Question

simple random sample of size n is drawn from a population that is normally distributed. The sample mean, X. is found to be 111, and the sample standard deviation is found to be 10. a) Construct a 95% confidence interval about if the sample size, n, is 28. b) Construct a 95% confidence interval about if the sample size, n, is 11 c) Construct a 90% confidence interval about if the sample size, n, is 28 ) Could we have computed the confidence intervals in parts (a) (c) if the population had not been normally distributed? = Click the icon to view the table of areas under the t-distribution a) Construct a 95% confidence interval about if the sample size, n, is 28 ower bound Upper bound: Jse ascending order. Round to one decimal place as needed) ) Construct a 95% confidence interval about if the sample size, n, is 11 ower bound Upper bound: Jse ascending order. Round to one decimal place as needed.) ow does decreasing the sample size affect the margin of error, E? A. As the sample size decreases, the margin of error increases B. As the sample size decreases, the margin of error decreases Ae the camels era derrasee the mornin forror stove the camo ck to select your answer(s)

Answer 1

a)

The provided sample mean is X = 111 and the sample standard deviation is s = 10. The size of the sample is n = 28 and the required confidence level is 95%. The number of degrees of freedom are df = 28 -1 = 27, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 27 degrees of freedom is t = 2.052. The 95% confidence for the population mean y is computed using the following expression C1 = (x * x + tema) Therefore, based on the information provided, the 95% confidence for the population mean u is 2.052 x 10 112 2.052 x 10 ,111+ 28 28 = (111 - 3.878, 111 +3.878) = (107.122, 114.878)

b)

The provided sample mean is X = 111 and the sample standard deviation is s= 10. The size of the sample is n = 11 and the required confidence level is 95%. The number of degrees of freedom are df = 11-1= 10, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 10 degrees of freedom is t = 2.228. The 95% confidence for the population mean y is computed using the following expression CI = (- x + y ) Therefore, based on the information provided, the 95% confidence for the population mean u is CI= (111 - 2.225.110,111 + 2-V11") = (111 – 6.718, 111 +6.718) = (104.282, 117.718)

As sample size increases, margin of error decreases

c)

The provided sample mean is X = 111 and the sample standard deviation is s = 10. The size of the sample is n = 28 and the required confidence level is 90%. The number of degrees of freedom are df = 28 -1 = 27, and the significance level is a = 0.1. Based on the provided information, the critical t-value for a = 0.1 and df = 27 degrees of freedom is t = 1.703. The 90% confidence for the population mean u is computed using the following expression CI= (9 + ta Therefore, based on the information provided, the 90 % confidence for the population mean u is CI = (111 ---->28",111 + CL_( 1.703 x 10 11 1.703 x 10 V28) = (111 - 3.219, 111 +3.219) = (107.781, 114.219)

d)

No, the population needs to be normally distributed.