Question

Shopping on Black Friday. The day after Thanksgiving— called Black Friday—is one of the largest shopping days in the United States. Winthrop University researchers conducted interviews with a sample of 38 women shopping on Black Friday to gauge their shopping habits and reported the results in the International Journal of Retail and Distribution Management (Vol. 39, 2011). One question was, “How many hours do you usually spend shopping on Black Friday?” Data for the 38 shoppers are listed in the accompanying table.

a. Describe the population of interest to the researchers.

b. What is the quantitative variable of interest to the researchers?

c. Use the information in the table to estimate the population mean number of hours spent shopping on Black Friday with a 95% confidence interval.

d. Give a practical interpretation of the interval.

e. A retail store advertises that the true mean number of hours spent shopping on Black Friday is 5.5 hours. Can the store be sued for false advertising? Explain.

Short answer

a. The population of interest to the researchers is all United States women who shop on Black Friday.

b. The quantitative variable of interest to the researchers is the number of hours spends shopping on Black Friday.

c. The 95% confidence interval for the population mean is (5.203 hours,6.955 hours)

d. For shopping on Black Friday, there is 95% confidence that the true population mean number of hours lies between 5.203 hours and 6.955 hours.

e. It can be concluded that the true mean 5.5 hours is reasonable value for population mean

Step-by-step solution

Given information

Data for the 38 samples given in the following table

Population of interest to researchers

The population of interest to the researchers is all United States women who shop on Black Friday.

Quantitative variable of interest to the researchers

The quantitative variable of interest to the researchers is the number of hours spends shopping on Black Friday.

Calculation of confidence interval

Since, mean of the given 38 samples is 6.079 and standard deviation is 2.755.

Let the confidence level be 0.95.

$\begin{array}{c}\left(1-\alpha \right)=0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=0.025\end{array}$

From table, the required$z_{0.025}$ value for 95% confidence level is 1.96. Thus, $z_{0.025}=1.96$

The 95% confidence interval is obtained as shown below:

$\begin{array}{c}\overline{x}\pm z_{\frac{\alpha }{2}}\left({\sigma }_{\overline{x}}\right)=6.079\pm 1.96\left(\frac{2.755}{\sqrt{38}}\right)\\ =6.079\pm 1.96\left(0.4469\right)\\ =6.079\pm 0.8759\end{array}$

.$\left(6.079-0.8759,6.079+0.8759\right)=\left(5.203,6.955\right)$

Thus, the 95% confidence interval for the population mean is (5.203 hours, 6.955 hours).

Interpretation for confidence interval

For shopping on Black Friday, there is 95% confidence that the true population mean number of hours lies between 5.203 hours and 6.955 hours.

Explanation

No, the store cannot be used for false advertising because the 95% confidence interval contains the population mean as 5.5 hours. Thus, it can be concluded that the true mean 5.5 hours is reasonable value for population mean. That is, the population mean lies inside the confidence interval (5.203, 6.955).