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 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

Answer 1

Answer:

Step-by-step explanation:

Dont know

Answer 2

Given Information:

population size = N = 15

Confidence level = 90%

Required Information:

confidence interval = ?

Answer:

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.