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Chapter 11 Comparing Two Populations or Treatments

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Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.5 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed. Suppose we take a random sample of 30 men and a random sample of 25 women from their respective populations and calculate the difference in their heights (man’s height – woman’s height). If we did this many times, what would the distribution of differences be like? On the next slide we will investigate this distribution.

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6 71 Male Heights 65 Female Heights 65 71 M = 2.5 F = 2.3 Suppose we took repeated samples of size n = 25 from the population of female heights and calculated the sample means. We would have the sampling distribution of x F xMxM - x F Randomly take one of the sample means for the males and one of the sample means for the females and find the difference in mean heights. Doing this repeatedly, we will create the sampling distribution of (x M – x F ) Suppose we took repeated samples of size n = 30 from the population of male heights and calculated the sample means. We would have the sampling distribution of x M.

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Heights Continued... Describe the sampling distribution of the difference in mean heights between men and women. What is the probability that the difference in mean heights of a random sample of 30 men and a random sample of 25 women is less than 5 inches? The sampling distribution is normally distributed with 6

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Properties of the Sampling Distribution of x 1 – x 2 1. 2. and If the random samples on which x 1 and x 2 are based are selected independently of one another, then Mean value of x 1 – x 2 The sampling distribution of x 1 – x 2 is always centered at the value of 1 – 2, so x 1 – x 2 is an unbiased statistic for estimating 1 – 2. 3. In n 1 and n 2 are both large or the population distributions are (at least approximately) normal, x 1 and x 2 each have (at least approximately) normal distributions. This implies that the sampling distribution of x 1 – x 2 is also (approximately) normal. The variance of the differences is the sum of the variances.

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The properties for the sampling distribution of x 1 – x 2 implies that x 1 – x 2 can be standardized to obtain a variable with a sampling distribution that is approximately the standard normal (z) distribution. When two random samples are independently selected and n 1 and n 2 are both large or the population distributions are (at least approximately) normal, the distribution of is described (at least approximately) by the standard normal (z) distribution. We must know and 2 in order to use this procedure. If 1 and is unknown we must use t distributions.

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Two-Sample t Test for Comparing Two Populations Null Hypothesis: H 0 : 1 – 2 = hypothesized value Test Statistic: The appropriate df for the two-sample t test is The computed number of df should be truncated to an integer. whereand The hypothesized value is often 0, but there are times when we are interested in testing for a difference that is not 0. A conservative estimate of the P- value can be found by using the t- curve with the number of degrees of freedom equal to the smaller of (n 1 – 1) or (n 2 – 1).

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Two-Sample t Test for Comparing Two Populations Continued... Null Hypothesis: H 0 : 1 – 2 = hypothesized value Alternative Hypothesis:P-value: H a : 1 – 2 > hypothesized value Area under the appropriate t curve to the right of the computed t H a : 1 – 2 < hypothesized value Area under the appropriate t curve to the left of the computed t H a : 1 – 2 ≠ hypothesized value 2(area to right of computed t) if +t or 2(area to left of computed t) if -t

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Another Way to Write Hypothesis Statements: H 0 : 1 - 2 = 0 H a : 1 - 2 < 0 H a : 1 - 2 > 0 H a : 1 - 2 ≠ 0 H 0 : 1 = 2 H a : 1 < 2 H a : 1 > 2 H a : 1 ≠ 2 Be sure to define BOTH 1 and 2 ! When the hypothesized value is 0, we can rewrite these hypothesis statements:

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Two-Sample t Test for Comparing Two Populations Continued... Assumptions: 1)The two samples are independently selected random samples from the populations of interest 2)The sample sizes are large (generally 30 or larger) or the population distributions are (at least approximately) normal. When comparing two treatment groups, use the following assumptions: 1)Individuals or objects are randomly assigned to treatments (or vice versa) 2)The sample sizes are large (generally 30 or larger) or the treatment response distributions are approximately normal.

