Question

Problem 5. 1. (a) If a random sample of 40 items produces x = 128.4 and s = 20.6, what is the 95% confidence interval for ? Assume x is normally distributed for the population (5 points)

2. (b) Suppose the following data are selected randomly from a population of normally distributed values. Data : 0, 10 , 20 ,5 ,15 Construct a 95% confidence interval to estimate the population mean. (15 points)

Answer 1

Ans.

a) Point estimate, x = 128.4

Sample standard deviation, s = 20.6

Sample size, n = 40

Standard Error, se = s/(n^0.5) = 20.6/(40^0.5) = 3.257

t critical at 5% level of significance and 39 degrees of freedom, t = 2.0227

=> Margin of error, M = t*se = 2.0227*3.257 = 6.5882

Thus, 95% confidence interval of the population mean, CI = [x-M, x+M] = [128.4 - 6.5882, 128.4+6.5882]

=> CI = [121.8117, 134.9882]

b) Data, xi : 0, 10, 20, 5, 15

=> Mean, x = Sum(xi)/Sample Size(n) = 50/5 = 10

and Standard deviation, s = [Sum of (x - xi)^2 / (n-1)]^0.5 = 7.0710

Standard Error, se = s/(n^0.5) = 7.0710/(5^0.5) = 3.1622

t critical at 5% level of significance and 4 degrees of freedom, t = 2.7762

=> Margin of error, M = t*se = 2.7762*3.1622 = 8.7790

Thus, 95% confidence interval of the population mean, CI = [x-M, x+M] = [10 - 8.7790, 10+8.7790]

=> CI = [1.2209, 18.7790]