Question

il store chain with 2 locations is worried that Store #1 is slacking on the quality of their customer service. They want to run a hypothesis test that the percent of customers who think their service was pooris the same at both locations. They have collected the following data. Refer to Store #1 as Population 1 and Store #2 as Population 2, and answer the following questions leading up to the hypothesis test.

Store #1 (population 1)Customers surveyed = 200 Customers , Number that said customer service was ”poor” = 44

Store #2 (population 2) surveyed = 300 , Number that said customer service was ”poor” = 48

(a) What is the estimate of the proportion in population 1? .22

(b) What is the estimate of the proportion in population 2? .16

(c) What is the estimate of the difference between the two population proportions? .06

(d) What is the standard error of the difference between the two population proportions? .036

e) Develop a 90% confidence interval for the difference between the two population proportions. ..059

(f) Develop a 95% confidence interval for the difference between the two population proportions.

(g) Develop a 99% confidence interval for the difference between the two population proportions.

(h) What is the pooled estimate of p?

(i) What is the test statistic (z-value)?

(j) What is then-value?

k) Givenα=0.10, what is your conclusion?(

l) Givenα=0.05, what is your conclusion?

(m) Givenα=0.01, what is your conclusion?

Answer 1

a) The estimate of proportion is $\fn_jvn \overline{p_1}=44/200=0.22$

b) The estimate of proportion is $\fn_jvn \overline{p_2}=48/300=0.16$

c)  The estimate of the proportion in population is $\fn_jvn \overline{p_1}-\overline{p_2}=0.06$

d) Standard error of the differences is

$\fn_jvn SE=\sqrt{\frac{\overline{p_1}\left ( 1-\overline{p_1} \right )}{n_1}+\frac{\overline{p_2}\left ( 1-\overline{p_2} \right )}{n_2}}\\ SE=\sqrt{\frac{0.22\left ( 1-0.22 \right )}{200}+\frac{0.16\left ( 1-0.16\right )}{300}}\\ SE=0.0361$

e) The $\fn_jvn \left ( 1-\alpha \right )100\%$ confidence interval is

$\fn_jvn \left (\overline{p_1}-\overline{p_2} \right )\pm z_{1-\alpha /2}\sqrt{\frac{\overline{p_1}\left ( 1-\overline{p_1} \right )}{n_1}+\frac{\overline{p_2}\left ( 1-\overline{p_2} \right )}{n_2}}$

When $\fn_jvn \alpha =0.01$ , 90% confidence interval for the difference between the two population proportions is

$\fn_jvn 0.06\pm z_{1-0.1 /2}0.0161\\ \left (0.0335 ,0.0865 \right )$

f)When $\fn_jvn \alpha =0.05$ , 95% confidence interval for the difference between the two population proportions is

$\fn_jvn 0.06\pm z_{1-0.05 /2}0.0161\\ \left (0.0284, 0.0916\right )$

g)When $\fn_jvn \alpha =0.01$ , 99% confidence interval for the difference between the two population proportions is

$\fn_jvn 0.06\pm z_{1-0.01 /2}0.0161\\ \left (0.0185, 0.1015\right )$

h) Pooled estimate is $\fn_jvn \frac{44+48}{200+300}=0.184$