Chapter 11 Inferences on Two Samples 2010

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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 104

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 106

Parallel Example 1: Distinguish between Independent and Dependent Sampling 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. © 2010 Pearson Prentice Hall. All rights reserved 107

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 108

“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 110

Testing Hypotheses Regarding the Difference of Two Means Using a Matched-Pairs Design 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). © 2010 Pearson Prentice Hall. All rights reserved 111

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 112

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. © 2010 Pearson Prentice Hall. All rights reserved 114

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 126

Parallel Example 2: Testing a Claim Regarding Matched-Pairs Data 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. © 2010 Pearson Prentice Hall. All rights reserved 127

City Dallas Tampa Bay St. Louis Seattle San Diego Chicago New Orleans Phoenix Atlanta Orlando Hampton Inn 129 149 189 109 160 149 129 119 La Quinta 105 96 49 119 89 72 59 90 69 © 2010 Pearson Prentice Hall. All rights reserved 128

Solution 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 with 53 100 40 -10 71 77 70 39 50 = 51. 4 and sd = 30. 8336. © 2010 Pearson Prentice Hall. All rights reserved 129

Solution 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 © 2010 Pearson Prentice Hall. All rights reserved 132

Solution: Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 133

Solution: Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 134

Solution: P-Value Approach 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 = 2 P(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. © 2010 Pearson Prentice Hall. All rights reserved 135

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 137

Confidence Interval for 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. © 2010 Pearson Prentice Hall. All rights reserved 139

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. © 2010 Pearson Prentice Hall. All rights reserved 140

Parallel Example 4: Constructing a 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. © 2010 Pearson Prentice Hall. All rights reserved 141

Solution • We have already verified that the differenced data come from a population that is approximately normal with no outliers. • Recall • From Table VI with = 0. 10 and 9 degrees of freedom, we find t /2 = 1. 833. = 51. 4 and sd = 30. 8336. © 2010 Pearson Prentice Hall. All rights reserved 142

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 143

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 145

Sampling Distribution of the Difference of Two Means: Independent Samples with Population Standard Deviations Unknown (Welch’s t) 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 si is the sample standard deviation from population i. © 2010 Pearson Prentice Hall. All rights reserved 146

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

Parallel Example 1: Testing Hypotheses Regarding Two Means 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. © 2010 Pearson Prentice Hall. All rights reserved 164

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 166

Solution 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 © 2010 Pearson Prentice Hall. All rights reserved 168

Solution: Classical Approach Step 4: This is a right-tailed test with =0. 05. Since n 11=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 169

Solution: Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 170

Solution: P-Value Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 171

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 174

Constructing a (1 - ) 100% Confidence Interval for the Difference of Two Means 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: © 2010 Pearson Prentice Hall. All rights reserved 176

Parallel Example 3: Constructing a Confidence Interval for the Difference of Two Means 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. © 2010 Pearson Prentice Hall. All rights reserved 177

Solution • We have already verified that the populations are approximately normal and that there are no outliers. • Recall = 5. 702, s 1 = 0. 0497, 0. 0689. • From Table VI with = 0. 05 and 15 degrees of freedom, we find t /2 = 2. 131. =5. 6494 and s 2 = © 2010 Pearson Prentice Hall. All rights reserved 178

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 180

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 181

Objectives 1. Test hypotheses regarding two population proportions 2. Construct and interpret confidence intervals for the difference between two population proportions 3. Use Mc. Nemar’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 183

Sampling Distribution of the Difference between Two Proportions 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 © 2010 Pearson Prentice Hall. All rights reserved . 184

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

Hypothesis Test Regarding the Difference between Two Population Proportions 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 3. n 1 ≤ 0. 05 N 1 and n 2 ≤ 0. 05 N 2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment. © 2010 Pearson Prentice Hall. All rights reserved 188

Parallel Example 1: Testing Hypotheses Regarding Two Population Proportions 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. © 2010 Pearson Prentice Hall. All rights reserved 203

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 204

Solution 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 versus H 1: p 1 - p 2 > 0 or, equivalently, H 0: p 1 - p 2=0 Step 2: The level of significance is = 0. 05. © 2010 Pearson Prentice Hall. All rights reserved 205

Solution: Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 208

Solution: P-Value Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 209

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 211

Constructing a (1 - ) 100% Confidence Interval for the Difference between Two Population Proportions 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. 05 N 1 and n 2 ≤ 0. 05 N 2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment. © 2010 Pearson Prentice Hall. All rights reserved 213

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

Parallel Example 3: Constructing a Confidence Interval for the Difference between Two Population Proportions 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. © 2010 Pearson Prentice Hall. All rights reserved 215

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

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 218

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 Success Failure Treatment B Success f 11 f 12 Failure f 21 f 22 © 2010 Pearson Prentice Hall. All rights reserved 221

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 3. outcomes differ must be greater than or equal to 10. That is, f 12 + f 21 ≥ 10. © 2010 Pearson Prentice Hall. All rights reserved 222

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 223

Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 226

P-Value Approach 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). © 2010 Pearson Prentice Hall. All rights reserved 229

Parallel Example 4: Analyzing the Difference of Two Proportions from Matched-Pairs Data 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? © 2010 Pearson Prentice Hall. All rights reserved 233

Parallel Example 4: Analyzing the Difference of Two Proportions from Matched-Pairs Data The results of the survey are shown below: Detain Agree Tap Phone Disagree Agree 572 237 Disagree 224 450 Do the proportions who agree with each scenario differ significantly? Use the =0. 05 level of significance. © 2010 Pearson Prentice Hall. All rights reserved 234

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 235

Solution 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 Mc. Nemar’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. © 2010 Pearson Prentice Hall. All rights reserved 236

Solution: Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 239

Solution: P-Value Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 240

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 242

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 243

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 244

Parallel Example 5: Determining Sample Size 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. © 2010 Pearson Prentice Hall. All rights reserved 245

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 246

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 247

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 252

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 254

Fisher's F-distribution If 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 255

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 tdistribution, whose shapes depend upon their degrees of freedom. © 2010 Pearson Prentice Hall. All rights reserved 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 distribution and Student’s tdistribution, whose shape depends upon their degrees of freedom. 3. The total area under the curve is 1. © 2010 Pearson Prentice Hall. All rights reserved 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 tdistribution, 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 259

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 261

Parallel Example 1: Finding Critical Values for the F-distribution 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. © 2010 Pearson Prentice Hall. All rights reserved 263

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 265

Test Hypotheses Regarding Two Population Standard Deviations 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. © 2010 Pearson Prentice Hall. All rights reserved 267

Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways: Two-Tailed Left-Tailed Right-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 268

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 270

Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 271

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

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 279

Parallel Example 2: Testing Hypotheses Regarding Two Population Standard Deviations 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. © 2010 Pearson Prentice Hall. All rights reserved 280

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 282

Solution 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 © 2010 Pearson Prentice Hall. All rights reserved 283

Solution: Classical Approach 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 for the above calculation since this is the closest to the required degrees of freedom available from Table VIII. © 2010 Pearson Prentice Hall. All rights reserved 284

Solution: Classical Approach 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. © 2010 Pearson Prentice Hall. All rights reserved 285

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

Solution 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. © 2010 Pearson Prentice Hall. All rights reserved 288

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 294

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 295

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 298