Question

A large on-demand video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot survey of 10 executives indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. The estimate of the mean viewing time should be within one-quarter hour. The 95% level of confidence is to be used. How many executives should be surveyed?

Answer 1

Answer:

$n=(\frac{2.262(3)}{0.25})^2 =736.79 \approx 737$

So the answer for this case would be n=737 rounded up to the nearest integer

Step-by-step explanation:

We have the following info given:

$\bar X = 12$ the sample mean

$s = 3$ the sample deviation

$n = 10$ represent the sample size

The degrees of freedom are given by:

$df =n-1 =10-1=9$

The margin of error is given by this formula:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$    (a)

And on this case we have that ME =0.25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{t_{\alpha/2} s}{ME})^2$   (5)

The confidence level is 95% and the critical value for this case would be given by $t_{\alpha/2}=2.262$, replacing into formula (5) we got:

$n=(\frac{2.262(3)}{0.25})^2 =736.79 \approx 737$

So the answer for this case would be n=737 rounded up to the nearest integer