One operation of a steel mill is to cut pieces of steel info parts that are used in the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts must be cut to be within ± 0.005 inch of the length specified by the automobile company. The file Steel contains a sample of 100 steel parts. The file steel contains a sample of 100 steel parts. The measurement reported is the difference, in inches, between the actual length of the steel part, a measured by a laser measurement device, and the specified length of the steel part. For example, a value of –0.002 represents a steel shorter than the specified length. a. At the 0.005 level of significance, is there evidence that the mean difference is different from 0.0 inches? b. Construct a 95 % confidence interval estimate of the population mean. Interpret this interval. c. Compare the conclusions reached in (a) and (b). d. Because n = 100 , do you have to be concerned about the normally assumption needed for the t test and t interval?
One operation of a steel mill is to cut pieces of steel info parts that are used in the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts must be cut to be within ± 0.005 inch of the length specified by the automobile company. The file Steel contains a sample of 100 steel parts. The file steel contains a sample of 100 steel parts. The measurement reported is the difference, in inches, between the actual length of the steel part, a measured by a laser measurement device, and the specified length of the steel part. For example, a value of –0.002 represents a steel shorter than the specified length. a. At the 0.005 level of significance, is there evidence that the mean difference is different from 0.0 inches? b. Construct a 95 % confidence interval estimate of the population mean. Interpret this interval. c. Compare the conclusions reached in (a) and (b). d. Because n = 100 , do you have to be concerned about the normally assumption needed for the t test and t interval?
Solution Summary: The author explains how to perform the t-test using Minitab.
One operation of a steel mill is to cut pieces of steel info parts that are used in the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts must be cut to be within
±
0.005
inch of the length specified by the automobile company. The file Steel contains a sample of 100 steel parts. The file steel contains a sample of 100 steel parts. The measurement reported is the difference, in inches, between the actual length of the steel part, a measured by a laser measurement device, and the specified length of the steel part. For example, a value of –0.002 represents a steel shorter than the specified length.
a. At the 0.005 level of significance, is there evidence that the mean difference is different from 0.0 inches?
b. Construct a
95
%
confidence interval estimate of the population mean. Interpret this interval.
c. Compare the conclusions reached in (a) and (b).
d. Because
n
=
100
,
do you have to be concerned about the normally assumption needed for
the t test and t interval?
Definition Definition Method in statistics by which an observation’s uncertainty can be quantified. The main use of interval estimating is for describing a range that is made by transforming a point estimate by determining the range of values, or interval within which the population parameter is likely to fall. This range helps in measuring its precision.
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