Respuesta :
Answer:
The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).
Step-by-step explanation:
The net change in 7 students' scores on the exam after completing the course are:
S = {37 ,12 ,12 ,17 ,13 ,32 ,23}
Compute the sample mean and sample standard deviation as follows:
[tex]\bar x=\frac{1}{n}\sum x=\frac{1}{7}\times 146=20.857\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}}=\sqrt{\frac{1}{7}\times 622.8571}=10.189[/tex]
As the population standard deviation is not known, a t-interval will be formed.
Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:
[tex]t_{\alpha/2, (n-1)}=t_{0.20/2, (7-1)}=t_{0.10,6}=1.415[/tex]
*Use a t-table.
Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times\frac{s}{\sqrt{n}}[/tex]
[tex]=20.857\pm 1.415\times\frac{10.189}{\sqrt{7}}\\\\ =20.857\pm 5.4493\\\\=(15.4077, 26.3063)\\\\\approx (15.4,26.3)[/tex]
Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).
