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6166 UNIT 2 Section 2 Answers

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ANSWERS
Computations for this problem set will be done in SAS.
The SAS program can be found here. The output found here.
1. a. Use these data to test the research hypothesis that there is a difference in mean hay consumption for the two diets. Use a = 0.05.
Statistics Lower CL Upper CL Lower CL Upper CL Variable treatment N Mean Mean Mean Std Dev Std Dev Std Dev Std Err Minimum Maximum response Group_1 8 13.616 14.538 15.459 0.7285 1.1019 2.2426 0.3896 12.8 15.9 response Group_2 8 7.6276 8.375 9.1224 0.5911 0.894 1.8196 0.3161 6.8 9.5 response Diff (12) 5.0865 6.1625 7.2385 0.7346 1.0033 1.5824 0.5017 TTests Variable Method Variances DF t Value Pr > t response Pooled Equal 14 12.28 <.0001 response Satterthwaite Unequal 13.4 12.28 <.0001
The pvalue reported here is for the twosided hypothesis. With this very small pvalue, we would conclude that the hypothesis that the two groups have similar hay consumption levels should be rejected. Even a onesided hypothesis that Group_1 mean is greater than the Group_2 mean would result in rejection. The critical value from a ttable would be t(0.05,14)=1.761 and t(0.05,13)=1.771 respectively. The observed critical value of 12.28 is way into the rejection region. Note that the standard deviations for the two groups are not very different.
1. b. Provide an estimate of the amount of difference in the mean hay consumption of the two groups.
Our best estimate is the difference between the sample means of the two groups. In this case, 6.1625. An interval estimate can be given as the 95% confidence interval for the difference which is computed as (5.0865, 7.2385). Note that zero is not a part of this interval, another indication that the Group_1 consumption is significantly greater than the Group_2 consumption.
2. Using the data from problem 6.75, perform the Wilcoxon rank sum test. Do the ttest and Wilcoxon test give different results?
The output from SAS is duplicated below:
Wilcoxon Scores (Rank Sums) for Variable response Classified by Variable treatment Sum of Expected Std Dev Mean treatment N Scores Under H0 Under H0 Score  Group_1 8 100.0 68.0 9.521905 12.50 Group_2 8 36.0 68.0 9.521905 4.50 Wilcoxon TwoSample Test Statistic 100.0000 Normal Approximation Z 3.3082 OneSided Pr > Z 0.0005 TwoSided Pr > Z 0.0009 t Approximation OneSided Pr > Z 0.0024 TwoSided Pr > Z 0.0048 Z includes a continuity correction of 0.5.
We can test both the twosided alternative hypothesis or the oneside hypothesis that Group_1 median consumption is greater than that of Group_2. The test statistic for both tests is the set of ranks for Group_1, in this case T=100. With samples of size 8 we can use the Table 5 in the book to look up the critical values (TL=48,Tu=87) for the a=0.05 two sided test and (TL=52, Tu=84) for the onesided a=0.05 test. In both cases, T=100 is greater than Tu suggesting that we would reject the null hypothesis of no differences in group medians. In this case, both the Wilcoxon and ttests lead to the same conclusions.
3. What is the 95% confidence interval for the difference between the two means?
An interval estimate can be given as the 95% confidence interval for the difference which is computed as (5.0865, 7.2385). Note that zero is not a part of this interval, another indication that the Group_1 consumption is significantly greater than the Group_2 consumption.
4. How many dairy cows would be needed for each group if we wanted to know the average difference between the two groups to within plus or minus .5 kg with 95% confidence (Hint see page 314). Use the pooled variance estimate from the ttest as if it were the true variance.
We will assume the common variance for the two groups is the pooled variance. Note that the SAS output does not provide us with this number so we will have to compute it from its equation given on page 268 in the book.
Using the equation from page 314 we have
suggesting that we would need 32 cattle in EACH group to reach this level of precision in the average difference between the two groups.