There is not enough evidence to conclude that the standard deviation of weight is less for “state” quarters than it is for “traditional” quarters at the = 0.05 level of significance.

Solution

Section

Putting It Together: Which Method Do I Use?

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Objectives

1. Determine the appropriate hypothesis test to perform

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Objective 1

• Determine the Appropriate Hypothesis Test to Perform

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What parameter is addressed in the hypothesis?

• Proportion, p• or 2

• Mean,

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Proportion, p

Is the sampling Dependent or Independent?

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Proportion, p

Dependent samples:Provided the samples are obtained randomly and the total number of observations where the outcomes differ is at least 10, use the normal distribution with

z0 f12 f21 1

f12 f21

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Proportion, p

Independent samples:Provided for each sample and the sample size is no more than 5% of the populationsize, use the normal distribution with

nˆ p 1 ˆ p 10

z0 ˆ p 1 ˆ p 2

ˆ p 1 ˆ p 1n1

ˆ p x1 x2

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Provided the data are normally distributed, use the F-distribution with

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Is the sampling Dependent or Independent?

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Dependent samples:Provided each sample size is greater than 30 or the differences come from a population that is normally distributed, use Student’s t-distributionwith n-1 degrees of freedom with

t0 d d

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Independent samples:Provided each sample size is greater than 30 or eachpopulation is normally distributed, use Student’st-distribution

t0 x 1 x 2 1 2

© 2010 Pearson Prentice Hall. All rights reserved Chapter Inferences on Two Samples 11

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