# Assume that the drug cannot in crease the number of

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between treated and untreated lambs. Assume that the drug cannot in- crease the number of worms and, hence, use the alternative hypothesis that the mean for treated lambs is less than the mean for untreated lambs. Use α = . 05. b. Indicate the level of significance for this test. SOLUTION a. The calculations for the samples of treated and untreated sheep are sum- marized next. 8

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Drug-Treated Sheep Untreated Sheep j y 1 j = 200 j y 2 j = 280 j y 2 1 j = 6906 j y 2 2 j = 12 , 492 ¯ y 1 = 200 7 = 28 . 57 ¯ y 2 = 280 7 = 40 . 0 s 2 1 = 1 6 h 6906 - (200) 2 7 i s 2 2 = 1 6 h 12 , 492 - (280) 2 7 i = 1 6 = [6906 - 5714 . 29] = 1 6 [12 , 492 - 11 , 200] = 198.62 = 215.33 Under the assumption of equal population variances, the sample variances are combined to form an estimate of the common population standard deviation σ . This assumption appears reasonable based on the sample variances. s p = s ( n 1 - 1) s 2 1 + ( n 2 - 1) s 2 2 n 1 + n 2 - 2 = r 6(198 . 62) + 6(215 . 33) 12 = 14 . 39 . The test procedure for the research hypothesis that the treated sheep will have a mean infestation level ( μ 1 ) less than the mean level ( μ 2 ) for untreated sheep is as follows: H 0 : μ 1 - μ 2 = 0 (that is, no difference in the mean infestation levels) H a : μ 1 - μ 2 < 0 T.S. : t = ¯ y 1 - ¯ y 2 s p p 1 /n 1 + 1 /n 2 = 28 . 57 - 40 14 . 39 p 1 / 7 + 1 / 7 = - 1 . 49 R.R. : For α = . 05, the critical t -value for a one-tailed test with df = n 1 + n 2 - 2 = 12 can be obtained from Table 4 in the Appendix, using a = . 05. We will reject H 0 if t < - 1 . 782. Conclusion: Since the observed value of t , -1.49, does not fall in the rejec- tion region, we have insufficient evidence to reject the hypothesis that tere is no difference in the mean number of worms in treated and untreated lambs. b. Using Table 4 in the Appendix with t = - 1 . 49 and df = 12, we see the level of significance for this test is in the range . 05 < p < . 10. 9
3. Normal Populations. Dependent Samples (Paired data) ¤ § ¥ ƒ Test Statistic for Two Dependent Samples t = ¯ d - μ d s d n where degrees of freedom = n - 1. If the number of pairs of data is large ( n > 30), the number of degrees of freedom will be at least 30, so critical values will be z scores (Table A-2) instead of t scores (Table A-3). / £ ¡ ¢ EXAMPLE 6.7 Using a reaction timer similar to the one described in the Cooperative Group Activities of Chapter 5, subjects are tested for reaction times with their left and right hands. (Only right-handed subjects were used.) The results (in thou- sandths of a second) are given in the accompanying table. Use a 0.05 significance level to test the claim that there is a difference between the mean of the right- and left-hand reaction times. If an engineer is designing a fighter-jet cockpit and must locate the ejection-seat activator to be accessible to either the right or the left hand, does it make a difference which hand she chooses? Subject A B C D E F G H I J K L M N Right 191 97 116 165 116 129 171 155 112 102 188 158 121 133 Left 224 171 191 207 196 165 177 165 140 188 155 219 177 174 SOLUTION Using the traditional method of hypothesis testing, we will test the claim that there is a difference between the right- and left-hand reaction times. Because we are dealing with paired data, begin by finding the differences d = right - left.

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