Please answer question C, E, F and show steps. The article \"The Influence of Te
ID: 3254974 • Letter: P
Question
Please answer question C, E, F and show steps.
The article "The Influence of Temperature and Sunshine on the Alpha-Acid Contents of Hops"† reports the following data on yield (y), mean temperature over the period between date of coming into hops and date of picking (x1), and mean percentage of sunshine during the same period (x2) for the Fuggle variety of hop:
Use the following R Code to complete the regression analysis:
x1 = c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)
x2 = c(30,42,47,47,43,41,48,44,43,50,56,60)
y = c(210,110,103,103,91,76,73,70,68,53,45,31)
mod = lm(y~x1+x2)
summary(mod)
(b) What is the estimate for ? s = 24.45 correct
(_____, _____)
(f) Use the information in parts (b) and (e) to obtain a 95% PI for yield in a future experiment when x1 = 18.4 and x2 = 47. (Round your answers to two decimal places.)
(_____, _____)
Residual: y - = ______???Explanation / Answer
Answer:
The article "The Influence of Temperature and Sunshine on the Alpha-Acid Contents of Hops"† reports the following data on yield (y), mean temperature over the period between date of coming into hops and date of picking (x1), and mean percentage of sunshine during the same period (x2) for the Fuggle variety of hop:
Use the following R Code to complete the regression analysis:
x1 = c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)
x2 = c(30,42,47,47,43,41,48,44,43,50,56,60)
y = c(210,110,103,103,91,76,73,70,68,53,45,31)
mod = lm(y~x1+x2)
summary(mod)
(b) What is the estimate for ? s = 24.45 correct
(c) According to the model what is the predicted value for y when x1 = 18.4 and x2 = 47 and what is the corresponding residual? (Round your answers to four decimal places.)
Residual: y - y = 103-82.13845 =20.86155
=20.8616
(e) The estimated standard deviation of y when x1 = 18.4 and x2 = 47 is sy = 7.23. Use this to obtain the 95% CI for Y · 18.4, 47. (Round your answers to two decimal places.)
( 65.77, 98.50)
(f) Use the information in parts (b) and (e) to obtain a 95% PI for yield in a future experiment when x1 = 18.4 and x2 = 47. (Round your answers to two decimal places.)
=(24.45, 139.83)
R output:
lm(formula = y ~ x1 + x2)
Residuals:
Min 1Q Median 3Q Max
-41.730 -12.174 0.791 12.374 40.093
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 415.113 82.517 5.031 0.000709 ***
x1 -6.593 4.859 -1.357 0.207913
x2 -4.504 1.071 -4.204 0.002292 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 24.45 on 9 degrees of freedom
Multiple R-squared: 0.768, Adjusted R-squared: 0.7164
F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395
> newdata=data.frame(x1=18.4, x2=47)
> # confidence interval
> predict(mod, newdata, interval="confidence" , level=.95)
fit lwr upr
1 82.13845 65.77307 98.50383
> #prediction interval
> predict(mod, newdata, interval="predict", level=.95)
fit lwr upr
1 82.13845 24.44855 139.8284