It costs more to produce defective items - since they must be scrapped or reworked - than it does to produce non-defective items. This simple fact suggests that manufacturers should ensure the quality of their products by perfecting their production processes instead of depending on inspection of finished products (Deming, 1986). In order to better understand a particular metal stamping process, a manufacturer wishes to estimate the mean length of items produced by the process during the past 24 hours. (Give answer to nearest whole number.) a) How many parts should be sampled in order to estimate the population mean to within .1 millimeter (mm) with 90% confidence? Previous studies of this machine have indicated that the standard deviation of lengths produced by the stamping operation is about 2mm. 1083 Correct: Your answer is b) Time permits the use of a sample size no larger than 100. If a 90% confidence interval for the mean is constructed with n = 100, will it be wider or narrower than would have been obtained using the sample size determined in part a? cannot be determined narrower wider unchanged.
Accepted Solution
A:
Answer:1,082Step-by-step explanation:The sample size n in Simple Random Sampling is given by[tex] \bf n=\frac{z^2s^2}{e^2}[/tex]where Β z = 1.645 is the critical value for a 90% confidence level (*)s = 2 is the estimated population standard deviatione = 0.1 mm points is the margin of errorso Β [tex] \bf n=\frac{z^2s^2}{e^2}=\frac{(1.645)^2(2)^2}{(0.1)^2}=1,082.22\approx 1,083[/tex]rounded up to the nearest integer.So the manufacturer should test 1,083 parts.---------------------------------------------------------------------------------------(*)This is a point z such that the area under the Normal curve N(0,1) outside the interval [-z, z] equals 10% = 0.1It can be obtained in Excel withNORMINV(1-0.05,0,1)and in OpenOffice Calc withNORMINV(1-0.05;0;1)