We fail to reject h there is no evidence against the

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We fail to reject H 0 . There is no evidence against the hypothesis that the two varieties have the same mean yield. 3.(a) These are paired samples. The 90% confidence inter- val is 7 . 875 ± 1 . 753 * 14 . 96 16 , i.e. (1.314, 14.431). The t mutliplier 1.753 was obtained from the t distribution with df=15, since there were 16 pairs of data. (b) We are 90% confidence that the true increase in mean tissue sulfur lies between 1.314 and 14.431 ppm dry weight. The 90% confidence refers to the fact that if we conduct many experiments (such as this one) and use the method we have used to get a confidence interval from each of these experiments, then 90% of these confidence intervals would contain the true value. (c) The confidence interval does not contain 0, so we can reject the null hypothesis of no decrease in mean tissu sulfur at the α = 0 . 10 level. (d) We assumed sites were sampled at random. We don’t know much about that from the text. We would like to make sure that all plots were at about the same distance from the power plant, etc. We also made the assumption that the differences were normally distributed. We can assess this assumption with the normal scores plot on the right. It seems straight enough. 4. Levene transformed data: 2, 1, 3, 4 for the low-light group (mean 2.5, var 1.666) and 5, 2, 1, 0, 0, 2, 3, 5 for the high-light group (mean 2.25 and var 3.93). Pooled sd is 1.803. A t-test on these transformed data gives t = 2 . 5 - 2 . 25 1 . 803 1 / 4 + 1 / 8 = 0 . 226. Against the t distribution with df=10, we get a p-value > . 20. We
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