A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 14 hours, with a standard deviation of 2.0 hours. It is desired to estimate the mean viewing time within 30 minutes. The 98% level of confidence is to be used.


How many executives should be surveyed? (Round your z-score to 2 decimal places and round your final answer to the next whole number.)

Number of executives

play around with Z table stuff here:

http://davidmlane.com/hyperstat/z_table.html

To determine the number of executives that should be surveyed to estimate the mean viewing time within 30 minutes with a 98% level of confidence, we can use the formula for sample size calculation:

n = (Z * σ / E)^2

Where:
n = sample size
Z = z-score corresponding to the desired level of confidence (in this case, 98%)
σ = standard deviation of the population
E = maximum error margin (in this case, 30 minutes)

Given:
σ = 2.0 hours
E = 0.5 hours (30 minutes)
Z = the z-score corresponding to a 98% level of confidence (or 2.33 approximately; you can find this value in a standard normal distribution table or use statistical software)

Plugging the values into the formula:

n = (2.33 * 2.0 / 0.5)^2
n = (4.66 / 0.5)^2
n = 9.32^2
n ≈ 86.65

Rounding the answer to the next whole number, the number of executives that should be surveyed is approximately 87.