Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Solutions to Homework #9, 36-220 Due 16 November 2005 8.54 We are supposed to test whether “the true average percentage of organic matter in such soil is something other than 3%”. So our null hypothesis H 0 is that μ = 3. We need a two-sided test, since we want to tell whether the true percentage is different than 3%, whether the difference is positive or negative. So H 1 is that μ = 3. Because σ is not known, and n is comparatively small, only 30, we can’t use a z (standard normal) test. We can however use a t test, because we’re given that the individual measurements are normal. t = 2 . 481 - 3 0 . 295 = - 1 . 76 Since n = 30, there are 30 - 1 = 29 degrees of freedom. To compute the p -value, we need the probability that a t -distribution with 29 degrees of freedom would give a value as far from the origin as we got — Pr ( | T | ≥ | t | ). This is p = Pr ( T ≤ -| t | ) + Pr ( T ≥ | t | ) = 2Pr ( T ≥ | t | ) since the t -distribution is symmetric. Turning to appendix A.8, we see that, with 29 degrees of freedom, Pr ( T 1 . 7) = 0 . 05, and Pr ( T 1 . 8) = 0 . 041. Linear extrapolation between these values gives Pr ( T 1 . 76) = 0 . 0446. (Using statistical software gives 0 . 04456319.) That is, according to the null hypothesis, we should expect to see results this far out along one of the tails 4 . 46 percent of the time. Since we’re using a two-sided test, we need to double this probability, p = 0 . 0892. This is less than 0 . 1, so we’d reject the null at the 10% level. At that level, we do have evidence that the mean concentration of organic matter in the soil isn’t 3%. At the stricter α = 0 . 05 level, we don’t have enough evidence to reject the null that μ = 3. 8.58 The SAS includes a two-sided p -value, namely 0 . 0711. We want to perform a two-sided test — there will be a problem with our ciruits if the cut-on voltage of the diodes is either too high or too low — so this suits us just fine. Because 1
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0 . 0711 > 0 . 01 and > 0 . 05, we conclude that, at that significance level , cut-on voltages do not differ systematically from 0 . 60V; with those values of α , we do not need to adjust anything. But 0 . 0711 < 0 . 1, so the deviation from the null is significant at the less reliable 10% level. If that is our chosen value of α , we must adjust, because the average cut-on voltage is too high (the calculated test statistic t is positive).
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