Question

23. A study was conducted to find out the average number of months it takes a Tennessee job seeker to find a job in our tough economy. In a random sample of 25 job seekers, it took an average of 5.9 months and standard deviation of 0.8 months to find a job. Find the 90% confidence interval for average amount of time to find a job in Tennessee. A. 3.16 < C. 5.63 < < 8.63 < 6.17 B. 5.56 < D. 2.60 < < 6.23 < 9.20

Answer 1

The provided sample mean is X = 5.9 and the sample standard deviation is s=0.8. The size of the sample is n=25 and the required confidence level is 90%. The number of degrees of freedom are df = 25 -1 = 24, and the significance level is a = 0.1 Based on the provided information, the critical t-value for a = 0.1 and df = 24 degrees of freedom is te = 1.711 The 90% confidence for the population mean y is computed using the following expression CI = (x - Home 8 + Therefore, based on the information provided, the 90 % confidence for the population mean is CI = (5.9 - 1.71), 0.85.9 + 1.711.X 0.8) 25 = (5.9 -0.274,5.9 +0.274) = (5.626, 6.174)

Therefore, from the above calculations

90 % c.i 5.63 <  $\mu$ < 6.17

Answer : Option ( C )