155S9.4_3 Inferences from Dependent Samples. April 11, Key Concept. Chapter 9 Inferences from Two Samples. Key Concept

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1 MAT 155 Statistical Analysis Dr. Claude Moore Cape Fear Community College Chapter 9 Inferences from Two Samples 9 1 Review and Preview 9 2 Inferences About Two Proportions 9 3 Inferences About Two Means: Independent Samples 9 4 Inferences from Dependent Samples 9 5 Comparing Variation in Two Samples Key Concept In this section we develop methods for testing hypotheses and constructing confidence intervals involving the mean of the differences of the values from two dependent populations. With dependent samples, there is some relationship whereby each value in one sample is paired with a corresponding value in the other sample. Key Concept Notation for Dependent Samples Because the hypothesis test and confidence interval use the same distribution and standard error, they are equivalent in the sense that they result in the same conclusions. Consequently, the null hypothesis that the mean difference equals 0 can be tested by determining whether the confidence interval includes 0. There are no exact procedures for dealing with dependent samples, but the t distribution serves as a reasonably good approximation, so the following methods are commonly used. 1

2 Hypothesis Test Statistic for Matched Pairs Requirements 1. The sample data are dependent. 2. The samples are simple random samples. 3. Either or both of these conditions is satisfied: The number of pairs of sample data is large (n > 30) or the pairs of values have differences that are from a population having a distribution that is approximately normal. P values and Critical Values Use Table A 3 (t distribution). Confidence Intervals for Matched Pairs Data Set 3 in Appendix B includes measured weights of college students in September and April of their freshman year. Table 9 1 lists a small portion of those sample values. (Here we use only a small portion of the available data so that we can better illustrate the method of hypothesis testing.) Use the sample data in Table 9 1 with a 0.05 significance level to test the claim that for the population of students, the mean change in weight from September to April is equal to 0 kg. 2

3 TI input, test, and results. Enter data in L1 and L2. Set L3 = L1 L2 to calculate the difference. Use T Test on the difference, L3. Requirements are satisfied: samples are dependent, values paired from each student; although a volunteer study, we ll proceed as if simple random sample and deal with this in the interpretation; STATDISK displays a histogram that is approximately normal. P value = Since P value = > α = 0.05, fail to reject H 0. NOTE: Weights are Sept. (1st) to April (2nd) of school year; therefore, April Sept = difference for weight gain (+) or loss ( ). Weight gained = April weight Sept. weight µ d denotes the mean of the April Sept. differences in weight; the claim is µ d = 0 kg Step 1:claim is µ d = 0 kg Step 2:If original claim is not true, we have µ d 0 kg Step 3:H 0 : µ d = 0 kg original claim H 1 : µ d 0 kg Step 4:significance level is α = 0.05 Step 5:use the student t distribution Step 6:find values of d and s d differences are: 1, 1, 4, 2, 1 d = 0.2 and s d = 2.4 now find the test statistic Table A 3: df = n 1, area in two tails is 0.05, yields a critical value t = ±

4 Step 7: Because the test statistic does not fall in the critical region, we fail to reject the null hypothesis. We conclude that there is not sufficient evidence to warrant rejection of the claim that for the population of students, the mean change in weight from September to April is equal to 0 kg. Based on the sample results listed in Table 9 1, there does not appear to be a significant weight gain from September to April. The conclusion should be qualified with the limitations noted in the article about the study. The requirement of a simple random sample is not satisfied, because only Rutgers students were used. Also, the study subjects are volunteers, so there is a potential for a self selection bias. In the article describing the study, the authors cited these limitations and stated that Researchers should conduct additional studies to better characterize dietary or activity patterns that predict weight gain among young adults who enter college or enter the workforce during this critical period in their lives. The P value method: Using technology, we can find the P value of (Using Table A 3 with the test statistic of t = and 4 degrees of freedom, we can determine that the P value is greater than 0.20.) We again fail to reject the null hypothesis, because the P value is greater than the significance level of α =

