Suppose 10 accounts were randomly selected from all 382 accounts handled by the

Question

Suppose 10 accounts were randomly selected from all 382 accounts handled by the firm and then audited. The differences between the audit values and the corresponding book values were computed to test the following hypotheses about the true mean difference Ha: #50 The t-statistic was computed and equaled 2.14. Can we reject the null hypothesis that the true mean difference is $0 if we use a significance level of 0 , 05 ? a. Reject H 0 evidence supports H a b. Accept H 0 - evidence supports H 0 C. Fail to reject HO - not enough evidence to support H a

Explanation / Answer

Looking at the hypothesis we see that it is a two tailed test. Now for 5% level of significance and for n-1 = 9 degrees of freedom, we get from the t-distribution tables that:

P( -2.262 < t9 < 2.262 ) = 0.95

Therefore 2.262 is the critical value for the test here.

Now as the test statistic values lies in the non rejection region that is:

-2.262 < 2.14 < 2.262, therefore the test is insignificant and we cannot reject the nul hypothesis here.

Therefore c is the correct answer here.

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