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© 2010 Pearson Prentice Hall. All rights reserved Two Sample Hypothesis Testing for Means from Paired or Dependent Groups

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11-2 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.

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11-3 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.

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11-4 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 Difference of Two Means Using a Matched-Pairs Design

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11-5 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.

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11-6 Step 2: Select a level of significance, , based on the seriousness of making a Type I error.

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11-7 Step 3: Compute the test statistic which approximately follows Student’s t-distribution with n-1 degrees of freedom. The values of and s d are the mean and standard deviation of the differenced data.

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11-8 Step 4: Use Table VI to determine the P-value using n-1 degrees of freedom. P-Value Approach

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11-9 P-Value Approach Two-Tailed

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11-10 P-Value Approach Left-Tailed

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11-11 P-Value Approach Right-Tailed

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10-12 Step 5: If the P-value < , reject the null hypothesis. If the P-value ≥ α, fail to reject the null hypothesis P-Value Approach

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10-13 Step 6: State the conclusion in the context of the problem.

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11-14 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.

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11-15 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

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11-16 CityHampton InnLa Quinta Dallas129105 Tampa Bay14996 St. Louis14949 Seattle189149 San Diego109119 Chicago16089 New Orleans14972 Phoenix12959 Atlanta12990 Orlando11969

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11-17 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 s d = 30.8336. Solution

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11-18 No violation of normality assumption. Solution

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11-19 No outliers. Solution

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11-20 Step 1: We want to determine if the prices differ: H 0 : d = 0 versus H 1 : d 0 Step 2: The level of significance is =0.05. Step 3: The test statistic is Solution

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11-21 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 t 0 =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

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11-22 Step 5: Since the P-value is less than the level of significance =0.05, we reject the null hypothesis. Solution: P-Value Approach

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11-23 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

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11-24 Objective 3 Construct and Interpret Confidence Intervals for the Population Mean Difference of Matched-Pairs Data

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11-25 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

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11-26 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

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11-27 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

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11-28 We have already verified that the differenced data come from a population that is approximately normal with no outliers. Recall = 51.4 and s d = 30.8336. From Table VI with = 0.10 and 9 degrees of freedom, we find t /2 = 1.833. Solution

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11-29 Thus, 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

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