A pilot study has revealed a sample standard deviation of 16 for age of people in the work force. How large a sample must be chose to obtain an estimator of the mean that, with 95% confidence, will be correct to within 2 years?

Formula:

n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent 95% confidence, sd = 16, E = 2, ^2 means squared, and * means to multiply.

Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.

I hope this will help get you started.

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 5%? A previous study indicates that the proportion of left-handed golfers is 8%

To determine the required sample size, we can use the formula for the confidence interval:

Sample size (n) = (Z * σ / E)²

Where:
n = required sample size
Z = Z-score corresponding to the desired level of confidence
σ = population standard deviation
E = maximum error allowed in the estimate (margin of error)

In this case, we want a 95% confidence level, which corresponds to a Z-score of approximately 1.96 (from standard normal distribution table). The population standard deviation (σ) is given as 16, and we want the estimate to be within 2 years of the true mean (E = 2).

Plugging these values into the formula:

n = (1.96 * 16 / 2)²
n = 30.06²
n ≈ 904.81

Therefore, you should choose a sample size of at least 905 to obtain an estimator of the mean that, with 95% confidence, will be correct to within 2 years.