Question

The following measurements are the net change in 66 students' scores on the exm after completing the course:

6,16,19,12,15,146,16,19,12,15,14

Using these data, construct a 90% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal.

Step 1 of 4 :  

Calculate the sample mean for the given sample data. Round your answer to one decimal place.

Please show work for the whole problem not just step one

Answer 1

Solution :

Given that,

x	x2
6	36
16	256
19	361
12	144
15	225
146	21316
16	256
19	361
12	144
15	225
14	196
∑x=290	∑x2=23520

Mean ˉx=∑xn

=6+16+19+12+15+146+16+19+12+15+14/11

=290/11

=26.3636

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√23520-(290)211/10

=√23520-7645.4545/10

=√15874.5455/10

=√1587.4545

=39.8429

Degrees of freedom = df = n - 1 = 11 - 1 = 10

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,10 =2.797

Margin of error = E = t/2,df * (s /n)

= 2.797 * (39.8 / 11 )

= 21.7

Margin of error =21.7

The 90% confidence interval estimate of the population mean is,

- E < < + E

26.4 - 21.7 < < 26.4 + 21.7

4.7 < < 48.1

(4.7, 48.1 )