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Are women still paid less than men for comparable work? A study was carried out in which salary data was collected from a random sample of men and from a random sample of women who worked as purchasing managers and who were subscribers to Purchasing magazine. Annual salaries (in thousands of dollars) appear below (the actual sample sizes were much larger). Use =.05 to determine if there is convincing evidence that the mean annual salary for male purchasing managers is greater than the mean annual salary for female purchasing managers. H 0 : 1 – 2 = 0 H a : 1 – 2 > 0 Men81698176 7469767965 Women78606761627371586848 Where 1 = mean annual salary for male purchasing managers and 2 = mean annual salary for female purchasing managers State the hypotheses: If we had defined 1 as the mean salary for female purchasing managers and 2 as the mean salary for male purchasing managers, then the correct alternative hypothesis would be the difference in the means is less than 0.

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Men Women 60 80 Salary War Continued... H 0 : 1 – 2 = 0 H a : 1 – 2 > 0 Assumptions: 1)Given two independently selected random samples of male and female purchasing managers. Men81698176 7469767965 Women78606761627371586848 Where 1 = mean annual salary for male purchasing managers and 2 = mean annual salary for female purchasing managers 2) Since the sample sizes are small, we must determine if it is plausible that the sampling distributions for each of the two populations are approximately normal. Since the boxplots are reasonably symmetrical with no outliers, it is plausible that the sampling distributions are approximately normal. Verify the assumptions Even though these are samples from subscribers of Purchasing magazine, the authors of the study believed it was reasonable to view the samples as representative of the populations of interest.

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Salary War Continued... H 0 : 1 – 2 = 0 H a : 1 – 2 > 0 Test Statistic: P-value =.004 =.05 Since the P-value < , we reject H 0. There is convincing evidence that the mean salary for male purchasing managers is higher than the mean salary for female purchasing managers. Men81698176 7469767965 Women78606761627371586848 Where 1 = mean annual salary for male purchasing managers and 2 = mean annual salary for female purchasing managers Compute the test statistic and P-value To find the P-value, first find the appropriate df. Now find the area to the right of t = 3.11 in the t- curve with df = 15. Truncate (round down) this value. What potential type error could we have made with this conclusion? Type I

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The Two-Sample t Confidence Interval for the Difference Between Two Population or Treatment Means The general formula for a confidence interval for 1 – 2 when 1)The two samples are independently selected random samples from the populations of interest 2)The sample sizes are large (generally 30 or larger) or the population distributions are (at least approximately) normal. is The t critical value is based on df should be truncated to an integer. whereand For a comparison of two treatments, use the following assumptions: 1) Individuals or objects are randomly assigned to treatments (or vice versa) 2) The sample sizes are large (generally 30 or larger) or the treatment response distributions are approximately normal.

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In a study on food intake after sleep deprivation, men were randomly assigned to one of two treatment groups. The experimental group were required to sleep only 4 hours on each of two nights, while the control group were required to sleep 8 hours on each of two nights. The amount of food intake (Kcal) on the day following the two nights of sleep was measured. Compute a 95% confidence interval for the true difference in the mean food intake for the two sleeping conditions. x 4 = 3924 s 4 = 829.67 x 8 = 4069.27 s 8 = 952.90 4-hour sleep 35854470306853382221479144353099 3187390138683869487836324518 8-hour sleep 49653918198749935220365335103338 4100579245473319333643044057 Find the mean and standard deviation for each treatment.

