A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that sigma is ​$2.15. Construct the 90​% and 99​% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. 18.41 19.11 18.67 18.97 21.97 17.26 18.69 16.73 21.97 16.43 21.97 21.66 18.77 17.93 15.47 16.44 The​ 90% confidence interval is ______________

Answer:[ 17.8941, 19.6621]Step-by-step explanation:Given:widths : 18.41 19.11 18.67 18.97 21.97 17.26 18.69 16.73 21.97 16.43 21.97 21.66 18.77 17.93 15.47 16.44Total number of observations = 16Mean of the sample = [tex]\frac{\textup{Sum of observations}}{\textup{Total number of observations}}[/tex] = [tex]\frac{\textup{300.45}}{\textup{16}}[/tex] = 18.7781For 90% confidence level, z value = 1.645Therefore,Margin of error, E = [tex]z\frac{\sigma}{\sqrt n}[/tex]= [tex]1.645\times\frac{2.15}{\sqrt{16}}[/tex]= 0.884 thus,Confidence interval = Mean ± EorConfidence interval = [18.7781 - 0.884 , 18.7781 + 0.884  ]= [ 17.8941, 19.6621]