(a). Compute the estimated standard error of the model s (b). If appropriate, pr
ID: 3276468 • Letter: #
Question
(a). Compute the estimated standard error of the model s
(b). If appropriate, predict the yield when fertilizer=10. If inappropriate, explain why.
(c). Conduct a hypothesis test to determine whether yield (y) increases as the amount of fertilizer (x) increases. Use = 0.1.
(d). Find a 95% confidence interval for 1and interpret it.
(e). Find a 95% prediction interval for the yield of the plot with fertilizer=6.
(f). Compute coefficient of determination (R^2) and interpret it.
(g). Calculate the correlation coefficient (r)
Question 3. A researcher is looking for the relationship between yields of corn and amounts of fertilizer. The table below contains data from 10 plots. Plot Yield12 13 13 14 15 15 14 16 17 18 Fertilizer 2 233 4 45 5 66 1 23 45 6 7 8910 Note: SSzz-20, ss,,-23, ssy,-32.1. Least squares line is: y = 10.1 + 1.15xExplanation / Answer
a. Using formula, SSE=SSy/SS^2xy/SSx=32.1-(23)^2/20=5.65
Standard error of estimate, Se=sqrt (SSE/n-2)=sqrt (5.65/10-2)=0.84
b. Noting at the values of fertilizer, one can see that it doesnot include 10. Thus, th eregression equation is not obtained using the value. If the value of independent variable is not included in computation of regression equation or is not close to any of the values in the data set, the value if used to to cocmpute the response variable would result into extrapolation. Thus, predicting yield when fertilizer=10 is inappropriate.
c. The linear correlation coefficient, r=Sxy/sqrt (Sxx*Syy)=23/sqrt (20*32.1)=0.91
Assume p denote the population correlation coefficient for the variables yield and fertilizer. The hypotheses are as follows:
H0:p=0 (yield and fertilizer are linearly correlated)
H1:p=/=0 (yield and fertilizer are positively correlated)
Significance level is 0.01.
Test statistic: t=r/sqrt {(1-r^2)/(n-2)}=0.91/sqrt {(1-0.91^2)/(10-2)}=6.21
p value at n-2=10-2=8 degrees of freedom is: 0.0002.
Conclusion: Reject null hypothesis if p value is less than alpha=0.01. Here, p value is less than 0.01, therefore, reject null hypothesis and conclude that yield increases as amount of fertilizer increases.
d. The 95% confidence interval for beta1 is: b1+-talpha/2,n-2 (Se/sqrt SSxx)=1.15+-2.306(0.84/sqrt 20)=(0.7169,1.5831)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.