The accounts of a company show that on average, accounts receivable are $97.45.

Question

The accounts of a company show that on average, accounts receivable are $97.45. An auditor checks a random sample of 81 of these accounts, finding a sample mean of $93.71 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $97.45 at a=0.05

a is alpha

Below are spots to fill the answer in. Please explain step by step.

Question 1: What is the decision rule? Fill in only one of the following statements.

If the hypothesis is one tailed:

   Reject H0 if _____ ______ ________
If the hypothesis is two tailed:
   Reject H0 if ____ < _____ or ______ < _____

Question 2: What is the test statistic?

Question 3: What is the p-value? Fill in one of the following statements.

If the Z table is appropriate, p value =

If the t table is appropriate, _____ < p-value < ______

For the hypothesis stated above... Question 1 what is the decision rule? Fill in only one of the following statements. If the hypothesis is one tailed: Reject Ho if It the hypothesis is two tailed: Reject Ho if Question 2 What is the test statistic? Question 3 What is the p-value? Fill in only one of the following statements. If the Z table is appropriate, p-value If the t table is appropriate,

Explanation / Answer

1. The hypotheses are as follows:

H0:mu=97.45 (mean accounte receivable is $97.45)

H1:mu=/=97.45 (mean accounte receivable is different from 97.45)

Therefore, from the sign of alternative hypothesis, it is clear that the test is two-tailed.

For sample size, N=81 (N>50), use student's t model assuming the normal population assumption is met safely. The population standard deviation is unknown, therefore, use student's t model to find 1-sample t test staistic.

The significance level is alpha=0.05, the t critical at 80 degrees of freedom [df=N-1] and alpha/2=+-1.99

Therefore, reject H0, observed t<-1.99 or observed t>1.99.

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2. Test statistic, t=(Xbar-mu)/(s/sqrt N), where, Xbar is sample mean, mu is population mean, s is sample standard deviation, and N is sample size.

=(93.71-97.45)/(40.56/sqrt 81)

=-0.83 (ans)

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3. P value is: 0.1<p<0.5

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