Question

One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.

Construct a 99% confidence interval for the population mean hours spent watching television per month.

Fill in the blank: Round to two decimal places. ( , )

Answer 1

Answer: (143.07, 158.93)

Step-by-step explanation:

The formula to find the confidence interval is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

where n= sample size

$\overline{x}$ = Sample mean

z* = critical z-value (two tailed).                

$\sigma$ = Population standard deviation

We assume that the underlying population distribution is normal.

As per given , we have

n= 108

$\overline{x}=151$

$\sigma=32$

Critical value for 99% confidence level = 2.576  (By using z-table)

Then , the 99% confidence interval for the population mean hours spent watching television per month :-

$151\pm (2.576)\dfrac{32}{\sqrt{108}}$

$151\pm (2.576)\dfrac{32}{10.3923048454}$

$151\pm (2.576)(3.07920143568)$

$151\pm (7.93202289831)\approx151\pm7.93\\\\=(151-7.93,\ 151+7.93)\\\\=(143.07,\ 158.93 )$

Hence, the required 99% confidence interval for the population mean hours spent watching television per month. = (143.07, 158.93)