e s. A random sample of 12 2-week old chickens as used to examine the relationah
ID: 3375854 • Letter: E
Question
e s. A random sample of 12 2-week old chickens as used to examine the relationahip between weight gain (grams) and anount of lysine ingested (grams). The amount of lysine ingested is the explanatory variable (X) and the weight gain tx) is the response variable. The mean of the explanatory variable is o.166. 1inear regression model was fitted to the data. The results are shownm below. Round al1 answers to three decimal places it necessary Regression Analysis: weight gain versus lysine ingested Source sine ingested 28.3579 2.35926.52 0.0o0 2871 1.1435 2.02 0.434 Lack-ot-TLE .0500 Rodel Saasty .03403 72-6269.8 beetficienta Regression Equatien weight gnin 12.S1+35.63 lysine ingested a. Lysine ingested ranges from .09 to 23 grans. What is the predicted weight gain for .20 grams of lysine ingested? b. Consider a hypothesis test H0A-0 vs H.:gz0. what is the p-value for the test? Conclusion: There (is/is not) sufficient evidence to conclude A is not 0 c. What is the Pearson correlation coefficient? d, What is the 95% prediction interval for an individual y-value at x- .20Explanation / Answer
Part (a)
From the regression analysis summary given, the equation of estimated regression line is:
ycap = 12.31 + 35.83x.
Substituting in the above equation, x = 0.2, predicted weight gain at 0.2 gm of lysine ingested
= 12.31 + (35.83 x 0.2) = 19.48 grams ANSWER
Part (b)
From the regression analysis summary given, against the ‘lysine ingested’ coefficient, t-value is5.15 which is the test statistic for testing ?1 = 0. Also given is that the p-value = 0.000 indicating it is very small. So, null hypothesis is rejected. Thus,
There is sufficient evidence to suggest that ?1 is NOT zero. ANSWER
Part (c)
Pearson correlation coefficient = 0.8522 ANSWER [vide the given analysis summary, rsqure = 0.7262]
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