The accounts of a company show that on average, accounts receivable are $91.44.

Question

The accounts of a company show that on average, accounts receivable are $91.44. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $87.29 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $91.44 at -0,001? For the hypothesis stated above... Question 1What is the decision rule? Fill in only one of the following statements If the hypothesis is one tailed: Reject Ho if If the hypothesis is two tailed: Reject Ho if or Question 2 What is the test statistic? What is the p-value? FHill in only gne of the following statements If the Z table is appropriate, If the t table is appropriate, Question 3 p-value p-value

Explanation / Answer

Answer in detail as below:

1.

Ha: Mu = 91.44
Ho: Mu !=91.44

Reject H0 if test statistic Z is less than -3.1 or more than 3.1

2.

Test statistic
= (Xbar -Mu)/(sigma/sqrt(N))
= (87.29-91.44)/(40.56/sqrt(49))
= -0.72

3.

The P-Value is 0.471525.

FYI since Z is between 3.1 and -3.1 and not beyond them we fail to reject null hypothesis

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