In: Statistics and Probability
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
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.
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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 |