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

© 2010 Pearson Prentice Hall. All rights reserved 11-104 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

© 2010 Pearson Prentice Hall. All rights reserved 11-106 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-107 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

© 2010 Pearson Prentice Hall. All rights reserved 11-108 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

© 2010 Pearson Prentice Hall. All rights reserved 11-110 “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.

© 2010 Pearson Prentice Hall. All rights reserved 11-111 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

© 2010 Pearson Prentice Hall. All rights reserved 11-112 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-114 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-126 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-127 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

© 2010 Pearson Prentice Hall. All rights reserved 11-128 CityHampton InnLa Quinta Dallas129105 Tampa Bay14996 St. Louis14949 Seattle189149 San Diego109119 Chicago16089 New Orleans14972 Phoenix12959 Atlanta12990 Orlando11969

© 2010 Pearson Prentice Hall. All rights reserved 11-129 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

© 2010 Pearson Prentice Hall. All rights reserved 11-132 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

© 2010 Pearson Prentice Hall. All rights reserved 11-133 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 -t 0.025 =-2.262 and t 0.025 =2.262. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-134 Step 5: Since the test statistic, t 0 =5.27 is greater than the critical value t.025 =2.262, we reject the null hypothesis. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-135 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

© 2010 Pearson Prentice Hall. All rights reserved 11-137 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

© 2010 Pearson Prentice Hall. All rights reserved 11-139 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

© 2010 Pearson Prentice Hall. All rights reserved 11-140 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

© 2010 Pearson Prentice Hall. All rights reserved 11-141 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

© 2010 Pearson Prentice Hall. All rights reserved 11-142 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

© 2010 Pearson Prentice Hall. All rights reserved 11-143 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

© 2010 Pearson Prentice Hall. All rights reserved 11-145 Objectives 1.Test hypotheses regarding the difference of two independent means 2.Construct and interpret confidence intervals regarding the difference of two independent means

© 2010 Pearson Prentice Hall. All rights reserved 11-146 Suppose that a simple random sample of size n 1 is taken from a population with unknown mean  1 and unknown standard deviation  1. In addition, a simple random sample of size n 2 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 (n 1 ≥ 30, n 2 ≥ 30), then approximately follows Student’s t-distribution with the smaller of n 1 -1 or n 2 -1 degrees of freedom where is the sample mean and s i 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)

© 2010 Pearson Prentice Hall. All rights reserved 11-148 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 (n 1 ≥ 30, n 2 ≥ 30). Testing Hypotheses Regarding the Difference of Two Means

© 2010 Pearson Prentice Hall. All rights reserved 11-163 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-164 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

© 2010 Pearson Prentice Hall. All rights reserved 11-166 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

© 2010 Pearson Prentice Hall. All rights reserved 11-168 Step 1: We want to determine whether state quarters weigh more than traditional quarters: H 0 :  1 =  2 versus H 1 :  1 >  2 Step 2: The level of significance is  =0.05. Step 3: The test statistic is Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-169 Solution: Classical Approach Step 4: This is a right-tailed test with  =0.05. Since n 1 - 1=17 and n 2 -1=15, we will use 15 degrees of freedom. The corresponding critical value is t 0.05 =1.753.

© 2010 Pearson Prentice Hall. All rights reserved 11-170 Step 5: Since the test statistic, t 0 =2.53 is greater than the critical value t.05 =1.753, we reject the null hypothesis. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-171 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 t 0 =2.53. That is, P-value = P(t > 2.53) ≈ 0.01. Solution: P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-173 Step 6: There is sufficient evidence at the  =0.05 level to conclude that the state quarters weigh more than the traditional quarters. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-174 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:

© 2010 Pearson Prentice Hall. All rights reserved 11-176 A simple random sample of size n 1 is taken from a population with unknown mean  1 and unknown standard deviation  1. Also, a simple random sample of size n 2 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 (n 1 ≥30 and n 2 ≥30), a (1-  )  100% confidence interval about  1 -  2 is given by Lower bound: Upper bound: Constructing a (1-  )  100% Confidence Interval for the Difference of Two Means

© 2010 Pearson Prentice Hall. All rights reserved 11-177 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

© 2010 Pearson Prentice Hall. All rights reserved 11-178 We have already verified that the populations are approximately normal and that there are no outliers. Recall = 5.702, s 1 = 0.0497, =5.6494 and s 2 = 0.0689. From Table VI with  = 0.05 and 15 degrees of freedom, we find t  /2 = 2.131. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-180 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

© 2010 Pearson Prentice Hall. All rights reserved 11-181 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 n 1 +n 2 -2 degrees of freedom. The disadvantage to this test statistic is that it requires that the population variances be equal.

