The HFE (y) of an electrical system is thought to be modeled by a multiple linea

Question

The HFE (y) of an electrical system is thought to be modeled by a multiple linear regression model on the Emitter-RS (x1), the Base-RS (x2) and the Emitter to Base-RS (x3) as seen on the left. Fit a multiple linear regression model to these data using Excel.

a.Use Excel to develop a multiple linear regression model of this data. In your final model, what will the numeric value of the constant be? Express your answer as an integer.

b.Use Excel to develop a multiple linear regression model of this data. In your final model, what will the numeric value of B1_hat be? Express your answer to three decimal places.

c.Use Excel to develop a multiple linear regression model of this data. In your final model, what will the numeric value of B2_hat be? Express your answer to three decimal places with no leading zero.

d.Use Excel to develop a multiple linear regression model of this data. In your final model, what will the predicted response, y_hat, when x1 = 14.62, x2 = 226, and when x3 = 7? Express your answer to three decimal places and use the values you got for the previous parts of this question (i.e., values to 3 decimal places)

e.Use Excel to develop a multiple linear regression model of this data. determine P(15.457 < B3 < 21.018). Express your answer to two decimal places with no leading zero.

f.Use Excel to develop a multiple linear regression model of this data. What is the residual for the first data point (x1 = 14.62, x2 = 226, x3 = 7)? Express your answer to one decimal place.

Emitter-RS Base-RS Emitter to HFE Base-RS 14.620 15.630 14.620 15.000 14.500 15.250 16.120 15.130 15.500 15.130 15.500 16.120 15.130 15.630 15.380 14.380 15.500 14.250 14.500 14.620 226.00 22000 217.40 220.00 226.50 224.10 220.50 223,50 217.60 228.50 230.20 226.50 226.60 225.60 229.70 234.00 230.00 224.30 240,50 223.70 7.000 3.375 6.375 6.000 7.625 6.000 3.375 6.125 5.000 6.625 5.750 3.750 6.125 5.375 5.875 8.875 4.000 8.000 10.870 7.375 128.40 52.62 113.90 98.01 139.90 102.60 48.14 109.60 82.68 112.60 97.52 59.06 111.80 89.09 101.00 171.90 66.80 157.10 208.40 133.40

Explanation / Answer

a) constant - 47.174

b)b1_hat = -9.735

c)b2_hat = 0.42828

d)

y_hat, when x1 = 14.62, x2 = 226, and when x3 = 7

y^ = 47.17399883 -9.7352022*x1 +0.428287026*x2 +18.23745475*x3

= 47.17399883 -9.7352022*14.62 +0.428287026*226 +18.23745475*7

=129.300393

e) b3^ = 18.23745475

se(b3^) = 1.311801711

P(15.457 < B3 < 21.018)

=P(15.457-18.237 < B3 -18.237 < 21.018-18.237)

==P((15.457-18.237)/1.3118 < (B3 -18.237)/1.3118 <(21.018-18.237)/1.3118)

=P(-2.11922< t< 2.1199878)

df = n-k-1 = 20 -3-1 = 16

P(t < -2.11922) = 0.0250

hence required value = 1-2*0.025 = 0.95

f) residual = yi - y^ = 128.4 - 129.300393 = -0.900393

Please rate my solution

x1 x2 x3 y 14.62 226 7 128.4 15.63 220 3.375 52.62 14.62 217.4 6.375 113.9 15 220 6 98.01 14.5 226.5 7.625 139.9 15.25 224.1 6 102.6 16.12 220.5 3.375 48.14 15.13 223.5 6.125 109.6 15.5 217.6 5 82.68 15.13 228.5 6.625 112.6 15.5 230.2 5.75 97.52 16.12 226.5 3.75 59.06 15.13 226.6 6.125 111.8 15.63 225.6 5.375 89.09 15.38 229.7 5.875 101 14.38 234 8.875 171.9 15.5 230 4 66.8 14.25 224.3 8 157.1 14.5 240.5 10.87 208.4 14.62 223.7 7.375 133.4

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