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 13 hours, with a standard deviation of 2.5 hours. It is desired to estimate the mean viewing time within two-quarter hour. The 98 percent level of confidence is to be used.
How many executives should be surveyed? (Round up your answer to the next whole number.)
Number of executives
To determine the number of executives that should be surveyed, we can use the formula for sample size calculation in a confidence interval for the population mean:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score for the desired level of confidence (98% level of confidence corresponds to a Z-score of 2.33)
σ = standard deviation of the population (given as 2.5 hours)
E = margin of error (two-quarter hour, or 0.5 hours)
Plugging in the values into the formula:
n = (2.33 * 2.5 / 0.5)^2
n = (5.825)^2
n ≈ 33.9
Since we need to round up to the next whole number, the number of executives that should be surveyed is approximately 34.
To determine the number of executives that should be surveyed, we need to use the formula for sample size calculation in estimating a population mean with a specified level of confidence.
The formula for calculating the sample size for estimating the population mean is:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level
σ = standard deviation of the population
E = margin of error
In this case, the desired confidence level is 98%, which corresponds to a Z-score of 2.33 (since the Z-score corresponds to the area under the normal distribution curve).
The margin of error (E) is given as two-quarter hour, which is equivalent to 0.5 hours.
Substituting the values into the formula, we get:
n = (2.33 * 2.5 / 0.5)^2
n = (5.825)^2
n ≈ 33.93
Since the sample size must be a whole number, we need to round up to the nearest whole number. So, the number of executives that should be surveyed is 34.