标题: Reading 10: Sampling and Estimation - LOS e ~ Q11-12 [打印本页]
作者: mayanfang1 时间: 2009-1-9 11:00 标题: [2009] Session 3- Reading 10: Sampling and Estimation - LOS e ~ Q11-12
Q11. Frank Grinder is trying to introduce sampling into the quality control program of an old-line manufacturer. Grinder samples 38 items and finds that the standard deviation in size is 0.019 centimeters. What is the standard error of the sample mean?
A) 0.00204.
B) 0.00615.
C) 0.00308.
Q12. Frank Grinder is trying to introduce sampling into the quality control program of an old-line manufacturer. Currently, each item is individually inspected to make sure it meets size tolerances. For all items manufactured during August, the standard deviation of size was 0.02 centimeters. If Grinder takes a sample of 30 items and finds a standard deviation of size of 0.019 centimeters, what is the standard error of the sample mean?
A) 0.00200.
B) 0.00600.
C) 0.00365.
作者: mayanfang1 时间: 2009-1-9 11:00
答案和详解如下:
Q11. Frank Grinder is trying to introduce sampling into the quality control program of an old-line manufacturer. Grinder samples 38 items and finds that the standard deviation in size is 0.019 centimeters. What is the standard error of the sample mean?
A) 0.00204.
B) 0.00615.
C) 0.00308.
Correct answer is C)
If we do not know the standard deviation of the population (in this case we do not), then we estimate the standard error of the sample mean = the standard deviation of the sample / the square root of the sample size = 0.019 / √38 = 0.00308 centimeters.
Q12. Frank Grinder is trying to introduce sampling into the quality control program of an old-line manufacturer. Currently, each item is individually inspected to make sure it meets size tolerances. For all items manufactured during August, the standard deviation of size was 0.02 centimeters. If Grinder takes a sample of 30 items and finds a standard deviation of size of 0.019 centimeters, what is the standard error of the sample mean?
A) 0.00200.
B) 0.00600.
C) 0.00365.
Correct answer is C)
If we know the standard deviation of the population (in this case we do), then the standard error of the sample mean = the standard deviation of the population / the square root of the sample size = 0.02 / √30 = 0.00365 centimeters.
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