Respuesta :
Answer:
a) [tex] \hat p = \frac{X}{n}= \frac{12}{80}= 0.15[/tex]
b) [tex]0.15 - 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.072[/tex]
[tex]0.15 + 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.228[/tex]
And the 95% confidence interval would be given (0.072;0.228).
c) For this case since the confidence interval not contains the value 0.06 or 6% and since the lower limit for the confidence interval is higher than 0.06 (0.072>0.06), we have enough statistical evidence to support the conclusion that the true proportion of items returned is higher than 0.06 or 6% at a significance of 5%.
Step-by-step explanation:
Assumign the following question for the problem:
a. Construct a point estimate of the proportion of items returned for the population of sales transactions at the Houston store.
For this case the best estimate for the true proportion is given by the sample proportion:
[tex] \hat p = \frac{X}{n}= \frac{12}{80}= 0.15[/tex]
b. Construct a 95% confidence interval for the porportion of returns at the Houston store.
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.15 - 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.072[/tex]
[tex]0.15 + 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.228[/tex]
And the 95% confidence interval would be given (0.072;0.228).
c. Is the proportion of returns at the Houston store significantly different from the returns for the nation as a whole? Provide statistical support for your answer.
For this case since the confidence interval not contains the value 0.06 or 6% and since the lower limit for the confidence interval is higher than 0.06 (0.072>0.06), we have enough statistical evidence to support the conclusion that the true proportion of items returned is higher than 0.06 or 6% at a significance of 5%.
Using the sample proportion, it is found that the point estimate is of 0.15 = 15%.
What is a sample proportion?
A sample proportion is given by the number of desired outcomes divided by the number of total outcomes. It can also serve as the point estimate for the population proportion.
In this problem, 12 out of the 80 items sold were returned, hence:
12/80 = 0.15 = 15%.
The point estimate is of 0.15 = 15%.
More can be learned about point estimates at https://brainly.com/question/24651197