© 2010 Pearson Prentice Hall. All rights reserved 11-183 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-184 Suppose that a simple random sample of size n 1 is taken from a population where x 1 of the individuals have a specified characteristic, and a simple random sample of size n 2 is independently taken from a different population where x 2 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

© 2010 Pearson Prentice Hall. All rights reserved 11-185 The standardized version of is then written as which has an approximate standard normal distribution. Sampling Distribution of the Difference between Two Proportions

© 2010 Pearson Prentice Hall. All rights reserved 11-188 To test hypotheses regarding two population proportions, p 1 and p 2, we can use the steps that follow, provided that: 1.the samples are independently obtained using simple random sampling, 2. and and 3.n 1 ≤ 0.05N 1 and n 2 ≤ 0.05N 2 (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 Difference between Two Population Proportions

© 2010 Pearson Prentice Hall. All rights reserved 11-203 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

© 2010 Pearson Prentice Hall. All rights reserved 11-204 We must first verify that the requirements are satisfied: 1.The samples are simple random samples that were obtained independently. 2. x 1 =338, n 1 =800, x 2 =292 and n 2 =750, so 3.The sample sizes are less than 5% of the population size. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-205 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, H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 or, equivalently, H 0 : p 1 - p 2 =0 versus H 1 : p 1 - p 2 > 0 Step 2: The level of significance is  = 0.05. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-208 Step 5: Since the test statistic, z 0 =1.33 is less than the critical value z.05 =1.645, we fail to reject the null hypothesis. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-209 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 z 0 =1.33. That is, P-value = P(Z > 1.33) ≈ 0.09. Solution: P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-211 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

© 2010 Pearson Prentice Hall. All rights reserved 11-213 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.n 1 ≤ 0.05N 1 and n 2 ≤ 0.05N 2 (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

© 2010 Pearson Prentice Hall. All rights reserved 11-214 Provided that these requirements are met, a (1-  )  100% confidence interval for p 1 -p 2 is given by Lower bound: Upper bound: Constructing a (1-  )  100% Confidence Interval for the Difference between Two Population Proportions

© 2010 Pearson Prentice Hall. All rights reserved 11-215 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

© 2010 Pearson Prentice Hall. All rights reserved 11-216 We have already verified the requirements for constructing a confidence interval for the difference between two population proportions in the previous example. Recall Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-218 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

© 2010 Pearson Prentice Hall. All rights reserved 11-221 Testing a Hypothesis Regarding the Difference of Two Population Proportions: Dependent Samples To test hypotheses regarding two population proportions p 1 and p 2, where the samples are dependent, arrange the data in a contingency table as follows: Treatment A Treatment B SuccessFailure Successf 11 f 12 Failuref 21 f 22

© 2010 Pearson Prentice Hall. All rights reserved 11-222 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 and 2.the total number of observations where the outcomes differ must be greater than or equal to 10. That is, f 12 + f 21 ≥ 10.

© 2010 Pearson Prentice Hall. All rights reserved 11-223 Step 1: Determine the null and alternative hypotheses. H 0 : the proportions between the two populations are equal (p 1 = p 2 ) H 1 : the proportions between the two populations differ (p 1 ≠ p 2 )

© 2010 Pearson Prentice Hall. All rights reserved 11-226 Step 4: Use Table V to determine the critical value. This is a two tailed test. However, z 0 is always positive, so we only need to find the right critical value z  /2. Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-229 Step 4: Use Table V to determine the P-value.. Because z 0 is always positive, we find the area to the right of z 0 and then double this area (since this is a two-tailed test). P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-233 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

© 2010 Pearson Prentice Hall. All rights reserved 11-234 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 Tap Phone AgreeDisagree Agree572237 Disagree224450

© 2010 Pearson Prentice Hall. All rights reserved 11-235 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

© 2010 Pearson Prentice Hall. All rights reserved 11-236 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 H 0 : the proportions between the two populations are equal (p T = p D ) H 1 : the proportions between the two populations differ (p T ≠ p D ) Step 2: The level of significance is  = 0.05. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-239 Step 5: Since the test statistic, z 0 =0.56 is less than the critical value z.025 =1.96, we fail to reject the null hypothesis. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-240 Step 4: The P-value is two times the area under the normal curve to the right of the test statistic z 0 =0.56. That is, P-value = 2*P(Z > 0.56) ≈ 0.5754. Solution: P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-242 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

© 2010 Pearson Prentice Hall. All rights reserved 11-243 Objective 4 Determine the Sample Size Necessary for Estimating the Difference between Two Population Proportions within a Specified Margin of Error

© 2010 Pearson Prentice Hall. All rights reserved 11-244 Sample Size for Estimating p 1 -p 2 The sample size required to obtain a (1-  )  100% confidence interval with a margin of error, E, is given by rounded up to the next integer, if prior estimates of p 1 and p 2,, are available. If prior estimates of p 1 and p 2 are unavailable, the sample size is rounded up to the next integer.