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Assumptions: 1)Men were randomly assigned to two treatment groups Food Intake Study Continued... 4-hour sleep 35854470306853382221479144353099 3187390138683869487836324518 8-hour sleep 49653918198749935220365335103338 4100579245473319333643044057 x 4 = 3924 s 4 = 829.67 x 8 = 4069.27 s 8 = 952.90 Verify the assumptions. 2) The assumption of normal response distributions is plausible because both boxplots are approximately symmetrical with no outliers. 4000 4-hour 8-hour

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We are 95% confident that the true difference in the mean food intake for the two sleeping conditions is between -814.1 Kcal and 523.6 Kcal. Food Intake Study Continued... 4-hour sleep 35854470306853382221479144353099 3187390138683869487836324518 8-hour sleep 49653918198749935220365335103338 4100579245473319333643044057 x 4 = 3924 s 4 = 829.67 x 8 = 4069.27 s 8 = 952.90 Calculate the interval. Interpret the interval in context. Based upon this interval, is there a significant difference in the mean food intake for the two sleeping conditions? No, since 0 is in the confidence interval, there is not convincing evidence that the mean food intake for the two sleep conditions are different.

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Pooled t Test Used when the variances of the two populations are equal ( 1 = 2 ) Combines information from both samples to create a “pooled” estimate of the common variance which is used in place of the two sample standard deviations Is not widely used due to its sensitivity to any departure from the equal variance assumption When the population variances are equal, the pooled t procedure is better at detecting deviations from H 0 than the two-sample t test. P-values computed using the pooled t procedure can be far from the actual P-value if the population variances are not equal.

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Suppose that an investigator wants to determine if regular aerobic exercise improves blood pressure. A random sample of people who jog regularly and a second random sample of people who do not exercise regularly are selected independently of one another. Can we conclude that the difference in mean blood pressure is attributed to jogging? What about other factors like weight? One way to avoid these difficulties would be to pair subjects by weight then assign one of the pair to jogging and the other to no exercise.

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Summary of the Paired t test for Comparing Two Population or Treatment Means Null Hypothesis: H 0 : d = hypothesized value Test Statistic: Where n is the number of sample differences and x d and s d are the mean and standard deviation of the sample differences. This test is based on df = n – 1. Alternative Hypothesis:P-value: H a : d > hypothesized value Area to the right of calculated t H a : d < hypothesized value Area to the left of calculated t H a : d ≠ hypothesized value2(a rea to the right of t) if +t or 2(a rea to the left of t) if -t The hypothesized value is usually 0 – meaning that there is no difference. Where d is the mean of the differences in the paired observations

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Summary of the Paired t test for Comparing Two Population or Treatment Means Continued... Assumptions: 1.The samples are paired. 2.The n sample differences can be viewed as a random sample from a population of differences. 3.The number of sample differences is large (generally at least 30) or the population distribution of differences is (at least approximately) normal.

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Is this an example of paired samples? An engineering association wants to see if there is a difference in the mean annual salary for electrical engineers and chemical engineers. A random sample of electrical engineers is surveyed about their annual income. Another random sample of chemical engineers is surveyed about their annual income. No, there is no pairing of individuals, you have two independent samples

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Is this an example of paired samples? A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to volunteers, company researchers weigh each person. After a month of using the drug, each person’s weight is measured again. Yes, you have two observations on each individual, resulting in paired data.

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Can playing chess improve your memory? In a study, students who had not previously played chess participated in a program in which they took chess lessons and played chess daily for 9 months. Each student took a memory test before starting the chess program and again at the end of the 9-month period. Student123456789101112 Pre-test510610640675600550610625450720575675 Post-test850790850775700775700850690775540680 Difference -340-180-210-100 -225-90-225-240-5535-5 H 0 : d = 0 H a : d < 0 Where d is the mean memory score difference between students with no chess training and students who have completed chess training First, find the differences pre-test minus post-test. State the hypotheses. If we had subtracted Post-test minus Pre-test, then the alternative hypothesis would be the mean difference is greater than 0.

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Playing Chess Continued... Student123456789101112 Pre-test510610640675600550610625450720575675 Post-test850790850775700775700850690775540680 Difference -340-180-210-100 -225-90-225-240-5535-5 H 0 : d = 0 H a : d < 0 Assumptions: 1) Although the sample of students is not a random sample, the Verify assumptions Where d is the mean memory score difference between students with no chess training and students who have completed chess training investigator believed that it was reasonable to view the 12 sample differences as representative of all such differences. 2) A boxplot of the differences is approximately symmetrical with no outliers so the assumption of normality is plausible.

