Question

(1 point) A random sample of n measurements was selected from a population with unknown mean u and standard deviation o. Calculate a 99% confidence interval for u for each of the following situations: (a) n= 105, X = 25.8, s = = 2.92 sus (b) n = 95, x = 96, s = 4.13 su < (c) n = 85, X = 63.6, s = 2.31 su < (d) n = 115, x = 102.1, s = 2.95 su < Nate Varaonorn nortial aredit on this problem

Answer 1

Solution:

1)

The provided sample mean is X = 25.8 and the sample standard deviation is s = = 2.92. The size of the sample is n = 105 and the required confidence level is 99%. The number of degrees of freedom are df = 105 – 1= 104, and the significance level is a 0.01. = = 0.01 and df = 104 Based on the provided information, the critical t-value for a degrees of freedom is tc = 2.624. The 99% confidence for the population mean y is computed using the following expression CI= Ñ _ tc Xs „x + te X s ✓n vn Therefore, based on the information provided, the 99 % confidence for the population mean u is CI= 25.8 2.624 x 2.92 , 25.8 + 2.624 x 2.92 V105 V105 = (25.8 – 0.748, 25.8 + 0.748) = (25.052, 26.548)

2)

The provided sample mean is à = 96 and the sample standard deviation is s = - 4.13. The size of the sample is n = 95 and the required confidence level is 99%. The number of degrees of freedom are df = 95 – 1 = 94, and the significance level is 0.01. a = 0.01 and df = 94 Based on the provided information, the critical t-value for a = degrees of freedom is to 2.629. The 99% confidence for the population mean u is computed using the following expression CI = à te xs vn à + te xs vn Therefore, based on the information provided, the 99 % confidence for the population mean u is 2.629 x 4.13 CI= 96 2.629 x 4.13 95 96 + 2 V95 = - (96 – 1.114, 96 + 1.114) = (94.886, 97.114)

3)

2.31. The provided sample mean is 7 = 63.6 and the sample standard deviation is s The size of the sample is n = 85 and the required confidence level is 99%. The number of degrees of freedom are df = 85 – 1= 84, and the significance level is 0.01. a = 0.01 and df = 84 Based on the provided information, the critical t-value for a = degrees of freedom is te = 2.636. The 99% confidence for the population mean u is computed using the following expression CI= X te XS vn X + te Xs vn terminal Therefore, based on the information provided, the 99 % confidence for the population mean μis 2.636 x 2.31 CI = 63.6 63.6 + 2.636 x 2.31 785 785 (63.6 – 0.66, 63.6 + 0.66) = = (62.94, 64.26)

4)

The provided sample mean is X = 102.1 and the sample standard deviation is s = 2.95. The size of the sample is n = = 115 and the required confidence level is 99%. 115 – 1= 114, and the significance level The number of degrees of freedom are df is a = 0.01. 0.01 and df = 114 Based on the provided information, the critical t-value for a = degrees of freedom is tc = 2.62. The 99% confidence for the population mean u is computed using the following expression CI = ( Ž (x te xs vn X + te xs vn 9 Therefore, based on the information provided, the 99 % confidence for the population mean u is CI = 102.1 2.62 x 2.95 V115 102.1 + 2.62 x 2.95 V115 - (102.1 – 0.721, 102.1 +0.721) (101.379, 102.821)