A researcher conducts an experiment to examine the relationship between the weight gain (wtgn, gm) of chickens, whose diets had been supplemented by different amounts of lysine, and the amount of lysine (gm) ingested. A random sample of 12 chicks was  selected for the study. The R output is given below. Call: lm(formula = wtgn ~ lysine, data = lys)   Coefficients:             Estimate Std. Error t value Pr(>|t|) (Intercept)    12.57       1.18   10.65  8.9e-07 lysine         35.48       6.89    5.15  0.00043 Residual standard error: 1.02 on 10 degrees of freedom Multiple R-squared:  0.726,    Adjusted R-squared:  0.699  F-statistic: 26.5 on 1 and 10 DF,  p-value: 0.000432         95% CI               95% PI    fit  lwr  upr        fit  lwr  upr   16.8 15.9 17.8       16.8 14.4 19.3   The results for a test of Ho:β1=0Ho:β1=0 versus Ha:β1≠0Ha:β1≠0 show that: (pick one) a. the null hypothesis can be rejected because t=5.15, p=0.00043.   b. the null hypothesis cannot be rejected because t=10.65, p=8.9e-07.   c. the null hypothesis cannot be rejected because t=5.15, p=0.00043. d. the null hypothesis can be rejected because t=10.65, p=8.9e-07.

Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter38: Achievement Review—section Three
Section: Chapter Questions
Problem 6AR
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A researcher conducts an experiment to examine the relationship between the weight gain (wtgn, gm) of chickens, whose diets had been supplemented by different amounts of lysine, and the amount of lysine (gm) ingested. A random sample of 12 chicks was  selected for the study. The R output is given below.

Call:
lm(formula = wtgn ~ lysine, data = lys)

 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)    12.57       1.18   10.65  8.9e-07
lysine         35.48       6.89    5.15  0.00043

Residual standard error: 1.02 on 10 degrees of freedom
Multiple R-squared:  0.726,    Adjusted R-squared:  0.699 
F-statistic: 26.5 on 1 and 10 DF,  p-value: 0.000432

 

      95% CI               95% PI
   fit  lwr  upr        fit  lwr  upr
  16.8 15.9 17.8       16.8 14.4 19.3

 

The results for a test of Ho:β1=0Ho:β1=0 versus Ha:β1≠0Ha:β1≠0 show that: (pick one)

a. the null hypothesis can be rejected because t=5.15, p=0.00043.

 

b. the null hypothesis cannot be rejected because t=10.65, p=8.9e-07.

 

c. the null hypothesis cannot be rejected because t=5.15, p=0.00043.

d. the null hypothesis can be rejected because t=10.65, p=8.9e-07.

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