Answer:
c. 0.62%
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 30 - 1 = 29
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 29degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.6991
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}}[/tex]
Sample of 30, standard deviation of 2%.
Then
[tex]M = T\frac{s}{\sqrt{n}} = 1.6991\frac{2}{\sqrt{30}} = 0.62[/tex]
0.62%, so option c.
