Respuesta :
Answer:
The critical value used is [tex]T_c = 2.132[/tex]
The 90% confidence interval for the average net change in a student's score after completing the course is (14.357, 22.843).
Step-by-step explanation:
Before building the confidence interval, we need to find the sample mean and the sample standard deviation.
Sample mean:
[tex]\overline{x} = \frac{16+21+22+12+22}{5} = 18.6[/tex]
Sample standard deviation:
[tex]s = \sqrt{\frac{(16-18.6)^2+(21-18.6)^2+(22-18.6)^2+(12-18.6)^2+(22-18.6)^2}{4}} = 4.45[/tex]
Confidence interval:
We have the standard deviation for the sample, so the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So
df = 5 - 1 = 4
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 2.132. The critical value used is [tex]T_c = 2.132[/tex]
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.132\frac{4.45}{\sqrt{5}} = 4.243[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 18.6 - 4.243 = 14.357
The upper end of the interval is the sample mean added to M. So it is 18.6 + 4.243 = 22.843.
The 90% confidence interval for the average net change in a student's score after completing the course is (14.357, 22.843).
