11.

A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that is $1.83. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. 18.99 18.14 20.05 16.78 16.21 21.26 17.29 20.17 o 21.72 19.42 15.89 21.07 19.48 18.62 16.22 21.03 - O The 90% confidence interval is ($D$D) i (Round to two decimal places as needed.) View an example | All parts showing X A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that is $2.50. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. 23.76 17.37 16.55 19.18 17.75 14.92 18.11 2251 o 20.87 18.26 18.93 18.52 16.32 18.77 20.04 22.46 (@GTEw) Find the left and right endpoints and form the confidence interval for the population mean p as shown below. Use the formula n and n is the sample size. g E=2z, T to find the margin of error, where z, is the z-score corresponding to an area of , is the population standard deviation Left endpoint: x-E Right endpoint: X+E While technology or the formula can be used to find the confidence intervals, this example uses technology. Determine the 90% confidence interval, rounding to two decimal places. (17.99,20.05) Now use technology to determine the 99% confidence interval, rounding to two decimal places. (17.41,20.63) View an example | Al parts showing X Arandom sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that o is $2.50. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. 23.76 17.37 16.55 19.18 17.75 14.92 18.11 22.51 m 20.87 18.26 18.93 18.52 16.32 18.77 20.04 22.46 Left endpoint; x-E Right endpoint: x+E While technology or the formula can be used to find the confidence intervals, this example uses technology. Determine the 90% confidence interval, rounding to two decimal places. (17.99,20.05) Now use technology to determine the 99% confidence interval, rounding to two decimal places. (17.41,20.63) When a large number of samples is collected and a % confidence interval is created for each sample, approximately c% of these intervals will contain the population mean p. Note that p is a fixed value predetermined by the population; it is either in the interval or not. Use this information to interpret the results. The 90% interval is (17.99,20.05). The 99% confidence interval is (17.41,20.63). Notice that the 99% confidence interval is wider. Determine the minimum sample size required when you want to be 95% confident that the sample mean is within one unit of the population mean and 6 = 12.6. Assume the population is normally distributed. A 95% confidence level requires a sample size of |:| (Round up to the nearest whole number as needed.) Question Viewer View an example | All parts showing X Determine the minimum sample size required when you want to be 95% confident that the sample mean is within one unit of the population mean and 6 = 21.4. Assume the population is normally distributed. The sample size required to estimate the population mean, with a level of confidence (1 - o) 100% with a specified margin z.0 of error, E, is given by n= ? Identify the margin of error and standard deviation. E =1 c =214 Because a 95% confidence interval is requested o =0.05. Use a standard normal table to find the critical value. Zy = Zpps5/2 = Zpo25 ~ 1.96 2 Z,0 Substitute the critical value found above, the margin of error, and the standard deviation into n = [T} . Since you cannot have View an example | All parts showing X Determine the minimum sample size required when you want to be 95% confident that the sample mean is within one unit of the population mean and =21.4. Assume the population is normally distributed. Co00 ) N o || Z0.05/2 = Zp.025 =~ 1.96 Z.0 C 2 Substitute the critical value found above, the margin of error, and the standard deviation into n = [ E ] . Since you cannot have a part of a person, round n up to the nearest integer. 2 z.C n= E 196+21.4)2 1 ~ 1760 Therefore, a 95% confidence level with a margin of error of one unit requires a sample size of 1760