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Given Information:population size = N = 15Confidence level = 90%Required Information:confidence interval = ?Answer:confidence interval = (3.88, 4.26)Step-by-step explanation:The confidence interval can be found byconfidence interval = μ ± (tα/2)*ο/√NWhere 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 toα = 100% - 99%α = 1%α = 0.01degree of freedom = N - 1degree of freedom = 15 - 1degree of freedom = 14From the t-table, the t-score for 99% confidence level and 14 DoF istα/2 = 2.977The mean of the population is given byMean = μ = ∑(xi)/Nμ = (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)/15μ = 61.05/15μ = 4.07The standard deviation of the population is given byο = √∑(xi - μ)²/(N - 1)ο = 0.25confidence interval = μ ± (tα/2)*ο/√Nconfidence interval = 4.07 ± (2.977)*0.25/√15confidence interval = 4.07 ± 0.19upper limit = 4.07 + 0.19 = 4.26lower limit = 4.07 - 0.19 = 3.88confidence 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....

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