A company receives a large shipment of bolts. The bolts will be used in an application that requires a torque of 100 J. Before the shipment is accepted, a quality engineer will sample 12 bolts and measure the torque needed to break each of them. The shipment will be accepted if the engineer concludes that fewer than 1% of the bolts in the shipment have a breaking torque of less than 100 J.

Question

a. If the 12 values are 107, 109, 111, 113, 113, 114, 114, 115, 117, 119, 122, 124, compute the sample mean and sample standard deviation.
b. Assume the 12 values are sampled from a normal population, and assume the the sample mean and standard deviation calculated in part (a) are actually the population mean and standard deviation. Compute the proportion of bolts whose breaking torque is less than 100 J. Will the shipment be accepted?
c. What if the 12 values had been 108, 110, 112, 114, 114, 115, 115, 116, 118, 120, 123, 140? Use the method outlined in parts (a) and (b) to determine whether the shipment would have been accepted.
d. Compare the sets of 12 values in parts (a) and (c). In which sample are the bolts stronger?
e. Is the method valid for both samples? Why or why not?

Answer 1

(a) The sample mean is 114.8 J and the sample standard deviation is 5.006 J.
(b) The z-score of 100 is (100 ? 114.8)/5.006 = ?2.96. The area to the left of z = ?2.96 is 0.0015. Therefore only 0.15% of bolts would have breaking torques less than 100 J, so the shipment would be accepted.
(c) The sample mean is 117.08 J; the sample standard deviation is 8.295 J. The z-score of 100 is (100?117.08)/8.295 =?2.06. The area to the left of z= ?2.06 is 0.0197. Therefore about 2% of the bolts would have breaking torquesless than 100 J, so the shipment would not be accepted.
(d) The bolts in part (c) are stronger. In fact, the weakest bolt in part (c) is stronger than the weakest bolt in part (a),the second-weakest bolt in part (c) is stronger than the second-weakest bolt in part (a), and so on.
(e) The method is certainly not valid for the bolts in part (c). This sample contains an outlier (140), so the normal distribution should not be used.