91影视

Data for the 38 samples given in the following table

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

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

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

Let the confidence level be 0.95.

$\begin{array}{c}\left(1鈭�伪\right)=0.95\\ 伪=0.05\\ \frac{伪}{2}=0.025\end{array}$

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:

$\begin{array}{c}\stackrel{炉}{x}卤z_{\frac{伪}{2}}\left({蟽}_{\stackrel{炉}{x}}\right)=6.079卤1.96\left(\frac{2.755}{\sqrt{38}}\right)\\ =6.079卤1.96\left(0.4469\right)\\ =6.079卤0.8759\end{array}$

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

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

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.

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).