A provincial politician would like to determine the average amount earned during summer employment by the provinces teenagers during the past summer’s vacation period. She wants to have 95% confidence that the sample mean is within $25 of the actual population mean. Bases on past studies, she knows the amount earned is normally distributed with a standard deviation of $400. What sample size must she take?

Question

A provincial politician would like to determine the average amount earned during summer employment by the provinces teenagers during the past summer’s vacation period. She wants to have 95% confidence that the sample mean is within $25 of the actual population mean. Bases on past studies, she knows the amount earned is normally distributed with a standard deviation of $400. What sample size must she take?

Answer 1

95% = mean ± 1.96 SEm

SEm = SD/√n

25 = 1.96 * 400/√n

Solve for n.

Answer 2

To calculate the required sample size, we need to use the formula for the confidence interval for a mean. The formula is as follows:

n = (Z * σ / E)^2

Where:
n = required sample size
Z = Z-score for the desired level of confidence (in this case, 95% confidence corresponds to a Z-score of 1.96, assuming a standard normal distribution)
σ = population standard deviation ($400 in this case)
E = margin of error (in this case, $25)

Plugging the values into the formula:

n = (1.96 * 400 / 25)^2

n ≈ 6,272.26

Since the sample size needs to be a whole number, you should round up the result to the nearest whole number. Therefore, the minimum sample size required is 6,273 (rounded up to the nearest whole number).

So, the politician should take a sample size of at least 6,273 in order to have a 95% confidence that the sample mean is within $25 of the actual population mean.