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*[tex]\sqrt{\frac{p*(1-p)}{n} } = 0.15 \pm 1.96 *\sqrt{\frac{0.15*(1-0.15)}{80} }[/tex],
CI(95%) = [0.0718 < p < 0.2282]
c) The hypothesis:
[tex]H_{0}[/tex]: The proportion of returns at the Houston store is not significantly different from the returns for the nation as a whole.
[tex]H_{a}[/tex]: The proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.
The test statistics:
Z = [tex]\frac{\hat{p} - p_{0}}{\sqrt{\frac{p*(1-p)}{n} }}[/tex], where [tex]p_{0}[/tex] 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.
