Bus. 5.60 When an audit must be conducted that involves a tedious examination of a large inventory, the audit may be very costly and time consuming if each item in the inventory must be examined. In...

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Extracted text: Bus. 5.60 When an audit must be conducted that involves a tedious examination of a large inventory, the audit may be very costly and time consuming if each item in the inventory must be examined. In such situations, the auditor frequently obtains a random sample of items from the complete inventory and uses the results of an audit of the sampled items to check the validity of the company's financial statement. A large company's financial statement claims an inventory that averages $600 per item. The following data are the auditor's assessment of a random sample of 75 items from the company's inventory. The values resulting from the audit are rounded to the nearest dollar. 303 547 1,368 493 984 507 148 2,546 738 83 2 135 274 74 1,472 399 1,784 71 751 136 571 147 282 2,039 1,909 748 188 548 1 280 102 618 129 1,324 1,428 469 102 454 1,059 939 303 600 234 514 17 551 293 1,395 7 28 2 973 506 511 812 1,290 685 447 11 35 252 1,526 464 5 67 99 67 259 7 67 248 3,215 3 33 41 a. Estimate the mean value of an item in the inventory using a 95% confidence interval. b. Is there substantial evidence (a = .01) that the mean value of an item in the inventory is less than $600? c. What is the target population for the above inferences? d. Would normal distribution-based procedures be appropriate for answering the above questions?


Extracted text: Question 6 5.60d(mod)a: Use the Anderson-Darling Normality Test to evaluate the normality of the sample based on a 5% significance level Hint: See L-20 pdf and lecture, page 10 (Exercise 5.65 (mod)) p-value=0.325>0.05=Fail to reject HO: There is insufficient evidence that the sample is NOT from a Normal population p-value=0.023<0.05=reject ho:="" there="" is="" significant="" evidence="" that="" the="" sample="" is="" not="" from="" a="" normal="" population="" p-value="0.687">0.05=Fail to reject HO: There is significant evidence that the sample is from a Normal population p-value<><0.05=reject h0: there is significant evidence that the sample is not from a normal population h0:="" there="" is="" significant="" evidence="" that="" the="" sample="" is="" not="" from="" a="" normal="">