STA 6166 – Fall 2008 PRINT Name ___________________A study is conducted to compare the effects of 4 types of feed for horses. The researcher also believes that the three breeds she works with may differ in terms of weight gains. She samples 12 colts from each breed, randomly assigning 3 to each diet. She measures weight gain on each colt over a 2 month period. Complete the following ANOVA table and conduct the following tests (at 0.05 significance level).Source df SS MS F F(.05)Feeds 3000 Breeds 1000 Interaction 1000 Error Total 35 10000 H0: No Feed/Breed Interaction HA: Interaction Present Reject / Don’t Reject H0H0: No Feed Main Effects HA: Feed Main Effects Reject / Don’t Reject H0H0: No Breed Main Effects HA: Breed Main Effects Reject / Don’t Reject H0A randomized block design is conducted to compare the output of three weaving looms (treatments) for a sample of 10 operators (blocks), where each operator’s output is measured on each loom. The Mean Square Error from the ANOVA is MSE= 500. Compute Bonferroni’s B, the minimum significant difference for concluding that two looms’ population means differ if their sample means differ by at lest B.A study is conducted to determine whether students’ first year GPA (Y) can be predicted by their ACT score (X). A random sample of n=120 freshmen from a small college were selected. The following EXCEL output gives the results of a simple linear regression on the data.Regression StatisticsMultiple R 0.2695R Square 0.0726Adjusted R Square 0.0648Standard Error 0.6231Observations 120ANOVA df SS MS FSignificance FRegression 1 3.5878 3.5878 9.2402 0.0029Residual 118 45.8176 0.3883Total 119 49.4055 CoefficientsStandardError t StatP-value Lower 95%Upper95%Intercept 2.11 0.3209 6.5880 0.0000 1.4786 2.7495ACT(X) 0.04 0.0128 3.0398 0.0029 0.0135 0.0641 Give the fitted equation for predicting GPA as function of ACT score, and prediction for student scoring 20 on the ACT. Test whether there is an association (positive or negative) between GPA and ACTo Null Hypothesis: Alternative Hypothesis:o Test Statistic:o P-valueWhat proportion of the variation in GPA is “explained” by ACT scores?A commercial real estate company is interested in the relationship between properties’ rental prices (Y), and the following predictors: building age, expenses/taxes, vacancy rates, and square footage. The results for a regression are given below.Regression StatisticsMultiple R 0.7647R Square 0.5847Standard Error 1.1369Observations 81ANOVA df SS MS F P-valueRegression 4 138.3269 34.5817 26.7555 0.0000Residual 76 98.2306 1.2925Total 80 236.5575 Coefficients Standard Error t Stat P-value Lower 95%Upper95%Intercept 12.2006 0.5780 21.1099 0.0000 11.0495 13.3517age -0.1420 0.0213 -6.6549 0.0000 -0.1845 -0.0995exp/tax 0.2820 0.0632 4.4642 0.0000 0.1562 0.4078vacancy 0.6193 1.0868 0.5699 0.5704 -1.5452 2.7839sqfoot 0.0000 0.0000 5.7224 0.0000 0.0000 0.0000 Can the company conclude that rental rate is associated with any of these predictors? Give the test statistic and P-value for testing:H0: Average rental rate is not associated with any of the 4 predictorsHA: Average rental rate is associated with at least one of the 4 predictors What proportion of variation in prices is “explained” by the 4 predictors? Controlling for all other factors, we conclude age is Positively / Negatively / Not associated with rental price. (Circle One)A study is conducted to compare 3 treatments for curing insomnia. A sample of 300 insomniacs are obtained, and 100 are assigned to treatment A, 100 to treatment B, and 100 to treatment C at random. The following table gives the observed counts of subjects who are able to fall asleep within 30 minutes and unable to fall asleep within 30 minutes.Treatment\Sleep Yes No TotalA 60 40 100B 50 50 100C 40 60 100Total 150 150 300Complete the following table for expected cell counts under the hypothesis of no association between treatment and sleep status.Treatment\Sleep Yes No TotalA B C Total Compute the chi-square statisticHow large will the Pearson chi-square statistic need to be to conclude that the treatment effects differ if we conduct the test at the = 0.05 significance
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