© 2010 Pearson Prentice Hall. All rights reserved 11-245 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

© 2010 Pearson Prentice Hall. All rights reserved 11-246 We have E=0.02 and z  /2 =z 0.025 =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

© 2010 Pearson Prentice Hall. All rights reserved 11-247 b)Without prior estimates of p 1 and p 2, 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 p 1 and p 2 reduces the number of mothers that need to be surveyed. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-252 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-254 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 n 1 : Sample size for population 1 n 2 : Sample size for population 2

© 2010 Pearson Prentice Hall. All rights reserved 11-255 Fisher's F-distribution If and and are sample variances from independent simple random samples of size n 1 and n 2, respectively, drawn from normal populations, then follows the F-distribution with n 1 -1 degrees of freedom in the numerator and n 2 -1 degrees of freedom in the denominator.

© 2010 Pearson Prentice Hall. All rights reserved 11-257 Characteristics of the F-distribution 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-258 Characteristics of the F-distribution 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-259 Characteristics of the F-distribution 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-261 is the critical F with n 1 – 1 degrees of freedom in the numerator and n 2 – 1 degrees of freedom in the denominator and an area of  to the right of the critical F.

© 2010 Pearson Prentice Hall. All rights reserved 11-263 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

© 2010 Pearson Prentice Hall. All rights reserved 11-265 NOTE: 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-267 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

© 2010 Pearson Prentice Hall. All rights reserved 11-268 Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways: Two-TailedLeft-TailedRight-Tailed H 0 :  1 =  2 H 1 :  1 ≠  2 H 1 :  1 <  2 H 1 :  1 >  2 Note:  1 is the population standard deviation for population 1 and  2 is the population standard deviation for population 2.

© 2010 Pearson Prentice Hall. All rights reserved 11-270 Step 3: Compute the test statistic which follows Fisher’s F-distribution with n 1 -1 degrees of freedom in the numerator and n 2 -1 degrees of freedom in the denominator.

© 2010 Pearson Prentice Hall. All rights reserved 11-271 Step 4: Use Table VIII to determine the critical value(s) using n 1 -1 degrees of freedom in the numerator and n 2 -1 degrees of freedom in the denominator. Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-275 Step 5: Compare the critical value with the test statistic: Classical Approach Two-TailedLeft-TailedRight-Tailed If or, reject the null hypothesis. If, reject the null hypothesis. If, reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 11-279 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.

© 2010 Pearson Prentice Hall. All rights reserved 11-280 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

© 2010 Pearson Prentice Hall. All rights reserved 11-282 Step 1: The researcher wants to know if “state” quarters have a standard deviation weight that is less than “traditional” quarters. Thus H 0 :  1 =  2 versus H 1 :  1 <  2 This is a left-tailed test. Step 2: The level of significance is  = 0.05. Solution

© 2010 Pearson Prentice Hall. All rights reserved 11-283 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

© 2010 Pearson Prentice Hall. All rights reserved 11-284 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 n 1 -1=18-1=17 degrees of freedom in the numerator and n 2 -1=16-1=15 degrees of freedom in the denominator. Thus, Note: we used the table value F 0.05,15,15 for the above calculation since this is the closest to the required degrees of freedom available from Table VIII. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-285 Step 5: Since the test statistic F 0 = 0.52 is greater than the critical value F 0.95,17,15 =0.42, we fail to reject the null hypothesis. Solution: Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-286 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. Solution: P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-287 Step 5: Since the P-value is greater than the level of significance,  = 0.05, we fail to reject the null hypothesis. Solution: P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 11-288 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

© 2010 Pearson Prentice Hall. All rights reserved 11-294 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

© 2010 Pearson Prentice Hall. All rights reserved 11-295 Proportion, p Independent samples: Provided for each sample and the sample size is no more than 5% of the population size, use the normal distribution with where

© 2010 Pearson Prentice Hall. All rights reserved 11-298 Mean,  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-distribution with n-1 degrees of freedom with

© 2010 Pearson Prentice Hall. All rights reserved 11-299 Mean,  Independent samples: Provided each sample size is greater than 30 or each population is normally distributed, use Student’s t-distribution

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

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