STA 6166 – Exam 2 – Fall 2006 PRINT Name _________________A researcher is comparing the breaking strenghts of two types of aluminum tubes. She samples 12 tubes of each type and measures pressure necessary to bend the tubes. Labelling 1 as the true mean for all tubes of type 1, and 2 as the true mean for all tubes of type 2, conduct the test of H0: 1-2 = 0 vs HA: 1-2 ≠ 0 at the = 0.05 significance level, based on the following output (assume population variances are equal):Independent Samples Test1.534 .229 8.574 22 .000 8.87316 1.03485 6.72702 11.019308.574 21.174 .000 8.87316 1.03485 6.72216 11.02416Equal variancesassumedEqual variancesnot assumedyF Sig.Levene's Test forEquality of Variancest df Sig. (2-tailed)MeanDifferenceStd. ErrorDifference Lower Upper95% ConfidenceInterval of theDifferencet-test for Equality of MeansTest Statistic _________________________Reject H0 if the test statistic falls in the range(s) ________________________P-value _____________________________Conclude (Circle One) Do Not Conclude 1-2 ≠ 0 1-2 <0 1-2 > 0A study is conducted to compare 4 formulations of a new drug in terms of the availability of the drug in the bloodstream over time. Ten healthy subjects are selected and each subject receives each drug in random order in a randomized block design. Theresearcher conducts the appropriate F-test for testing for formulation differences. If thetest is conducted at the =0.05 significance level, he will conclude formulation differences exist if the F-statistic falls in what range?A study is conducted to compare golf ball distances among 4 brands of golf balls. A mechanical driver is set up and hits 12 balls of each brand, and the distance travelled (meters) is measured. The F-test determines that there are differences among the brands. A follow-up comparison based on Tukey’s method yields the following table:Multiple ComparisonsDependent Variable: yTukey HSD-21.40796* 2.51529 .000 -28.1238 -14.6921-.00121 2.51529 1.000 -6.7171 6.7146-12.43549* 2.51529 .000 -19.1513 -5.719621.40796* 2.51529 .000 14.6921 28.123821.40675* 2.51529 .000 14.6909 28.12268.97247* 2.51529 .005 2.2566 15.6883.00121 2.51529 1.000 -6.7146 6.7171-21.40675* 2.51529 .000 -28.1226 -14.6909-12.43428* 2.51529 .000 -19.1501 -5.718412.43549* 2.51529 .000 5.7196 19.1513-8.97247* 2.51529 .005 -15.6883 -2.256612.43428* 2.51529 .000 5.7184 19.1501(J) group234134124123(I) group1234MeanDifference(I-J) Std. Error Sig. Lower Bound Upper Bound95% Confidence IntervalThe mean difference is significant at the .05 level.*. Clearly state what can be said of all pairs of brands at the 0.05 experimentwise error rate.Brands:1 vs 2: 1 vs 3: 1 vs 4: 2 vs 3: 2 vs 4: 3 vs 4:A study is conducted to compare whether incidence of muscle aches differs among athletes exposed to 5 types of pain medication. A total of 500 people who are members of a large fitness center are randomly assigned to one of the medications. After a lengthy workout, eachis given a survey to determine presence/absence of muscle pain. For the 5 groups: 25, 38, 32, 40, and 35 are classified as having muscle pain, respectively. The following output gives the results for the Pearson chi-square statistic for testing (=0.05):H0: True incidence rate of muscle pain doesn’t differ among medicationsHA: Incidence rates are not all equal Test Statistic _________________________Reject H0 if the test statistic falls in the range(s) ________________________P-value _____________________________Conclude (Circle One): Medication effects not all equal No differences in effectsGive the expected number of incidences of muscle pain for each medication under H0:In a 2-Factor ANOVA, measuring the effects of 2 factors (A and B) on a response (y), there are 3 levels each for factors A and B, and 4 replications per treatment combination. Give the values of of the F-statistic for the AB interaction for which we will conclude the effects of Factor A levels depend on Factor B levels and vice versa (=0.05):Chi-Square Tests6.150a4.188Pearson Chi-Square Value dfAsymp. Sig.(2-sided)A simple linear regression model is fit, relating plant growth over 1 year (y) to amount of fertilizer provided (x). Twenty five plants are selected, 5 each assigned to each of the fertilizer levels (12, 15, 18, 21, 24). The results of the model fit are given below:Can we conclude that there is an association between fertilizer and plant growth at the 0.05 significance level? Why (be very specific).Give the estimated mean growth among plants receiving 20 units of fertilizer.The estimated standard error of the estimated mean at 20 units is46.0450)1820(2511.22Give a 95% CI for the mean at 20 units of fertilizer.Coefficientsa8.624 1.810 4.764 .000.527 .098 5.386 .000(Constant)xModel1BStd. ErrorUnstandardizedCoefficientstSig.Dependent Variable: ya.A multiple regression model is fit, relating salary (Y) to the following predictor variables: experience (X1, in years), accounts in charge of (X2) and gender (X3=1 if female, 0 if male). The following ANOVA table and output gives the results for fitting the model. Conduct all tests at the 0.05 significance level:Y = 0 + 1X1 + 2X2 + 3X3 + ANOVA df SS MS F P-valueRegression 3 2470.4 823.5 76.9 .0000Residual 21 224.7 10.7Total 24 2695.1 CoefficientsStandardError t Stat P-valueIntercept 39.58 1.89 21.00 0.0000experience 3.61 0.36 10.04 0.0000accounts -0.28 0.36 -0.79 0.4389gender -3.92 1.48 -2.65 0.0149Test whether salary is associated with any of the predictor variables:H0: HA: Not all i = 0 (i=1,2,3)Test Statistic _________________________Reject H0 if the test statistic falls in the range(s) ________________________P-value _____________________________Conclude (Circle One) Set-up the predicted value (all numbers, no symbols) for a male employee with 4 years of experience and 2 accounts.The following tables give the results for the full model, as well as a reduced model, containing only expereience.Test H0: 2 = 3 = 0 vs HA: 2 and/or 3 ≠ 0Complete Model: Y = 0 + 1X1 + 2X2 + 3X3 + ANOVA df SS MS F P-valueRegression 3 2470.4 823.5 76.9 .0000Residual 21 224.7 10.7Total 24 2695.1 Reduced Model: Y = 0 + 1X1 + df SS MS F P-valueRegression 1 2394.9
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