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.
