A researcher claims that the average age of a woman before she has her first child is greater than the 1990 mean age of 26.4 years, on the basis of data from National Vital Statistics Report, Vol. 48, No. 14. She wants to obtain a simple random sample of women who gave birth to their first child in 1999. Assuming that the standard deviation is 6.4, what sample size is required to get a margin of error of 0.5 years for a 95% confidence interval?

the sample size must be 629.

Step-by-step explanation:

In order to determine the sample size n, we must use the formula:

n = (z_ (α/2) * δ / ε) ²

where

n is the sample size z_ (α/2) is the z-value δ is the standard deviation ε is the margin of error

Step 1:

Determine the z-value:

Since this is at confidence level of 95%, therefore,

α = 1 - 95%

= 0.05

Thus,

z_ (α/2) = z_ (0.05/2) = z_ (0.025)

Looking at the table of standard normal probabilities, we find the z_ (0.025) = 1.96

Step 2:

Calculate n:

Therefore,

n = (z_ (α/2) * δ / ε) ²

n = ((1.96*6.4) / 0.5) ²

n = 629

Therefore, the sample size must be 629.