Respuesta :
Answer:
The critical value is [tex]T_c = 2.5706[/tex].
The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).
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{6+16+19+12+15+14}{6} = 13.67[/tex]
Sample standard deviation:
[tex]s = \sqrt{\frac{(6-13.67)^2+(16-13.67)^2+(19-13.67)^2+(12-13.67)^2+(15-13.67)^2+(14-13.67)^2}{5}} = 4.4121[/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, we have. This is the sample size subtracted by 1. So
df = 6 - 1 = 5
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 2.5706, that is, the critical value is [tex]T_c = 2.5706[/tex]
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.5706\frac{4.4121}{\sqrt{6}} = 4.63[/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 13.67 - 4.63 = 9.04.
The upper end of the interval is the sample mean added to M. So it is 13.67 + 4.63 = 18.30.
The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).
