It is known that the population variance (σ²) is 125. At 95% confidence, what sample size should be taken so that the margin of error does not exceed 3? A) 52 B) 53 C) 54 D) 55

Question

Quizplus · Accepted Answer

We can use the formula for margin of error:

ME = z* (σ / sqrt(n))

where ME is the margin of error, z* is the critical value for the confidence level (in this case, 95% confidence corresponds to z*=1.96), σ is the population standard deviation (which we don't know, but we have the population variance, so we can take the square root to get σ=√125=11.18), and n is the sample size we want to determine.

We want the margin of error to be at most 3, so we can set up the following inequality:

3 ≤ 1.96* (11.18 / sqrt(n))

Solving for n, we get:

n ≥ (1.96*11.18 / 3)² = 53.79

Since we need a whole number for the sample size, we round up to the nearest integer and get n=54.

Therefore, the best choice is C) 54 because it is the smallest sample size that satisfies the requirements.

It Is Known That the Population Variance (σ2) Is 125

It Is Known That the Population Variance (σ²) Is 125