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# A3 - 19 20 21 22 23 24 25 26 D The mean square of errors is...

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Unformatted text preview: 19. 20. 21. 22. 23. 24. 25. 26. D. The mean square of errors is found by ﬁrst ﬁnding the SSE: 5*(3.5 + 2.7 + 6.4) = 63. The MSE is then 63/(n—k) :2 63/15 = 4.2. B. This is a one—tailed test with an alternative hypothesis of H1 : g; > 1 thus we go from 2 the test statistic to. positive inﬁnity. The F distribution is approximately centered at 1 so the test statistic is deﬁnitely right of the center of the F distribution and then the p—value is from the test statistic to the extreme right. This corresponds to option (b). B. This is a one—tailed test on the left end of the t—distribution. Thus the appropriate critical value is —1.706. The test statistic of —1.988 is more extreme (further to the left) than the critical value so we reject the null hypothesis and can conclude that drug A is more effective than drug B. - C. The test statistic would change since the numerator in the test statistic formula would now need to subtract -1. Since the test statistic has changed, the p—value(s) in the output would change as well. However, the critical values would stay the same since they are on the standardized distribution, found only by the degrees of freedom and level of signiﬁcance. B. You need to show that the After nitrate level is 50 ppm less than the Before nitrate level or H1 : MB > MA + 50 and this is matched pairs since the same ﬁrms were sampled, so this corresponds to H1 : up > 50. And then back into the null hypothesis. E. Since this is matched pairs, you will need to find the matched pairs differences for each observation. These correspond to 50, 54, 56, 53, 58, 47. Plug these into your calculator to ﬁnd the sample mean (53) and sample standard deviation (4). The test statistic is then found by 3:}? =' 1.837. B. We have a one—tailed test on the right end of the t—distribution with Tip — 1 degrees of freedom. Since a is not given to us we are to assume it is equal to .05. Thus the correct critical value is 2.015. The test statistic given to us of 2.25 is more extreme than the critical value so we reject the null hypothesis, this implies ﬁrms have reduced their nitrate levels by more than 50 ppm. E. The appropriate sample size is found by taking the z—critical value of 1.960 and using the . 2 2 . sample Size formula: W = 50.09 and we always round sample Size answers up and thus we need to sample 51 people. 405 ...
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