An engineer at a semiconductor company wants to model the relationship between t

Question

An engineer at a semiconductor company wants to model the relationship between the gain or hFE of a device (y) and three parameters: emitter-RS (x1), base-RS (x2), and emitter-to-base-RS (x3). The data are shown below:

(a) Fit a multiple linear regression to the data.

(b) Predict hFE when x1 = 14, x2 = 220, and x3 = 5.

(c) Estimate ?2 using the multiple regression of y on x1, x2, and x3.

(d) Test whether the emitter-to-base-RS (x3) is significant in this linear model at the level 0.05.

(e) Compute a 90% prediction interval for the observed gain and a 90% confident interval for mean response with the three regressors at x1 = 15.0, x2 = 220.0, and x3 = 6.0.

Explanation / Answer

we use R-software to make the answer

the table of regression analysis is

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) -51.472 315.080 -0.163 0.87241   
x1 6.989 21.882 0.319 0.75382   
x2 -0.426 1.675 -0.254 0.80269   
x3 23.544 7.668 3.071 0.00777 **

and

Analysis of Variance Table

Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x1 1 16645.5 16645.5 25.9882 0.0001311 ***
x2 1 7091.4 7091.4 11.0716 0.0045935 **
x3 1 6038.8 6038.8 9.4282 0.0077708 **
Residuals 15 9607.5 640.5   

a ) from the regression table we get the fitted multiple regression line as

y=-51.472+6.989*x1-0.426*x2+23.544*x3

b) hence Predict of hFE when x1 = 14, x2 = 220, and x3 = 5

y=-51.472+6.989*14-0.426*220+23.544*5=70.375

c) Here we know that the Estimate 2 is the MSE of the ANOVA fitting.From the table of ANOVA we get the Estimated 2 as MSE=640.5

d) from the table of regression analysis we can test of each coefficient by noting p-value. We know the if p-value < significance level we reject H0( i.e. hypothesis is euality of zero that coefficient) otherwise accept H0

here we can note that the p-value corresponding x3 is 0.00777 <0.05 =significance level(given)

so we reject H0 and comment that there is significant role of x3 in the regression of predicting y

e) we are submitting the prediction and condifence interval of all 19 observation at 0.95 level

confidence interval

fit lwr upr
1 119.24555 98.87915 139.61196
2 43.51395 20.29948 66.72842
3 108.19408 80.94935 135.43882
4 100.91348 81.62391 120.20304
5 132.90879 112.15568 153.66190
6 100.91426 87.31864 114.50987
7 46.72570 19.09955 74.35186
8 103.27412 89.60708 116.94117
9 81.88657 54.40288 109.37026
10 112.91615 98.67274 127.15957
11 94.17708 74.33235 114.02182
12 52.99876 26.53168 79.46583
13 101.95357 88.25767 115.64947
14 88.21626 70.82784 105.60468
15 165.29436 127.48418 203.10453
16 53.06043 13.85752 92.26334
17 140.92761 115.19810 166.65712
18 203.34499 164.93189 241.75809
19 129.05429 110.09975 148.00882

prediction interval

fit lwr upr
1 119.24555 61.585862 176.9052
2 43.51395 -15.212203 102.2401
3 108.19408 47.761271 168.6269
4 100.91348 43.625289 158.2017
5 132.90879 75.111379 190.7062
6 100.91426 45.284317 156.5442
7 46.72570 -13.880020 107.3314
8 103.27412 47.626684 158.9216
9 81.88657 21.345653 142.4275
10 112.91615 57.124357 168.7079
11 94.17708 36.699584 151.6546
12 52.99876 -7.087477 113.0850
13 101.95357 46.299034 157.6081
14 88.21626 31.539926 144.8926
15 165.29436 99.419790 231.1689
16 53.06043 -13.623281 119.7441
17 140.92761 81.162592 200.6926
18 203.34499 137.122521 269.5675
19 129.05429 71.878035 186.2305

here we require  regressors at x1 = 15.0, x2 = 220.0, and x3 = 6.0. that is the 4th row of the data hence from the table the prediction interval is (43.625289 ,158.2017)

and the condidence interval is (81.62391, 120.20304)

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