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Playing Chess Continued... Student123456789101112 Pre-test510610640675600550610625450720575675 Post-test850790850775700775700850690775540680 Difference -340-180-210-100 -225-90-225-240-5535-5 H 0 : d = 0 H a : d < 0 Test Statistic: Where d is the mean memory score difference between students with no chess training and students who have completed chess training P-value ≈ 0df = 11 =.05 Since the P-value < , we reject H 0. There is convincing evidence to suggest that the mean memory score after chess training is higher than the mean memory score before training. Compute the test statistic and P-value. State the conclusion in context.

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Paired t Confidence Interval for d When 1.The samples are paired. 2.The n sample differences can be viewed as a random sample from a population of differences. 3.The number of sample differences is large (generally at least 30) or the population distribution of differences is (at least approximately) normal. the paired t interval for d is Where df = n - 1

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Playing Chess Revisited... Student123456789101112 Pre-test510610640675600550610625450720575675 Post-test850790850775700775700850690775540680 Difference -340-180-210-100 -225-90-225-240-5535-5 We are 90% confident that the true mean difference in memory scores before chess training and the memory scores after chess training is between -201.5 and -87.69. Compute a 90% confidence interval for the mean difference in memory scores before chess training and the memory scores after chess training.

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Large-Sample Inferences Concerning the Difference Between Two Population or Treatment Proportions

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Investigators at Madigan Army Medical Center tested using duct tape to remove warts versus the more traditional freezing treatment. Suppose that the duct tape treatment will successfully remove 50% of warts and that the traditional freezing treatment will successfully remove 60% of warts. Some people seem to think that duct tape can fix anything... even remove warts! Let’s investigate the sampling distribution of p freeze - p tape

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.6.1.5 p freeze = the true proportion of warts that are successfully removed by freezing p freeze =.6 p tape = the true proportion of warts that are successfully removed by using duct tape p tape =.5 Suppose we repeatedly treated 100 warts using the traditional freezing treatment and calculated the proportion of warts that are successfully removed. We would have the sampling distribution of p freeze Suppose we repeatedly treated 100 warts using the duct tape method and calculated the proportion of warts that are successfully removed. We would have the sampling distribution of p tape. p freeze - p tape Doing this repeatedly, we will create the sampling distribution of (p freeze – p tape ) Randomly take one of the sample proportions for the freezing treatment and one of the sample proportions for the duct tape treatment and find the difference.

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Properties of the Sampling Distribution of p 1 – p 2 If two random samples are selected independently of one another, the following properties hold: 1. This says that the sampling distribution of p 1 – p 2 is centered at p 1 – p 2 so p 1 – p 2 is an unbiased statistic for estimating p 1 – p 2. 2. 3. If both n 1 and n 2 are large (that is, if n 1 p 1 > 10, n 1 (1 – p 1 ) > 10, n 2 p 2 > 10, and n 2 (1 – p 2 ) > 10), then p 1 and p 2 each have a sampling distribution that is approximately normal, and their difference p 1 – p 2 also has a sampling distribution that is approximately normal. When performing a hypothesis test, we will use the null hypothesis that p 1 and p 2 are equal. We will not know the common value for p 1 and p 2. Since the value for p 1 and p 2 are unknown, we will combine p 1 and p 2 to estimate the common value of p 1 and p 2 Use:

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Summary of Large-Sample z Test for p 1 – p 2 = 0 Null Hypothesis: H 0 : p 1 – p 2 = 0 Test Statistic: Alternative Hypothesis: P-value: H a : p 1 – p 2 > 0 area to the right of calculated z H a : p 1 – p 2 < 0 area to the left of calculated z H a : p 1 – p 2 ≠ 0 2( area to the right of z) if +z or 2( area to the left of z) if -z Use:

