Question

The accounts of a company show that on average, accounts receivable are $98.64. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $93.69 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $98.64 at a-0.01? For the hypothesis stated above... Question 1 What is the decision rule? Fill in only one of the following statements. Ifthe hypothesis isone tailed Reject H, if If the hypothesis is two tailed Reject Ho if t stat 2.682 or t stat 2.682 Question 2 What isthe test statistic? 0.854 Question 3 What is the p value? Fill in only one of the following statements Ifthe Ztable is appropriate, p-value Ifthe t table is appropriate, 0.2 p-value

QUESTION 1?

Answer 1

Test hypothesis are Ho: 98.64 Given x-93.69, s-40.56 df-n1-49-1-48 Critical value tc-taa-±2.682 (two tailed) Test statistic P-value -P0.854,48,2)-0.3973 n -49 -93.69-98.64 - -0.854 s/Vn 40.56/V49 Using t value table at df 48,0 854 is between 1,0478 and 0 8492 and its probabilities are 0.3 and 0.4 Hence 0.3 < p-value <0.4

Answers Questionl Test hypothesis is two tailed Reject Ho if t-stat <-2.682 (or) t-stat >2.682 Question2 Test statistic0.854 Question3] Hence 0.3<p-value <0.4