A random sample of 18 subjects was asked to perform a given

A random sample of 18 subjects was asked to perform a given task. The time in seconds it took each of them to complete the task is recorded below: 31, 47, 43, 40, 49, 38, 33, 29, 30, 46, 31, 29, 45, 28, 40, 29, 28, 40 If we assume that the completion times are normally distributed, find a 95% confidence interval for the true mean completion time for this task. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list o formulas.) What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?

Solution

Since the number of samples is =18 (<30), we will go for t-tst. It will be a two tailed t-test because the deviation could be on both sides. The confidence interval is given by :

x +/- s*t/ n^0.5 where x= sample mean; s= sample standard deviation; t= t-value for particular degree of freedom; n= number of observation; df= n-1; thus degree of freedom in this case = 18-1=17;
x= mean of (31,47,43, ...28,40) = 656/18=36.444;
s= sample stadnard distribution, you can calculate this using any simple s.d. calculator, which comes out ot be s=7.429;
t value = 2.11 (from 2 tailed t-test for df=17)
Thus CI = 36.444 +/- 7.429*2.11 / (17)^0.5 = 36.444 +/- 3.801 = [40.241 , 32.639]

Hence the 95% confidence interval for true mean completion time for the task is [32.639, 40.241]