9.10 A researcher is studying the effects of cocaine dose on a hamster\'s heart
ID: 3312094 • Letter: 9
Question
9.10 A researcher is studying the effects of cocaine dose on a hamster's heart rate. She plans on administering doses of 1,2. 3 10 milligrams of cocaine and then 354 LEAST-SQUARES REGRESSION METHODS recording the hamster's heart rate. She uses 10 hamsters for each dose level of co- caine, for a total of 100 hamsters used in the experiment. Let y, represent the heart rate for the i th rat and let 11 represent the cocaine dose for the i th rat. The results of the experiment are summarized below. Cocaine Dose 413.2 446.5 491.6 571.1 843.7 917.5 1,043.9 1,214.6 ,424.5 17,602.9 20,166.4 25,293.5 9,1414 1,371.7 84,617.8 109.612.7 148,548.7 203,401.5 10 (a) Compute the least-squares estimates of the slope and intercept for regressing y on I (b) Compute the least-squares estimates of the slope and intercept for regressing y on w where w2 (c) The true relationship between yi and w, is given by Vi = 42 + wi + ei, where ~ Normal(0, 64). what is ? w (d) Using your answers to (a) and (b), compute the predicted values of y for each level of x for the two models. Then, compute the estimated MSE for the two models. If you were to perform this experiment repeatedly, what would be the mean MSE using the model in (b)? (e) For the model in (a), compare the predicted values of y with the observed mean of y at each level of (f) Does there seem to be a trend in the difference of the predicted and observed means? For which values of r is y underestimated by the predicted values?Explanation / Answer
(a)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-74.26 -46.70 -24.28 23.76 111.28
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 200.123 50.963 3.927 0.0057 **
x 111.310 8.058 13.814 2.46e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 73.07 on 7 degrees of freedom
Multiple R-squared: 0.9646, Adjusted R-squared: 0.9596
F-statistic: 190.8 on 1 and 7 DF, p-value: 2.46e-06
Hence, the model is:
y = 200.123 + 111.310 * x
(b)
Call:
lm(formula = y ~ w)
Residuals:
Min 1Q Median 3Q Max
-17.849 -13.934 -8.553 0.226 65.573
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 414.6690 14.3307 28.94 1.52e-08 ***
w 10.0961 0.2735 36.91 2.78e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 27.77 on 7 degrees of freedom
Multiple R-squared: 0.9949, Adjusted R-squared: 0.9942
F-statistic: 1363 on 1 and 7 DF, p-value: 2.783e-09
Hence, the model is:
y = 414.6690 + 10.0961 * w
(c)
The value is equal to 64 since Wi follows N(0,64)
(d)
The predicted values in model (a) are:
1 2 3 4 6 7 8
311.4328 422.7427 534.0526 645.3624 867.9822 979.2920 1090.6019
9 10
1201.9118 1313.2216
The predicted values under model (b) is:
1 2 3 4 6 7 8
424.7651 455.0533 505.5335 576.2059 778.1269 909.3756 1060.8163
9 10
1232.4492 1424.2742
The MSE in part a is = 4152.323
The MSE in part b = 599.752
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