The ages of a group of 142 randomly selected adult females have a standard deviation of 18.1 years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so let σ=18.1 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 99% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is

We want the sample mean to be within one-half year of the population mean, so we set x-µ=0.5. We are looking for a 99% confidence interval, so we set Z=2.7578. We are told to use σ=18.1. Plugging those values into the formula, we get:

2.5758 = 0.5(18.1/√n)

We can rearrange to solve for n:

((2.5758-18.1)/0.5)2 = n

Plugging that into our calculator, we get n = 964.003. Since we can't have a fraction of a person in our sample, it would be safest to round up to n=965. (But since .003 is so small, I'd also accept 964 as an answer.)Step-by-step explanation:

AFOKE88 AFOKE88 · Answer 2 · 2021-12-14T09:45:29+01:00

The required sample size for the given population distribution is; n = 8696 female ages

We are given;

Standard deviation; σ = 18.1

Confidence level; CL = 99%

Now, formula to find the margin of error is;

E = z(σ/√n)

Where;

E is margin of error

z is critical value at confidence level

σ is standard deviation

n is required sample size

Now we are told that the sample mean is within one-half year of the population mean.

Thus;

E = 0.5

z value at 99% Confidence level is;

z = 2.576

Thus, Making n the subject of the formula is;

n = (zσ/E)²

n = (2.576 × 18.1/0.5)²

n = 8695.78

Approximating to a whole number gives;

n = 8696 female ages

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