Question

Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test Ho: Mi = u2 against H: M1 < u2 with known variances (j = 11 and 02 = 6. Suppose that sample sizes nj = 10 and n2 = 16 and that Æj = 14.4 and 12 = 19.8. 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 Hi is 4 units less than 12? (c) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if p1 is 4 units less than H2 ? Assume that a = 0.05. (a) The null hypothesis is not rejected. The P-value is 0.077 Round your answer to four decimal places (e.g. 98.7654). (b) The power is 0.278 Round your answer to two decimal places (e.g. 98.76). (c) ni = n2 = Round your answer up to the nearest integer. the absolute tolerance is +/-1 Statistical Tables and Chart
Can you complete my answer and verify it thank you.

Answer 1

Part 1:

$(a):~Test~statistic=z=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}=-1.4255\\\\ p-value=P(Z<-1.4255|Z\sim N(0,1))=0.077\\\\ (R~code:~round(pnorm(-1.4255),3)).\\\\ Part~(a)~is~correct.\\\\ (b)~We~reject~H_0~at~5\%~level~if~\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}<-z_{0.05}=-1.645\\\\ Power=P\left( \frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\leq -1.645|\mu_1-\mu_2=-4\right )\\\\ =P\left( \frac{\bar{x}_1-\bar{x}_2+4}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\leq -1.645+\frac{4}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}|\mu_1-\mu_2=-4\right )\\\\ \left(Since~\frac{\bar{x}_1-\bar{x}_2+4}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\sim N(0,1) \right )\\\\ =0.278\\\\ R~code:~round(pnorm(-1.645+4/sqrt(11^2/10+6^2/16)),3)\\\\ Part~(b)~is~correct.\\\\$

$(c)\\ Power=P\left( \frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\leq -1.645|\mu_1-\mu_2=-4\right )=0.95=\Phi(1.645)\\\\ or,~P\left( \frac{\bar{x}_1-\bar{x}_2+4}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\leq -1.645+\frac{4}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}|\mu_1-\mu_2=-4\right )=\Phi(1.645)\\\\ or,~-1.645+\frac{4}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}=1.645\Rightarrow \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}=4/(2*1.645)=1.2158\\\\ \frac{1}{\sqrt{n}}=0.0970~(assume,~n_1=n_2=n,~say)\Rightarrow n=(1/0.0970)^2=106~(approx).$

Part 2:

(a) Minitab output:

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 8 1.140 0.110 0.039
2 8 1.0700 0.0900 0.032

Difference = mu (1) - mu (2)
Estimate for difference: 0.0700
95% lower bound for difference: -0.0185
T-Test of difference = 0 (vs >): T-Value = 1.39 P-Value = 0.093 DF = 14
Both use Pooled StDev = 0.1005

Here p-value=0.093.

Decision is correct (since p-value<0.10).

(b)

$95\%~CI~for~\mu_1-\mu_2:~-0.0378\leq \mu_1-\mu_2\leq 0.1778.$

R code:

n1=8
m1=1.14
s1=0.11
n2=8
m2=1.07
s2=0.09
s=sqrt(7*(s1^2+s2^2)/14)
LB=m1-m2-qt(0.975,14)*s*sqrt(2/8)
UB=m1-m2+qt(0.975,14)*s*sqrt(2/8)
round(LB,4)
round(UB,4)