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 99% confidence level. what does the confidence interval tell about the population of all college students in the state? 4.0, 3.1, 3.8, 4.5, 3.0, 4.4, 3.5, 4.6, 4.2, 4.1, 4.6, 3.9, 3.5, 4.0, 3.9

Given Information:

population size = N = 15

Confidence level = 90%

Required Information:

confidence interval = ?

confidence interval = (3.88, 4.26)

Step-by-step explanation:

The confidence interval can be found by

confidence interval = μ ± (tα/2)*ο/√N

Where μ is the mean of population, ο is standard deviation of population, N is the population size and tα/2 is the t score corresponding to

α = 100% - 99%

α = 1%

α = 0.01

degree of freedom = N - 1

degree of freedom = 15 - 1

degree of freedom = 14

From the t-table, the t-score for 99% confidence level and 14 DoF is

tα/2 = 2.977

The mean of the population is given by

Mean = μ = ∑(xi)/N

μ = (4.0+3.1+3.8+4.5+3.0+4.4+3.5+4.6+4.2+4.1+4.6+3.9+3.5+4.0+3.9)/15

μ = 61.05/15

μ = 4.07

The standard deviation of the population is given by

ο = √∑(xi - μ)²/(N - 1)

ο = 0.25

confidence interval = μ ± (tα/2)*ο/√N

confidence interval = 4.07 ± (2.977)*0.25/√15

confidence interval = 4.07 ± 0.19

upper limit = 4.07 + 0.19 = 4.26

lower limit = 4.07 - 0.19 = 3.88

confidence interval = (3.88, 4.26)

We are 90% confident that the student evaluation ratings of courses are from 3.88 to 4.26 that were obtained at one university in a state.

Please complete the question..

Step-by-step explanation: