Question

In: Statistics and Probability

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent."...

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 95​% confidence level. What does the confidence interval tell about the population of all college students in the​ state? 4.0​, 3.0​, 4.0​, 4.8​, 3.1​, 4.5​, 3.3​, 4.9​, 4.5​, 4.4​, 4.3​, 3.8​, 3.4​, 4.0​, 3.9

Solutions

Expert Solution

Given : n = 15, = 0.05

From the data: = 3.99, s = 0.5922

Since population standard deviation is unknown, the tcritical (2 tail) for = 0.05, for df = n -1 = 14, is 2.145

The Confidence Interval is given by ME, where

The Lower Limit = 3.99 - 0.33 = 3.66

The Upper Limit = 3.99 + 0.33 = 4.32

The 95% Confidence Interval is (3.66 , 4.32)

From the Interval we can be 95% confident that the population of students have a true evaluation mean score which is between the limits 3.66 to 4.32.

____________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.

# X Mean (x - mean)2
1 4 3.99 0.00
2 3 3.99 0.98
3 4 3.99 0.00
4 4.8 3.99 0.66
5 3.1 3.99 0.79
6 4.5 3.99 0.26
7 3.3 3.99 0.48
8 4.9 3.99 0.83
9 4.5 3.99 0.26
10 4.4 3.99 0.17
11 4.3 3.99 0.10
12 3.8 3.99 0.04
13 3.4 3.99 0.35
14 4 3.99 0.00
15 3.9 3.99 0.01
n 15
Sum 59.9
Average 3.99
SS 4.9095
Variance = SS/n-1 0.350678571
Std Dev 0.5922

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