Question

In a study of academic procrastination, the authors of a paper reported that for a sample of 451 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.74 hours and the standard deviation of study times was 3.10 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university.

Required:
Construct a 95% confidence interval to estimate μ, the mean time spent studying for the final exam for students taking introductory psychology at this university.

Answer 1

The 95% confidence interval to estimate μ is between 7.45 hours and 8.03 hours.

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 - 0.95)/(2) = 0.025`

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

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error 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.

`M = 1.96(3.1)/(√(451)) = 0.29`

The lower end of the interval is the sample mean subtracted by M. So it is 7.74 - 0.29 = 7.45 hours

The upper end of the interval is the sample mean added to M. So it is 7.74 + 0.29 = 8.03 hours

The 95% confidence interval to estimate μ is between 7.45 hours and 8.03 hours.