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Another Way to Write Hypothesis statements: H 0 : p 1 - p 2 = 0 H a : p 1 - p 2 > 0 H a : p 1 - p 2 < 0 H a : p 1 - p 2 ≠ 0 Be sure to define both p 1 & p 2 ! H 0 : p 1 = p 2 H a : p 1 > p 2 H a : p 1 < p 2 H a : p 1 ≠ p 2

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Assumption: 1)The samples are independently chosen random samples or treatments were assigned at random to individuals or objects Summary of Large-Sample z Test for p 1 – p 2 = 0 Continued... 2) Both sample sizes are large n 1 p 1 > 10, n 1 (1 – p 1 ) > 10, n 2 p 2 > 10, n 2 (1 – p 2 ) > 10 Since p 1 and p 2 are unknown we must use p 1 and p 2 to verify that the samples are large enough.

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Investigators at Madigan Army Medical Center tested using duct tape to remove warts. Patients with warts were randomly assigned to either the duct tape treatment or to the more traditional freezing treatment. Those in the duct tape group wore duct tape over the wart for 6 days, then removed the tape, soaked the area in water, and used an emery board to scrape the area. This process was repeated for a maximum of 2 months or until the wart was gone. The data follows: Do these data suggest that freezing is less successful than duct tape in removing warts? Treatmentn Number with wart successfully removed Liquid nitrogen freezing10060 Duct tape10488

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Duct Tape Continued... H 0 : p 1 – p 2 = 0 H a : p 1 – p 2 < 0 Assumptions: 1) Subjects were randomly assigned to the two treatments. Treatmentn Number with wart successfully removed Liquid nitrogen freezing10060 Duct tape10488 Where p 1 is the true proportion of warts that would be successfully removed by freezing and p 2 is the true proportion of warts that would be successfully removed by duct tape 2) The sample sizes are large enough because: n 1 p 1 = 100(.6) = 60 > 10 n 1 (1 – p 1 ) = 100(.4) = 40 > 10 n 2 p 2 = 100(.85) = 85 > 10 n 2 (1 – p 2 ) = 100(.15) = 15 > 10

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Duct Tape Continued... H 0 : p 1 – p 2 = 0 H a : p 1 – p 2 < 0 Treatmentn Number with wart successfully removed Liquid nitrogen freezing10060 Duct tape10488 P-value ≈ 0 =.01 Since the P-value < , we reject H 0. There is convincing evidence to suggest the proportion of warts successfully removed is lower for freezing than for the duct tape treatment.

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A Large-Sample Confidence Interval for p 1 – p 2 When 1)The samples are independently chosen random samples or treatments were assigned at random to individuals or objects a large-sample confidence interval for p 1 – p 2 is 2) Both sample sizes are large n 1 p 1 > 10, n 1 (1 – p 1 ) > 10, n 2 p 2 > 10, n 2 (1 – p 2 ) > 10

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The article “Freedom of What?” (Associated Press, February 1, 2005) described a study in which high school students and high school teachers were asked whether they agreed with the following statement: “Students should be allowed to report controversial issues in their student newspapers without the approval of school authorities.” It was reported that 58% of students surveyed and 39% of teachers surveyed agreed with the statement. The two samples – 10,000 high school students and 8000 high school teachers – were selected from schools across the country. Compute a 90% confidence interval for the difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement.

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Newspaper Problem Continued... p 1 =.58 p 2 =.39 1) Assume that it is reasonable to regard these two samples as being independently selected and representative of the populations of interest. 2) Both sample sizes are large enough n 1 p 1 = 10000(.58) > 10, n 1 (1 – p 1 ) = 10000(.42) > 10, n 2 p 2 = 8000(.39) > 10, n 2 (1 – p 2 ) = 8000(.61) > 10 We are 90% confident that the difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement is between.178 and.202. Based on this confidence interval, does there appear to be a significant difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement? Explain.

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