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

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

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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