Question

Question 2 of 5 < -/1 E View Policies Current Attempt in Progress Consider the hypothesis test H0: 1-2 against H1:1>2. Suppose that sample sizes are n 1 - 15 and n 2 - 15, that 1 = 7.6 and x 2 - 5.8 and that s 12 -4 and s 22-9. Assume that 012-022 and that the data are drawn from normal distributions. Use a -0.05 (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) if p 1 is 3 units greater than 2? (c) Assuming equal sample sizes, what sample size should be used to obtain B -0.05 if u 1 is 2.8 units greater than u 2 ? Assume that -0.05 o rejected. The P-value is Round your answer to three decimal places (e.g. (a) The null hypothesis 98.765). (b) The power is Round your answer to two decimal places (e.g. 98.76). (c) n 1 - 2 - Round your answer to the nearest Integer. Statistical Tables and Charts e Textbook and Media Save for Later Attempts: 0 of 3 used Submit Answer

Answer 1

a)

The provided sample means are shown below: X1 = 7.6 X9 = 5.8 Also, the provided sample standard deviations are: Si = 2 89 = 3 and the sample sizes are ni 15 and ne = 15 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha Hi > 2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 28. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this right-tailed test is te = 1.701, for a = 0.05 and df = 28. The rejection region for this right-tailed test is R={t:t> 1.701}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X, t= (n-1)33+ (13-14 ( + ) 7.6 - 5.8 (15-1)22+(15-1)32 15+15-2 = 1.934 +6+) (4) Decision about the null hypothesis Since it is observed that t = 1.934 >te = 1.701, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0317, and since p = 0.0317 <0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean 4, is greater than 112, at the 0.05 significance level.

b)

This is a right tailed test.

We will fail to reject the null (commit a Type II error) if we get a Z statistic less than 1.645.

This 1.645 Z-critical value corresponds to some X_d critical value ( X critical), such that

SE = (n) –1)s+(12 - 1)33 ( n + b) = (15-1) 22+(15-1)32 15-15-2 (1 + 1) =0.9309 nitne - 2

Xd - 0 0.9309 = 1.645

X = 1.53

So I will incorrectly fail to reject the null as long as a draw a sample mean difference that less than 1.53. To complete the problem what I now need to do is compute the probability of drawing a sample mean difference less than 1.53 given difference is 3 units. Thus, the probability of a Type II error is given by

$\beta=P(Z<\frac{1.53-3}{0.9309})=P(Z<-1.579)=0.0572$( using excel formula =NORM.S.DIST(-1.579,TRUE))

Power of the test = $1-\beta=1-0.0572=0.9428$

c) $n=(\frac{2*(Z_{alpha}+Z_{beta})^2(\frac{S_1^2+S_2^2}{2})}{\Delta^2})$

where $\Delta$ is the effectsize

$\Delta=2.8,\beta=0.05,\alpha=0.05$

$Z_{0.05}=-1.645$

$n=(\frac{2*(-1.645-1.645)^2(\frac{4+9}{2})}{(2.8)^2})\approx 18$

(a) The null hypothesis is rejected. The P-values 98.765). 0.032 Round your answer to three decimal places (e.g. (b) The power is 0.94 Round your answer to two decimal places (eg, 98.76), (c) n 1 - 2 - 18 Round your answer to the nearest integer.