5. A random sample of 12 2-week old chickens w between weight gain (grams) and a
ID: 3375859 • Letter: 5
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5. A random sample of 12 2-week old chickens w between weight gain (grams) and amount of lysine ingested (grams). The amount o Lysine ingested is the explanatory variable (X) and the weight gain (Y) is the response variable. The mean of the explanatory variable is 0.166. A linear regression model was fitted to the data. The results are shown below. Round all answers to three decimal places if necessary as used to examine the relationship Regression Analysis: weight gain versus lysine Ingested 128. 3579 28. 3579 26.52 ?. ysine ingestes i 28.3579 28.3579 26.52 . p 10.2971 1.4 2.82 0.434 0.4050 0.40s0 Total 1 .0500 Nadel Saasty 72.624 69.88% 61.86 Coet t coef T-Vaiue P-Value 19 10.50 0.000 ?ysinnangeted 35, ?? 6.36 5.25 0.0001-co Regreasiom Equarion veighe gain 12.51+35.83 lysise ingeated a. Lysine ingested ranges from .09 to .23 grams. What is the predicted weight gain for .20 grams of lysine ingested? b. Consider a hypothesis test 1,:A-0 vs Ha:A#0. what is the p-value for the test? conclusion: There s(is/is not) sufficient evidence to conelude B, is not 0 c. What is the Pearson correlation coefficient? d. What is the 95t prediction interval for an individual y-value at x.20.Explanation / Answer
Here dependent variable is weight gain and independent variable is lysine ingested.
Given that mean of X = 0.166
From the given output the regression equation is,
Y = 12.51 + 35.63*x
Now here we have to find y when x = 0.20
Y = 12.51 + 35.63*0.20 = 19.636
Now here we have to test,
H0 : B1 = 0 Vs H1 : B1 not= 0
Where B1 is population slope for independent variable.
Assume alpha = level of significance = 5% = 0.05
The test statistic = 5.15
P-value = 0.000
P-value < alpha
Reject H0 at 5% level of significance.
Conclusion : There is sufficient evidence to say that B1 is not 0.
Pearson correlation coefficient = sqrt(Rsq) = sqrt(72.62%) = sqrt(0.7262)
Correlation = 0.8522
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