In: Statistics and Probability
An article in Electronic Packaging and Production (2002, vol. 42) considered the effect of X-ray inspection of integrated circuits. The radiation dose (rads) were studied as a function of current (in milliamps) and exposure (in minutes).The data arein excel file uploaded to Moodle. Name of the file is “Assignment 4 Data”. Use a software (preferable MINITAB) to answer the following questions.
Part 1. Perform simple linear regression analysis with the variables, radiation dose and exposure time to answer the following questions. (Include the output in your pdf file.)
a) Determine response variable and find the fitted line. (Estimated regression line)
b) Predict the radiation dose when exposure time is 15 seconds.
c) Estimate the standard deviation of radiation dose.
d) What percentage of variability in radiation dose can be explained by the
exposure time?
e) Obtain 95% CI for the true slope of regression line.
*****Can you solve the problem above using Minitab and show the steps please?
X-ray | Inspection | Data | |
Rads | mA | Exposure Time | |
7,4 | 10 | 0,25 | |
14,8 | 10 | 0,5 | |
29,6 | 10 | 1 | |
59,2 | 10 | 2 | |
88,8 | 10 | 3 | |
296 | 10 | 10 | |
444 | 10 | 15 | |
592 | 10 | 20 | |
11,1 | 15 | 0,25 | |
22,2 | 15 | 0,5 | |
44,4 | 15 | 1 | |
88,8 | 15 | 2 | |
133,2 | 15 | 3 | |
444 | 15 | 10 | |
666 | 15 | 15 | |
888 | 15 | 20 | |
14,8 | 20 | 0,25 | |
29,6 | 20 | 0,5 | |
59,2 | 20 | 1 | |
118,4 | 20 | 2 | |
177,6 | 20 | 3 | |
592 | 20 | 10 | |
888 | 20 | 15 | |
1184 | 20 | 20 | |
22,2 | 30 | 0,25 | |
44,4 | 30 | 0,5 | |
88,8 | 30 | 1 | |
177,6 | 30 | 2 | |
266,4 | 30 | 3 | |
888 | 30 | 10 | |
1332 | 30 | 15 | |
1776 | 30 | 20 | |
29,6 | 40 | 0,25 | |
59,2 | 40 | 0,5 | |
118,4 | 40 | 1 | |
236,8 | 40 | 2 | |
355,2 | 40 | 3 | |
1184 | 40 | 10 | |
1776 | 40 | 15 | |
2368 | 40 | 20 | |
using minitab>stat>Regression>Regreession 'we have
Regression Analysis: Rads versus Exposure Time
Analysis of Variance
Source DF Seq SS Contribution Adj SS Adj MS F-Value
P-Value
Regression 1 9403685 71.42% 9403685 9403685 94.94 0.000
Exposure Time 1 9403685 71.42% 9403685 9403685 94.94 0.000
Error 38 3763670 28.58% 3763670 99044
Lack-of-Fit 6 609 0.00% 609 101 0.00 1.000
Pure Error 32 3763062 28.58% 3763062 117596
Total 39 13167355 100.00%
Model Summary
S R-sq R-sq(adj) PRESS R-sq(pred)
314.713 71.42% 70.66% 4521319 65.66%
Coefficients
Term Coef SE Coef 95% CI T-Value P-Value VIF
Constant -2.2 67.3 (-138.3, 134.0) -0.03 0.974
Exposure Time 68.18 7.00 ( 54.02, 82.35) 9.74 0.000 1.00
Regression Equation
Rads = -2.2 + 68.18 Exposure Time
Fits and Diagnostics for Unusual Observations
Obs Rads Fit SE Fit 95% CI Resid Std Resid Del Resid HI
8 592.0 1361.5 107.0 (1145.0, 1578.0) -769.5 -2.60 -2.83
0.115517
39 1776.0 1020.6 77.7 ( 863.2, 1177.9) 755.4 2.48 2.67
0.060981
40 2368.0 1361.5 107.0 (1145.0, 1578.0) 1006.5 3.40 4.02
0.115517
Obs Cook’s D DFITS
8 0.44 -1.02250 R
39 0.20 0.68022 R
40 0.76 1.45392 R
a) response variable = radiation dose
the fitted line is Rads = -2.2 + 68.18 Exposure Time
b) the radiation dose when exposure time is 15 seconds is -2.2 + 68.18*15 =1020.5
c) the standard deviation of radiation dose. is 314.713
d) 71.42 percentage of variability in radiation dose can be explained by the
exposure time
e) 95% CI for the true slope of regression line is ( 54.02, 82.35)