Question

Thirty-three college freshmen were randomly selected for an on-campus survey at their university. The participants' mean GPA was 2.5, and the standard deviation was 0.5. What is the margin of error, assuming a 95% confidence level? (Assume a normal distribution.)

Answer 1

Answer:
`\pm0.1706`

Explanation:

Given : Sample size : n= 33

Critical value for significance level of
`\alpha:0.05` :
`z_(\alpha/2)= 1.96`

Sample mean :
`\overline{x}=2.5`

Standard deviation :
`\sigma= 0.5`

We assume that this is a normal distribution.

Margin of error :
`E=\pm z_(\alpha/2)(\sigma)/(√(n))`

i.e.
`E=\pm (1.96)(0.5)/(√(33))=\pm0.170596102837\approx\pm0.1706`

Hence, the margin of error is
`\pm0.1706`