Groups of 10- day-old chicks were randomly assigned to seven treatment groups in which a basal diet was supplemented with 0, 50, 100, 150, 200, 250, or 300 micrograms/kilogram (g/kg) of biotin. The table gives the average biotin intake (x) in micrograms per day and the average weight gain (y) in grams per day.6

In the MINITAB printout, the second-order polynomial model
E(y) = β0 + β1x + β2x2
is fitted to the data. Use the printout to answer the questions.
a. What is the fitted least-squares line?
b. Find R2 and interpret its value.
c. Do the data provide sufficient evidence to conclude that the model contributes significant information for predicting y?
d. Find the results of the test of H0: β2 = 0. Is there sufficient evidence to indicate that the quadratic model provides a better fit to the data than a simple linear model does?
e. Do the residual plots indicate that any of the regression assumptions have been violated? Explain.
MINITAB output for Exercise 13.25
Regression Analysis: y versus x, x-sq
The regression equation is
y = 8.59 + 3.82 x - 0.217 x-sq

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Transcribed Image Text:

| 0134543 4633 w8 17 2 2 2 2 2 n- 4164550 1003165 266792 | 0 0 0 0 0 0 0 505050 11223 Predictor Constant Coef 8.585 3.8208 -0.21663 SE Coef 1.641 0.5683 0.04390 5.23 6.72 -4.93 0.006 0.003 0.008 X-sq S 1.83318 R-Sq = 94 . 4% R-Sq (adj ) 91。5% = Analysis of Variance Source Regression Residual Error Total DF MS 224.75 112.37 33.44 0.003 13.44 238.19 3.36 DF 1 Source Seq sS 142.92 81.83 X-sq Residuals versus the Fitted Values (response is y) 10 2 14 16 18 20 22 2426 Fitted Value Normal Probability Piot of the Residuals 40 30 20 10 Residual