The accounts of a company show that on average, accounts receivable are $93.18.

Question

The accounts of a company show that on average, accounts receivable are $93.18. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $89.34 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $93.18 at =0.05? 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 H if If the hypothesis is two tailed: Reject Hoif or Question2 What is the test statistic? Question 3What is the p-value? Fill in only ang of the following statements. f the Z table is appropriate, p-value If the t table is appropriate, p-value

Explanation / Answer

Question 1.

Here hypothesis is two- tailed test

Reject H0 if 95% confidence interval doesn't consist value of population mean.

95% CI = x +- t48, 0.05 (s/ sqrt(n)

= 89.34 +- 2.01 * (40.56/ sqrt(49))

= 89.34 +- 11.65

= (77.69, 100.99)

so we reject H0 if H < 77.69 or H > 100.99

Question 2

t = (x - H)/ s/ sqrt(n) = (89.34 - 93.18)/ (40.56/ sqrt(49)) = -3.84/ 5.79 = 0.66

Question 3

p - value =

if the t table is appropriate, 1.0 < p- value < 0.5

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