Question

According to the University of Nevada Center for Logistics Management, 6% of all mer-

Answer 1

Answer:

a) The point estimate of the proportion of items returned for the population of

sales transactions at the Houston store = 12/80 = 0.15

b) The 95% confidence interval for the proportion of returns at the Houston store = [0.0718 < p < 0.2282].

c) Yes.

We set an hypothesis and construct a test statistics. The test statistics result gives us:

Z calculated  = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

Step-by-step explanation:

a) Point estimate of the proportion = number of returned items/ total items sold = 12/80 = 0.15.

b) By formula of confident interval:

CI(95%) = p ± Z*$\sqrt{\frac{p*(1-p)}{n} } = 0.15 \pm 1.96 *\sqrt{\frac{0.15*(1-0.15)}{80} }$,

CI(95%) = [0.0718 < p < 0.2282]

c) The hypothesis:

$H_{0}$: The proportion of returns at the Houston store is not significantly different from the returns  for the nation as a whole.

$H_{a}$: The proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

The test statistics:

Z = $\frac{\hat{p} - p_{0}}{\sqrt{\frac{p*(1-p)}{n} }}$, where $p_{0}$ is the proportion of nation returns.

Z calculated  = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.