Question

(1 point) (Give answers to at least two decimal places. For simplicity, use the standard normal distribution bacause the samples are both large.) In order to compare the means of two populations, independent random samples of 93 observations are selected from each population, with the following results: Sample 1 Sample 2 s1 = 170 s2 = 195 (a) Use a 98 % confidence interval to estimate the difference between the population means ( ) - Test the null hypothesis: Ho : (μ,-H2-0 versus the alternative hypothesis: Ha : (μί-12-0. Using 0.02, give the following: (b) α the test statistic z The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that (μ 1-12)-0. B. We can reject the null hypothesis that (M1-M2-0 and conclude that (A-μ2) 0. (c) Test the null hypothesis: Ho : (A-12) = 28 versus the alternative hypothesis: Ha : (μι-μ2) * 28. Using α 0.02, give the following: the test statistic z- The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that (M-12)-28. B. We can reject the null hypothesis that (μ,-H2-28 and conclude that(M1-H2-28.

Answer 1

$\dpi{100} x_1 = 3, \ s_1 = 170,\ n_1=93$

$\dpi{100} x_2 = 1, \ s_2 = 195,\ n_2=93$

a) At $\dpi{100} \alpha$ = 0.05, critical value, $\dpi{100} z_{\alpha/2}$ = 2.33

98% confidence interval for the difference between population means:

$\dpi{100} \left ((\bar x_1-\bar x_2)\ \pm\ z_{\alpha /2}\sqrt{\frac{s^2_1}{n_1}+\frac{s_2^2}{n_2}} \right )$

$\dpi{100} \left ((3-1)\ \pm\ 2.33\sqrt{\frac{170^2}{93}+\frac{195^2}{93}} \right )$

$\dpi{100} \left (-60.504,\ 64.504 \right )$

$\dpi{100} \textbf{-60.504}<\mu_1-\mu_2<\textbf{64.504}$

b) Null and alternative hypothesis:

$\dpi{100} H_0:\mu_1-\mu_2 = 0$

$\dpi{100} H_a:\mu_1-\mu_2 \neq 0$

$\dpi{100} z = \frac{\bar x_1 - \bar x_2}{\sqrt{s_1^2/n_1 &plus; s_2^2/n_2}} = \frac{ 3 - 1}{\sqrt{ 170^2/93 &plus; 195^2/93}} =\textbf{ 0.075}$

As z = 0.075 < $\dpi{100} z_{\alpha /2}$ = 2.33, we fail to reject the null hypothesis.

Final conclusion: Answer A. There is not sufficient evidence.

c) Null and alternative hypothesis:

$\dpi{100} H_0:\mu_1-\mu_2 = 28$

$\dpi{100} H_a:\mu_1-\mu_2 \neq 28$

$\dpi{100} z = \frac{(\bar x_1 - \bar x_2)-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1 &plus; s_2^2/n_2}} = \frac{ (3 - 1)-28}{\sqrt{ 170^2/93 &plus; 195^2/93}} =\textbf{ -0.969}$

As |z| = 0.969 < $\dpi{100} z_{\alpha /2}$ = 2.33, we fail to reject the null hypothesis.

Final conclusion: Answer A. There is not sufficient evidence.