13.
Let
denote the mean class size at McClatchy and Grant High school
respectively.
b)
![Here, 21 = 26.430, $1 = 4.0200, n1 = 41, 22 = 24.510, S2 = 3.1500, n2 = 61 and associated degrees of freedom s ni 122 df = =](//img.homeworklib.com/questions/33be12e0-0d58-11eb-b8be-0962995d7307.png?x-oss-process=image/resize,w_560)
14.
Let us consider the difference
d = speeds before - speeds after
a)
![](//img.homeworklib.com/questions/34600770-0d58-11eb-a4b8-3fa60cd30636.png?x-oss-process=image/resize,w_560)
b) Out of 100 repetitive samples drawn, the paired difference in
speeds will lie between the confidence intreval for 99 samples.
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Hypotheses : To test null hypothesis H H1 - H2 = 0 against alternative hypothesis H1 H1-1270 This is a two tailed test as the alternative hypothesis contains '#' sign. Here, 21 = 26.4300, $1 = 4.0200, n1 = 41, 72 = 24.5100, S2 = 3.1500, n2 = 61 The test statistic can be written as: (21-22-0 ta + 112 which under H, follows a t distribution with 72 df. where 4.022 41 df= + 3.15) (2.132) + 61-1 72 1-1 T-1 41-1 and standard error for the difference between means si 82 SE + 4.022 41 + 3.152 61 0.746204 ni n2 Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tstat | > to.025,df = +0.025,72 = 1.993464 0.4 - 0.3- 0.1- Nowo Value of the test statistic : The value of the test statistic is (26.4300 - 24.5100) tstat = 2.573024 2.57 4.02002 V 3.1500 + 41 61 associated degrees of freedom = 72 The critical value = tcritical = $t0,025,72 =+ 1.993464 +1.993 P-value = P(|t72) > tobs) = 2* P(t72 > 2.573024) = 2* (1 -0.993929) = 0.012142 -0.0121 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.993, so we reject Ho at 0.05 level of significance Hence, we can conclude that there is significant difference between two population means.
Here, 21 = 26.430, $1 = 4.0200, n1 = 41, 22 = 24.510, S2 = 3.1500, n2 = 61 and associated degrees of freedom s ni 122 df = = 72 + ni-1 n2-1 The Standard error is 3.152 SEE + 4.022 V 41 + 0.746 ni n2 61 The critical value is tcritical = t 1-0.9 10.9,72 = to.05,72 = 1.666294 ~1.666 The margin of error is 82 4.022 ME=t1-09 -,72 * + =t0.050,72 * + 3.152 61 1.666 * 4.022 V 41 + 3.152 61 = 1.243395 1.243 ni n2 41 a 90 percent confidence interval for the differnce in population means (ui - U2 ) is = či - 22 + *t 1-0.900 ,72 ni n2 = 21 - 22 + ME = (26.430 – 24.510) +1.243395 = 1.920 +1.243395 = (0.676605, 3.163395) – (0.68, 3.16)
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