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OSU BUSMGT 2320 - Comparisons of Two Populations (mu) Additional Practice Solutions

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Comparisons [µ] - Additional Practice Problems - Solutions (1) If some natural relationship exists between each pair of observations that provides a logical reason to compare the first observation of sample 1 with the first observation of sample 2, the second observation of sample 1 with the second observation of sample 2, and so on, the samples are referred to as: a. matched samples - definition b. independent samples c. weighted samples d. random samples (2) True or False. A researcher is curious about the effect of sleep on students’ test performances. He chooses 50 students and gives each two tests: one given after four hours of sleep and one after eight hours of sleep. The test the researcher should use would be matched pairs t-test. Each student is measured twice. A student’s general understanding of the subject matter, analytical skills, etc. will remain the same for both tests. (3) Consider the following hypothesis test: a. What is the appropriate test statistic (“Z” or “t”)? If “t”, also indicate the degrees of freedom. b. What are the value(s) of the critical point(s)? Express your answer to 3 decimal places. The value will depend on access to computer or printed table. If using printed table, and df are not in table, use next lower available df. I’m using df = 40. t-criticals = ±t with 52 df (row) and 0.025 tail area (column). t-criticals = ±2.021 Ho: 1  2 = 1.7 H1: 1  2 ≠ 1.7 n1 = 20 n2 = 35 X1= 25.2 X2= 20.4 s1= 4s2= 7 = 0.05 Use the Satterthwaite Approximation Method since 20625.316492122ss. The test statistic is 2221212121)()(nsnsxxtobs ~ t(52) df: 1122222121212222121nnsnnsnsns=   1353549116201635492016222= 52.99. Use 52 df. Note: If you do not have access to statistical computing software, and the calculation of Δ is not reasonable, substitute the smaller of n1 – 1 and n2 – 1 for the df. In this case you would substitute 19. This is far fewer df than the actual 52, so use Δ if possible.c. What is the 95% Confidence Interval estimate of µ1  µ2. Express your answer to 3 decimal places.   22212121nsnstXX =    35492016021.24.202.25 = (1.802, 7.798) d. “Reject” / “Do not reject” null hypotheses? Reject H0 since the interval does not capture 1.7. e. Conclusion? (4) The number of degrees of freedom associated with the t test, when the data are gathered from a matched pairs experiment with 10 pairs, is: a. 10 b. 20 c. 9 – There will be 10 differences. One standard deviation will be calculated from the sample data based on those 10 differences. The df = nD – 1 = 10 – 1. d. 18 There is statistically significant evidence at the 5% level that the difference between the means of population1 and population 2 does not equal 1.7. Since the 95% confidence interval contains values that are all > 1.7, it is logical to conclude that the mean of population 1 > mean of population 2 by more than 1.7.Use the following narrative to answer questions (5) and (6). Automobile insurance appraisers examine cars that have been involved in accidental collisions to assess the cost of repairs. An insurance executive is concerned that different appraisers produce significantly different assessments. In an experiment 10 cars that have recently been involved in accidents were shown to two appraisers. Each assessed the estimated repair costs. These results are shown below. Car Appraiser 1 Appraiser 2 D 1 1650 1400 250 2 360 380 -20 3 640 600 40 4 1010 920 90 5 890 930 -40 6 750 650 100 7 440 410 30 8 1210 1080 130 9 520 480 40 10 690 770 -80 Sum = 540 (5) What does D1 equal? a. 1650 – 360, the range of repair cost assessments for Appraiser 1 b. 816, the average repair cost assessment for Appraiser 1 c. 762 – 816, the difference in average repair cost assessments for Appraiser 1 and Appraiser 2 d. 1650 – 1400, the difference in Appraiser 1’s repair cost assessment for car 1 and Appraiser 2’s repair cost assessment for car 1. The 2 appraisers’ cost assessments are matched to the car. (6) Can the executive conclude at the 5% significance level that the appraisers differ in their assessments? Solution: H0: µD = 0 H1: µD ≠ 0 (difference in assessments, no suggestion that one is higher than the other) Test Statistic: 80.1107746.940540DDDobsnsxt P-value (two-sided test): Using the t table with 9 df, 2P(t ≥ 1.80) is between 2(.05) and 2(.10). Using Minitab: t(9)-1.8 1.80.0527000.05270 Since P-value > 0.05, (or -2.262 < tobs = 1.80 < 2.262) there is not statistically significant evidence of a difference, on average, in repair cost assessments made by the two appraisers. Excel Data Analysis: These calculations would usually be done with technology help: 5410540Dx 7746.949)5480()54250(22DDss t* = 2.262 -2.262 α/2 = 0.025 α/2 = 0.025t-Test: Paired Two Sample for Means Appraiser 1 Appraiser 2 Mean 816 762 Variance 153560 105506.6667 Observations 10 10 Pearson Correlation 0.982375856 Hypothesized Mean Difference 0 df 9 t Stat 1.801780416 P(T<=t) one-tail 0.05254704 t Critical one-tail 1.833112933 P(T<=t) two-tail 0.10509408 t Critical two-tail 2.262157163 StatTools Output: Hypothesis Test (Paired-Sample) Appraiser 1 - Appraiser 2 Sample Size 10 Sample Mean 54 Sample Std Dev 94.77458637 Hypothesized Mean 0 Alternative Hypothesis <> 0 Standard Error of Mean 29.97035572 Degrees of Freedom 9 t-Test Statistic 1.8018 p-Value 0.1051 Null Hypoth. at 10% Significance Don't Reject Null Hypoth. at 5% Significance Don't Reject Null Hypoth. at 1% Significance Don't Reject Minitab Output: (7) Two types of new cars, Type A and Type B, are being considered for purchase by a fleet manager for a government agency. An important criterion in the purchase decision is gas mileage. The fleet manager has reason to think that Type Paired T-Test and CI: Appraiser 1, Appraiser 2 Paired T for Appraiser 1 - Appraiser 2 N Mean StDev SE Mean Appraiser 1 10 816 392 124 Appraiser 2 10 762 325 103 Difference 10 54.0 94.8 30.0 95% CI for mean difference: (-13.8, 121.8) T-Test of mean difference = 0 (vs not = 0): T-Value


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OSU BUSMGT 2320 - Comparisons of Two Populations (mu) Additional Practice Solutions

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