Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming the sample is from a normal population.(a) 90 percent confidence. (Round your answers to 4 decimal places.)The 90% confidence interval is from to_____________.(b) 95 percent confidence. (Round your answers to 4 decimal places.)The 95% confidence interval is from to_____________.(c) 99 percent confidence. (Round your answers to 4 decimal places.)The 99% confidence interval is from to_____________.(d) Describe how the intervals change as you increase the confidence level.A. The interval gets narrower as the confidence level increases.B. The interval gets wider as the confidence level decreases.C. The interval gets wider as the confidence level increases.D. The interval stays the same as the confidence level increases

Answer & Step-by-step explanation:The confidence interval formula is: I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)] alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90% σ= standard deviation. In this case 5 mean= 37n= number of observations. In this case, 15(a)Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645Then, the confidence interval (90%): I 90%(μ)= 37+- [1.645*(5/sqrt(15))]I 90%(μ)= 37+- [2.1236]I 90%(μ)= [37-2.1236;37+2.1236] I 90%(μ)= [34.8764;39.1236] (b)Z(alpha/2)= Z(2.5%)= 1.96Then, the confidence interval (90%): I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]I 95%(μ)= 37+- [2.5303]I 95%(μ)= [37-2.5303;37+2.5303] I 95%(μ)= [34.4697;39.5203] (c)Z(alpha/2)= Z(0.5%)= 2.5758Then, the confidence interval (90%): I 99%(μ)= 37+- [2.5758*(5/sqrt(15)) I 99%(μ)= 37+- [3.3253]I 99%(μ)= [37-3.3253;37+3.3253] I 99%(μ)= [33.6747;39.3253] (d)C. The interval gets wider as the confidence level increases.