Question

2. A mill cuts pieces of steel that are used as the base for the front seats of automobiles. Steel is cut with a diamond saw and parts are required to be between plus or minus 0.005 inches as specified by the car manufacturing company. The reported measurements for 100 pieces of steel is the difference in inches between the actual length of the parts as measured by a laser apparatus and the specified measurement for the steel parts. For example, an observation of -0.002 represents a piece of steel that is 0.002 inches shorter than the specified length. A random sample of 100 pieces is selected and returns an average of -0.00023 with a standard deviation for the sample of 0.0017 inches. to. Construct a 95% confidence interval for the mean difference between the current length of the steel parts, and the specified length. b. What assumptions should you make about the population distribution in part (a)? Explain. c. Would a value of 0.008 be unusual? Explain. d. Construct a 99% confidence interval for the mean difference between the current length of the steel parts, and the specified length. (20 points)

Answer 1

Result:

2. A mill cuts pieces of steel that are used as the base for the front seats of automobiles. Steel is cut with a diamond saw and parts are required to be between plus or minus 0.005 inches as specified by the car manufacturing company. The reported measurements for 100 pieces of steel is the difference in inches between the actual length of the parts as measured by a laser apparatus and the specified measurement for the steel parts. For example, an observation of -0.002 represents a piece of steel that is 0.002 inches shorter than the specified length. A random sample of 100 pieces is selected and returns an average of -0.00023 with a standard deviation for the sample of 0.0017 inches. to. Construct a 95% confidence interval for the mean difference between the current length of the steel parts, and the specified length.

n=100

Mean = -0.00023

SD=0.0017

At 95% level table value t at 100-1 = 99 DF is 1.9842

=(-0.000567, 0.000107)

95% confidence interval for the mean difference =(-0.000567, 0.000107).

Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation	0.0017
Sample Mean	-0.00023
Sample Size	100
Confidence Level	95%
Intermediate Calculations
Standard Error of the Mean	0.000170
Degrees of Freedom	99
t Value	1.9842
Interval Half Width	0.000337
Confidence Interval
Interval Lower Limit	-0.000567
Interval Upper Limit	0.000107

b. What assumptions should you make about the population distribution in part (a)? Explain.

We assume mean difference between the current length of the steel parts is normally distributed.

c. Would a value of 0.008 be unusual? Explain.

The value 0.008 is unusual because this value does not fall inside the 95% confidence interval (-0.000567, 0.000107).

d. Construct a 99% confidence interval for the mean difference between the current length of the steel parts, and the specified length. (20 points)

At 99% level table value t at 100-1 = 99 DF is 2.6264

=(-0.000676, 0.000216)

99% confidence interval for the mean difference =(-0.000676, 0.000216).

Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation	0.0017
Sample Mean	-0.00023
Sample Size	100
Confidence Level	99%
Intermediate Calculations
Standard Error of the Mean	0.000170
Degrees of Freedom	99
t Value	2.6264
Interval Half Width	0.000446
Confidence Interval
Interval Lower Limit	-0.000676
Interval Upper Limit	0.000216