Question

In order to compare the means of two populations, independent random samples of 395 observations are selected from each population, with the results found in the table to the right. Complete parts a through e below. Sample 2 x2 = 5,250 2-210 Sample 1 X,5,279 1-140 a. Use a 95% confidence interval to estimate the difference between the population means (μ1-μ2) . Interpret the confidence The confidence interval is Round to one decimal place as needed.) Interpret the confidence interval. Select the correct answer below. A. We are 95% confident that each of the population means is contained in the confidence interval B. ° C. 0 D. We are 95% confident that the difference between the population means falls outside of the confidence interval. We are 95% confident that the difference between the population means falls in the confidence interval. We are 95% confident that each of the population means falls outside of the confidence interval. b. Test the null hypothesis Ho: (버-u2)-0 versus the alternative hypothesis Hai (버-u2) 40, Give the significance level of the test, and interpret the result. Use α=0.05. What is the test statistic?

Answer 1

a)$CI=(\overline{x_1}-\overline{x_2})\pm Z_{\frac{\alpha }{2}}(\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}})$

for 95% confidence

$Z_{\frac{\alpha }{2}}=1.96$

$CI=(5279-5250)\pm 1.96\sqrt{\frac{140^2}{395}+\frac{210^2}{395}}$

$CI=29\pm 24.89=(4.1,53.9)$

Option C is right for the interpretation

b)

Test statistic:$Z=\frac{\overline{X_1}-\overline{X_2}}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}=\frac{(5279-5250)}{\sqrt{\frac{140^2}{395}+\frac{210^2}{395}}}=2.28$

P value for the two-tailed test = 0.0226

Since the P value is less than the significance level of the test, we can reject the null hypothesis.

The option C is right for interpretation

c) This would be a right-tailed test now. Test statistic remain same

P(Z>2.28) =1- P(Z<2.28) =1- 0.9887 = 0.0113

Hence option D is right

d)

Test statistic:$Z=\frac{\overline{X_1}-\overline{X_2}-(\mu _d)}{\sqrt{\frac{S_1^2}{n_1}&plus;\frac{S_2^2}{n_2}}}=\frac{(5279-5250)-22}{\sqrt{\frac{140^2}{395}&plus;\frac{210^2}{395}}}=0.55$

P value for the two-tailed test = 0.5823

Since the P value is greater than the significance level of the test, we cannot reject the null hypothesis.

Option B is right for interpretation

e) Option D is right