Question

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 98​% confidence level. What does the confidence interval tell about the population of all college students in the​ state?

3.8​, 3.1​, 3.7​, 4.9​, 2.9​, 4.1​, 3.7​, 4.3​, 4.7​, 4.3​, 4.2​, 3.9​, 3.0​, 4.0​, 3.6

Answer 1

Solution:

We are given a data of sample size n=15.

3.8​, 3.1​, 3.7​, 4.9​, 2.9​, 4.1​, 3.7​, 4.3​, 4.7​, 4.3​, 4.2​, 3.9​, 3.0​, 4.0​, 3.6

Using this, first we find sample mean() and sample standard deviation(s).

=   

= 3.88

Now ,

s=   

Using given data, find Xi- for each term.take squre for each.then we can easily find s.

s= 4.724

Point estimate = sample mean = = 3.88

sample standard deviation = s = 4.72

sample size = n = 15

Degrees of freedom = df = n - 1 = 14

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,14 = 2.624

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

= 2.624 * (4.72 / 15)

= 3.198

The 98% confidence interval estimate of the population mean is,

- E < < + E

3.88 - 3.198 < < 3.88 + 3.198

0.682 < < 7.078

(0.682 , 7.078)

Therefore, the 98% of the population of all college students in the​ state is between 0.682 and 7.078.