Question

XIII. (Hint: See 7.1) Consider the following statistics regarding class sizes at two high schools. McClatchy High Grant High n-41 n=61 x = 26.43 y = 24.51 =4.02 s=3.15 a) Use a = 0.05 to test the claim that the mean class size at McClatchy is different than at Grant High School. b) Construct a 90% Confidence Interval for uli - u2. XIV. (Hint: See 7.3) Consider the following typing speeds before and after taking a typing class. Before: 37 46 60 41 56 56 32 wpm After: 44 51 63 42 61 65 44 a. Construct a Confidence Interval with a 99% Confidence Level. b. State the meaning of the confidence interval within the context of the problem

Answer 1

13.

Let $\mu_1,\mu_2$ denote the mean class size at McClatchy and Grant High school respectively.

b)

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)

14.

Let us consider the difference

d = speeds before - speeds after

a)

b) Out of 100 repetitive samples drawn, the paired difference in speeds will lie between the confidence intreval for 99 samples.