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 95​% confidence level. What does the confidence interval tell about the population of all college students in the​ state?
3.63.6​,


2.92.9​,


3.83.8​,


4.54.5​,


3.23.2​,


3.93.9​,


3.33.3​,


4.64.6​,


4.14.1​,


4.34.3​,


4.44.4​,


3.93.9​,


3.23.2​,


4.24.2​,


3.83.8

Answer 1

Answer:

3.573 to 4.127

Step-by-step explanation:

Given

Sample size = 15

Mean = Sum of ratings/ sample size

Mean = 57.7/15

Mean = 3.85

Degree of freedom = sample size - 1

Degree of freedom = 15 - 1 = 14

df = 14

Then we calculate the standard deviation

(x - mean)² ||

(3.6 - 3.85)² || 0.0625

(2.9 - 3.85)² || 0.9025

(3.8 - 3.85)² || 0.0025

(4.5 - 3.85)² || 0.4225

(3.2 - 3.85)² || 0.4225

( 3.9 - 3.85)² || 0.0025

( 3.3 - 3.85)² || 0.3025

( 4.6 - 3.85)² || 0.5625

(4.1 - 3.85)² || 0.0625

(4.3 - 3.85)² || 0.2025

4.4 - 3.85)² || 0.3025

( 3.9 - 3.85)² || 0.0025

(3.2 - 3.85)² || 0.4225

( 4.2 - 3.85)² || 0.1225

( 3.8 - 3.85)² || 0.0025

Total || 3.7975

Variance = 3.7975/15 = 0.253167

Standard Deviation = √0.253167 = 0.50315703314174194

Standard Deviation = 0.5 ------- Approximated

The next step is to subtract the confidence level from 1, then divide by two.

i.e (1 - 0.95)/2 = 0.025

α = 0.025

Then we look up this answer to step in the t-distribution table.

For 14 degrees of freedom (df) and α = 0.025, my result is 2.145

The next step is to divide the sample standard deviation by the square root of the sample size.

0.5 / √15 = 0.129

Next is to multiply this result by step 2.145 (from the t table)

0.129 × 2.45 = 0.277

For the lower end of the range, subtract 0.277 from the sample mean.

3.85 – 0.277 = 3.573

Step 7: For the upper end of the range, add step 0.277 to the sample mean.

3.85 + 0.277 = 4.127