Question

Your instructor (Katie) believes that the average time of taking this exam is equal to 140 minutes. A sample of 4 students was taken and the following exam times were obtained. Assume the distribution of the population is normally distributed. Sample: 150, 150, 180, 170 (Sample Std. Dev = 15) The critical value for this test at a 0.1 level of significance is: O a. 1.285.-1.285 O b.2.353,-2.353 1.645,-1.645 OC. d. 1.638,-1.638

Answer 1

Given that the average time as $\mu$ = 140 and to test this a sample of n = 4 students we taken and following results were obtained.

The sample standard deviation obtained is s = 15, now based on the claim the hypotheses created are:

$Ho: \mu = 140$

$Ha: \mu \neq 140$

Based on the hypothesis it will be a two tailed test, but since the distribution is normal and the population standard deviation is unknown hence t-distribution is applicable for hypothesis testing.

Thus the critical score is obtailed using the excel formula for t-distribution which takes the degree of freedom and the significance level as the parameter. The degree fo freedom is calculated as df = n-1= 4-1= 3.

Now teh excel formula used is =T.INV.2T(0.10, 3), Thus the tc is obtained as:

b. -2.353, +2.353