an auditor wishes to test the assumption that the mean value of all accounts receivable in a given firm is $260.00. She will reject this claim on ly if it is clearly contradicted by the sample mean. the sample deviation of 36 accounts is $43, the sample mean is $250, and the levek if significance is 5%.
a) write out the null and alternative hypotheses.
b)calculate the A test statistic.
c)find the p-value both from A
d)what would be the conclusion of the auditor i.e., reject or not reject? justify answer with p-value
e)what would be the conclusion of the auditor if the level of significance is 1%(i.e., =.01) justify answer.
Ho: mean1 = mean2
Ha: mean1 ≠ mean2
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion from the Z score.
245.95 and 274.05
a) The null hypothesis (H0) is that the mean value of all accounts receivable is $260. The alternative hypothesis (Ha) is that the mean value is not equal to $260.
b) To calculate the t-test statistic, we can use the formula:
t = (sample mean - assumed population mean) / (sample deviation / sqrt(sample size))
In this case, the assumed population mean is $260, the sample mean is $250, the sample deviation is $43, and the sample size is 36.
t = ($250 - $260) / ($43 / sqrt(36))
t = (-10) / (43 / 6)
t ≈ -1.3953
c) To find the p-value, we need to determine the probability of obtaining a t-value as extreme as or more extreme than the calculated t-value (-1.3953) under the null hypothesis. We can use a t-distribution table or statistical software to find this probability.
Using a t-distribution table with 34 degrees of freedom (sample size - 1), and a two-tailed test at a 5% level of significance, we find that the p-value is approximately 0.1717.
d) The auditor should compare the p-value to the chosen level of significance (5%). Since the p-value (0.1717) is greater than the level of significance (0.05), the auditor does not have enough evidence to reject the null hypothesis. Therefore, the conclusion would be to not reject the claim that the mean value of all accounts receivable is $260.
e) If the level of significance is decreased to 1% (0.01), the auditor would compare the p-value (0.1717) to this new level of significance. Since the p-value is still greater than the level of significance, the conclusion remains the same: not enough evidence to reject the null hypothesis.