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 Number of subjects or observations included in a study. A large sample size typically provides more reliable results and better representation of the population. As sample size and width of confidence interval are inversely related, if the sample size is increased, the width of the confidence interval decreases.
28. (a) Under what conditions do we say that two random variables X and Y are
independent?
(b) Demonstrate that if X and Y are independent, then it follows that E(XY) =
E(X)E(Y);
(e) Show by a counter example that the converse of (ii) is not necessarily true.
1. Let X and Y be random variables and suppose that A = F. Prove that
Z XI(A)+YI(A) is a random variable.
30. (a) What is meant by the term "product measur"?
AND
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