Question

An educational research group wants to know how many hours college students spend studying outside of class per week. If they survey 100 students and find an average 10.5 hours of studying a week, with a standard deviation of 2.25, find a 98% confidence interval for the true average number of hours spent studying.

Answer choices:

10.5±.324

10.5±.225

None of these

10.5±.482

10.5±.524

Answer 1

Solution:
Given in the question
Sample of students (n) = 100
Sample mean (Xbar) = 10.5
Sample standard deviation(S) = 2.25
We need to construct 98% confidence interval which can be calculated as
Xbar +/- Zalpha/2 * S/sqrt(n)
Here Confidence level = 0.98
level of significance($\alpha$) = 1 - Confidence level = 1 - 0.98 = 0.02
$\alpha$ /2 = 0.02/2 = 0.01
Here we will use Z test as sample size is large enough
Z$\alpha$/2 = 2.33
So confidence interval is
10.5 +/- 2.33*2.25/sqrt(100)
10.5 +/- 0.524
So its correct answer is E i.e. 10.5 +/- 0.524