# STATS - 1 A rancher wants to compare the mean weight gain...

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1. A rancher wants to compare the mean weight gain of steers fed with two different feeds. She takes six pairs of steers, each pair being the same age and weight, and measures the weight gain of each pair over a two-week period. Weight gains are analyzed using the following MINITAB output. 2. Paired T-Test and Confidence Interval Paired T for Feed A – Feed B N Mean StDev SE Mean Feed A 6 39.33 9.00 3.68 Feed B 6 34.67 10.78 4.40 Difference 6 4.67 4.89 1.99 95% CI for mean difference: (-0.46, 9.79) T-Test of mean difference = 0 (vs 0): T-Value = 2.34 P-Value = 0.066 a) Define the parameter(s). b) Give the hypotheses. c) True or false: i) The null hypothesis can be rejected at α = .10. ii) If α = .10 had been chosen, it would be possible to have 90% confidence in the decision. iii) The null hypothesis can be rejected at α = .05. iv) If α = .05 had been chosen, it would be possible to have 95% confidence in the decision. v) These are independent samples. 2. A forester is interested in comparing the size of trees in two stands. As a quick way of sizing the trees, she measures the diameter at breast height (DBH). Then using MINITAB, she obtains the output that follows. Site N Mean StDev SE Mean 1 78 52.6 20.8 2.4 2 67 84.8 32.4 4.0 95% CI for mu (1) - mu (2): ( -41.4, -23.1) T-Test mu (1) = mu (2) (vs not =): T = XXXXX P = XXXXXX DF = 109 a) What is (are) the parameter(s) of interest? b) True or false: i) For this interval to be valid, it is necessary that dbh for each group is normal. ii) 95% of the time, the difference of the mean dbh for all the trees in the two stands will fall between -41.4 and -23.1. iii) 95% of the sample differences will fall between -41.4 and -23.1. iv) 95% of the differences of the dbh’s will fall between -41.4 and -23.1. v) You can be sure that the difference of the mean dbh for all the trees in the two stands is between -41.4 and -23.1. vi) It is possible that the difference of the mean dbh for all the trees in the two stands is never between -41.4 and -23.1. vii) 95% of all intervals created in this fashion will contain the difference in the mean dbh’s. viii) If this data are used to do a two-tailed test of the mean difference equaling 0, with a level of significance of 0.05, the null hypothesis would be rejected. ix) A 90% confidence interval calculated from this data would be narrower. 3. The vineyard manager of a winery is concerned with the possible differences in the sugar levels of grapes harvested from two different rootstocks, Freedom (1) and 5C (2). He decides to test to see if there is any difference in the mean sugar amount of the two. Doing so, he obtains the following output, some of which is unnecessary.

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