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: $\pm0.1706$

Step-by-step 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 : $E=\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $E=\pm (1.96)\dfrac{0.5}{\sqrt{33}}=\pm0.170596102837\approx\pm0.1706$

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