Question

In a study of academic procrastination, researchers reported that for a random sample of 41 undergraduate students preparing for a psychology exam, the mean time spent studying was 11.9 hours with a standard deviation of 4.5 hours. Compute a 95% confidence interval for μ, the mean time spent studying for the exam among all students taking this course.

Answer 1

The 95% confidence interval is
`10.5 < \mu <13.3`

Explanation:

From the question we are told that

The sample size is
`n = 41`

The sample mean is
`\= x = 11.9 \ hr`

The standard deviation is
`\sigma = 4.5`

For a 95% confidence interval the confidence level is 95%

Given that the confidence level is 95% then the level of significance can be mathematically represented as

`\alpha = 100 - 95`

`\alpha = 5 \%`

`\alpha = 0.05`

Next we obtain the critical values of
`(\alpha )/(2)` from the normal distribution table

The values is

`Z_{(\alpha )/(2) } = 1.96`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } * (\sigma )/( √(n) )`

substituting values

`E = 1.96 * ( 4.5 )/( √(41) )`

`E = 1.377`

The 95% confidence interval is mathematically represented as

`\= x - E < \mu < \= x - E`

substituting values

`11.9 - 1.377 < \mu <11.9 + 1.377`

`10.5 < \mu <13.3`