Question

In: Statistics and Probability

Consider the following data of experimental data measuring the lifetime of a wire as a function...

Consider the following data of experimental data measuring the lifetime of a wire as a function of temperature

Temp

200

200

200

200

200

200

Lifetime

5933

5404

4947

4963

3358

3878

Temp

220

220

220

220

220

220

Lifetime

1561

1494

747

768

609

777

Temp

240

240

240

240

240

240

Lifetime

258

299

209

144

180

184
  1. Sketch scatter plots of untransformed data and of ln(lifetime) vs 1/temp

  1. Estimate the lifetime of the wire at t=230 using the better model, be it untransformed or transformed.

Solutions

Expert Solution

Temp (x) Lifetime (y) ln(y) 1/x
200 5933 8.6883 0.0050
200 5404 8.5949 0.0050
200 4947 8.5065 0.0050
200 4963 8.5098 0.0050
200 3358 8.1191 0.0050
200 3878 8.2631 0.0050
220 1561 7.3531 0.0045
220 1494 7.3092 0.0045
220 747 6.6161 0.0045
220 768 6.6438 0.0045
220 609 6.4118 0.0045
220 777 6.6554 0.0045
240 258 5.5530 0.0042
240 299 5.7004 0.0042
240 209 5.3423 0.0042
240 144 4.9698 0.0042
240 180 5.1930 0.0042
240 184 5.2149 0.0042

a) Untransformed

Using Excel, insert scatter plot with temp on x-axis and lifetime on y-axis.

Right click on any point, select Add Trendline, choose Linear and tick Display Equation on Chart and Display R-square value on Chart

Transformed

Using Excel, insert scatter plot with 1/temp on x-axis and ln(lifetime) on y-axis.

Right click on any point, select Add Trendline, choose Linear and tick Display Equation on Chart and Display R-square value on Chart

b) Since R2 of transformed model is more (0.9543) than untranssformed model (0.8113), transformed model is better.

At lifetime = 230,

y = 3735.5x - 10.205

ln(lifetime) = 3735.5*(1/230) - 10.205

= 6.036

lifetime = exp(6.036) = 418.344

orchestra answered 1 year ago
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