From William E. Becker, Statistics for Business and Economics Using Microsoft Excel 97, S.R.B. Publishing, 1997, p. 290. Reproduced with permission of S.R.B. Publishing.
QUERY 8.1: A mill produces sheet metal that it says will average 0.20 inch in thickness, with deviations determined by the normal distribution. A truckload is received for which a sample of 23 different locational measurements yields a mean width of 0.192 inch, with a standard deviation of 0.0187 inch. Should you accept the shipment as having an average thickness of 0.20 inch?
ANSWER: The point estimate of the thickness of the sheet metal in the shipment is 0.192 inch. Because the population is normal, the distribution of sample average thickness is also normal but the t distribution must be used for probability calculations because sigma is unknown. A 95 percent confidence interval is constructed to reflect the randomness inherent in sampling, where t = 2.074 with 22 degrees of freedom. THIS 2.074 T VALUE IS OBTAINED IN EXCEL BY ENTERING "=TINV(0.05,22)" INTO ANY CELL, WHICH YIELDS 2.073875. As demonstrated in the textbook, the 95 percent confidence interval is then obtained as 0.1839 <µ >< 0.2001. The claimed 0.20 inch thickness is just inside this interval, suggesting that the true µ could be that high, as estimated by a 95 percent confidence interval.
IN GENERAL TO OBTAIN THE x VALUE FOR WHICH ANY GIVEN PERCENT OF THE VALUES IN A t DISTRIBUTION ARE LOWER, THIS "TINV" COMMAND IS USED, WHERE THE FIRST ENTRY IS DOUBLE THE SINGLE TAIL PERCENTAGE, THE SECOND IS DEGREES OF FREEDOM. IN QUERY 8.1, ON PAGE 290, NOTICE THAT 2.074 IS THE VALUE OF t FOR WHICH 2.5 PERCENT OF THE DISTRIBUTION IS LOWER BUT "TINV" REQUIRES THAT 0.05 BE ENTERED.
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