5 Confidence Interval method: Construct a 95% confidence interval estimate of μd, which is the mean of the April September weight differences of college students in their freshman year. = 0.2, s d = 2.4, n = 5, t a/2 = Find the margin of error, E Construct the confidence interval: We have 95% confidence that the limits of 2.8 kg and 3.2 kg contain the true value of the mean weight change from September to April. In the long run, 95% of such samples will lead to confidence interval limits that actually do contain the true population mean of the differences. Calculations with Paired Sample Data. In Exercises 5 and 6, assume that you want to use a 0.05 significance level to test the claim that the paired sample data come from a population for which the mean difference is μd = 0. Find (a), (b) sd, (c) the t test statistic, and (d) the critical values. 498/5. Car Mileage Listed below are measured fuel consumption amounts (in miles gal) from a sample of cars (Acura RL, Acura TSX, Audi A6, BMW 525i) taken from Data Set 16 in Appendix B. City fuel consumption Highway fuel consumption Recap In this section we have discussed: Requirements for inferences from matched pairs. Notation. Hypothesis test. Confidence intervals. TI input, test, and results. Enter data in L1 and L2. Set L3 = L1 L2 to calculator the difference. Use T Test on the difference, L3. P value = 6.12 E 5 Since P value = 6.12 E 5 < α = 0.05, reject H0. 5

6 Calculations with Paired Sample Data. In Exercises 5 and 6, assume that you want to use a 0.05 significance level to test the claim that the paired sample data come from a population for which the mean difference is μd = 0. Find (a), (b) s d, (c) the t test statistic, and (d) the critical values. 498/6. Forecast Temperatures Listed below are predicted high temperatures that were forecast before different days (based on Data Set 11 in Appendix B). Predicted high temperature forecast three days ahead Predicted high temperature forecast five days ahead /7. Confidence Interval Using the sample paired data in Exercise 5, construct a 95% confidence interval for the population mean of all differences, in this format: (city fuel consumption) ( fuel consumption). Data from Exercise 5: City fuel consumption Highway fuel consumption Statistically, there is a difference between City fuel consumption and Highway fuel consumption. Since L3 = L1 L2 yields a negative mean, L1 < L2. This means City fuel consumption is less than Highway fuel consumption. t = t = /8. Confidence Interval Using the sample paired data in Exercise 6, construct a 99% confidence interval for the population mean of all differences, in this format: (high temperature predicted three days ahead) (high temperature predicted five days ahead). Data from Exercise 6: Predicted high temperature forecast three days ahead Predicted high temperature forecast five days ahead /12. Are Flights Cheaper When Scheduled Earlier? Listed below are the costs (in dollars) of flights from New York (JFK) to San Francisco for US Air, Continental, Delta, United, American, Alaska, and Northwest. Use a 0.01 significance level to test the claim that flights scheduled one day in advance cost more than flights scheduled 30 days in advance. What strategy appears to be effective in saving money when flying? Flight scheduled one day in advance Flight scheduled 30 days in advance Statistically at the 99% confidence level, there is NO difference between Predicted high temperature forecast three days ahead and Predicted high temperature forecast five days ahead. Support the claim that flights scheduled one day in advance cost more than flights scheduled 30 days in advance. 6

7 499/14. Is Blood Pressure the Same for Both Arms? Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman (based on data from Consistency of Blood Pressure Differences Between the Left and Right Arms, by Eguchi, et al., Archives of Internal Medicine, Vol. 167). Use a 0.05 significance level to test for a difference between the measurements from the two arms. What do you conclude? Right arm Left arm /18. Self Reported and Measured Male Heights As part of the National Health and Nutrition Examination Survey, the Department of Health and Human Services obtained selfreported heights and measured heights for males aged All measurement are in inches. Listed below are sample results. a. Is there sufficient evidence to support the claim that there is a difference between selfreported heights and measured heights of males aged 12 16? Use a 0.05 significance level. b. Construct a 95% confidence interval estimate of the mean difference between reported heights and measured heights. Interpret the resulting confidence interval, and comment on the implications of whether the confidence interval limits contain 0. Reported height Measured height When comparing a one-tailed test and confidence interval for the same hypothesis, we must use the confidence level as 1-2α. For example, a one-tailed test with α = 0.05 is equivalent to the 1-2(0.05) = = 0.90 = 90% confidence interval. This is necessary because the confidence interval uses the same amount of area to the right and to the left. 500/20. Heights of Winners and Runners Up Listed below are the heights (in inches) of candidates who won presidential elections and the heights of the candidates who were runners up. The data are in chronological order, so the corresponding heights from the two lists are matched. For candidates who won more than once, only the heights from the first election are included, and no elections before 1900 are included. a. A well known theory is that winning candidates tend to be taller than the corresponding losing candidates. Use a 0.05 significance level to test that theory. Does height appear to be an important factor in winning the presidency? b. If you plan to test the claim in part (a) by using a confidence interval, what confidence level should be used? Construct a confidence interval using that confidence level, then interpret the result. Won Presidency: Runner Up: α = 0.05 α = 0.05 See pages 412 & 413. DO NOT reject H 0 because P value = > α = Box 4: There is not sufficient evidence to support the claim that winning candidates tend to be taller than the corresponding losing candidates. 7