Question

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a population standard deviation of three hours, what is the required sample size if the error should be less than a half hour with a 99% level of confidence? Select one: a. 196 b. 239 c. 15 d. 16

Answer 1

b. 239

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.05 = 0.995`, so
`z = 2.575`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

`M = 0.5, \sigma = 3`

`M = z*(\sigma)/(√(n))`

`0.5 = 2.575*(3)/(√(n))`

`0.5√(n) = 7.725`

`√(n) = 15.45`

`n = 238.7`

So a sample of at least 239 is required.

The correct answer is:

b. 239