Question

Exercise 3. An automotive supplier of interior parts places several electrical wires in a harness. A pull test measures the force required to pull spliced wires apart. A customer requires that each wire spliced into the harness must with-stand a pull force of 20 pounds. Let X equal the pull force required to pull 20 gauge wires apart. Assume that X ~ N(u, o). The following data give 20 observations of X: 28.8 20.8 24.4 30.1 25.6 26.4 27.7 24.4 25.1 24.6 23.9 22.1 22.5 27.6 28.1 26.3 28.2 22.2 26.3 24.4 @@ (i) Find point estimates for u and o. (ii) Find a 99% one-sided confidence interval for u that provides a lower bound for u. That is, first find a constant c such that p(x-con=4) 2099. Then obtain the desired one-sided confidence interval for u by computing ã and s from the given sample.

Answer 1

The computation table: 25.6 22.1 2625 20.8 24.4 X x- 28.8 3.325 11.05563 24.4 - 1.075 1.155625 30.1 4.625 21.39063 0.125 0.015625 26.4 0.925 0.855625 23.9 -1.575 2.480625 -3.375 11.39063 -2.975 8.850625 27.6 2.125 4.515625 28.1 6.890625 21.85563 27.7 2.225 4.950625 -1.075 1.155625 25.1 -0.375 0.140625 -0.875 0.765625 26.3 0.825 0.680625 28.2 7.425625 22.2 -3.275 10.72563 26.3 0.825 0.680625 24.4 -1.075 1.155625 2 x = 509.5 § (x,-8)2 =118.1375 Sample: The formula for calculating the sum of squares is S = Î (x,- 8) SS = [118.1375 Therefore, the sum of squares is approximately SS [118.1375 2725 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The formula for calculating the sample variance is 2---3-3) = (118.1375) = (118.1375) = 6.217763 sa 16.217763 Therefore, the sample variance is approximately sa 16.217763 - - - - - - - - - - - - - - - - - - - - 34 - The formula for calculating the sample standard deviation is o = 20-(118.1375) = o(118.1375) = 16.217763 = 2.493544 ON 2.493544 Therefore, the sample standard deviation is approximately 0 ~ 2.493544 -------------------- -------------------

The 99% confidence interval for the population mean: The Sample Size is n = 20. The Sample Mean is X = 25.475. The Sample standard deviation s = 2.493544. The formula for calculating the degrees of freedom is df = n-1 = 20-1 = 19 df = 19 Therefore, the degrees of freedom is df = 19 ----------------------------------------- The level of significance is a =0.01. The table values are calculated by using the Excel2010 function: = tinv(probability, degrees of freedom) tiv(0.01, 19) = 2.539483 40.1-*2.539483 c = 2.539483) - - - - - - The formula for calculating the confidence interval for the population mean is X-o = *--0018- in su = 8-how in su 2.493544 = 25.475 -2.539483 20 = 25.475 -2.539483(- 2.493544 SA = 25.475 -2.539483 4.472135955) 54 = 25.475-2.539483(0.557573455) 54 25.475-1.41594854 = 24.05905 54 X-4a11 54= 24.0591 Interpretation: We can be 99% one-sided confidence interval for that provides lower bound for u is X-1671 = 24.0591