Nielsen Media Research wants to estimate the mean amount of time (in min) that full-time college students spend watching television each weekday. Find the sample size necessary to estimate that mean with a 15-minmargin of error. Assume that a 96% confidence level is desired. Also assume that a pilot study showed that the standard deviation is estimated to be 112.2 min.
To find the sample size necessary to estimate the mean with a 15-minute margin of error, we can use the formula:
n = (Z * σ / E)^2
Where:
n = Sample size
Z = Z-score corresponding to the desired confidence level
σ = Standard deviation
E = Margin of error
In this case, we want a 96% confidence level, so the Z-score is determined from the standard normal distribution table. For a 96% confidence level, the Z-score is approximately 2.05.
Plugging in the values:
Z = 2.05
σ = 112.2 min
E = 15 min
n = (2.05 * 112.2 / 15)^2
n ≈ (230.01 / 15)^2
n ≈ 15.34^2
n ≈ 235.43
Therefore, the sample size necessary to estimate the mean with a 15-minute margin of error and a 96% confidence level is approximately 236.