Question: The accounts of a company show that on average, accounts receivable ar

Question

Question: The accounts of a company show that on average, accounts receivable are $89.84. An auditor checks a random sample of 81 of these accounts, finding a sample mean of $85.65 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $89.84 at =0.05?

For the hypothesis stated above ...

1) what is the decision rule? Fill in only one of the following statements

If hypothesis is one tailed : Reject Ho if ___________-

If hypothesis is two tailed : Reject Ho if ____________ < _________ or _______

2) what is the test statistic _______

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, _______ < p-value

Explanation / Answer

f hypothesis is two tailed : Reject Ho if z<-1.96 or Z>1.96

2) std error =std deviation/(n)1/2 =4.5067

hence test stat z=(X-mean)/std error =-0.9297

3)If the Z table is appropriate, p-value = 0.3525

please revert as sample size is >30 i hv considerec ztest, but some ppl consider t test too.

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