It is known that the population variance equals 484.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is
A) 190.
B) 74.
C) 189.
D) 75.

Question

Quizplus · Accepted Answer

We can use the formula for the margin of error:

Margin of Error = z*(sigma/sqrt(n))

where z is the z-score for the desired level of confidence (0.95), sigma is the population standard deviation (sqrt(484) = 22), and n is the sample size.

We want the margin of error to be 5 or less, so we can set up the inequality:

5 >= z*(22/sqrt(n))

Solving for n:

n >= (z*22/5)^2

Using the z-score corresponding to a 95% confidence level, which is about 1.96:

n >= (1.96*22/5)^2

n >= 74.05

We need to round up to the nearest whole number, so the answer is 75. Therefore, the best choice is D.

It Is Known That the Population Variance Equals 484