The article \"The Influence of Temperature and Sunshine on the Alpha-Acid Conten
ID: 3060863 • Letter: T
Question
The article "The Influence of Temperature and Sunshine on the Alpha-Acid Contents of Hops"t reports the following data on yield ( (xi), and mean percentage of sunshine during the same period (x2) for the Fuggle variety of hop mean temperature over the period between date of coming into hops and date of picking X1 16.7 17.4 18.4 16.8 18.9 17.1 17.3 18.2 21.3 21.2 20.7 18.5 30 42 47 47 43 4 484443 5056 60 210 110 103 103 91 76 73 70 68 53 45 31 x2 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(yMX1+x2) summary(mod) (a)According to the output, what is the least squares regression equation ybo+bix1+b2x2: (Round each value to 3 decimal places.) y = 415 114 -+-6.593 x1+-4.504 -V x2 (Hint: This is referred to as the residual standard error in R output) (b) what is the estimate for ? s-24.45 (c) According to the model what is the predicted value for y when X1 = 18.9 and X2 = 43 and what is the corresponding residual? (Round your answers to four decimal places.) Residual: y-y= 117.3374 X y 92.6626X (d) Test Ho: A = 2-0 versus Ha: either 1 or -0. From the output state the test statistie and the p-value. Round your test stat to one decimal place and your p-value to 4 decimal places.) f= 14.90 p-value= 0.0014 State the conclusion in the problem context. There is no suggestive evidence at least one of the explanatory variables is a significant predictor of the response. There is moderately suggestive evidence at least one of the explanatory variables is a significant predictor of the response There is slightly suggestive evidence at least one of the explanatory variables is a significant predictor of the response There is convincing evidence at least one of the explanatory variables is a significant predictor of the response (e) The estimated standard deviation of y when X1 = 18.9 and X2 = 43 is st = 8.20. Use this to obtain the 95% CI for . 18.9, 43. (Round your answers to two decimal places.) 2.26 X11.45 (f) Use the information in parts (b) and (e) to obtain a 95% PI for yield in a future experiment whenX1 18.9 and x2-43. (Round your answers to two decimal places.) 37.35 14797xExplanation / Answer
Result:
c).
predicted= 96.8563
residual: -5.8563
e). 95% CI = (78.30, 115.41)
f). 95% PI = (38.51, 155.20)
R code:
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)
mydata <- data.frame(x1,x2,y)
lmfit <-lm(y~x1+x2, data = mydata)
summary(lmfit)
anova(lmfit)
confint(lmfit)
newdata = data.frame(x1=18.9,x2=43)
predict(lmfit, newdata, interval="confidence")
predict(lmfit, newdata, interval="prediction")
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)
R output;
Call:
lm(formula = y ~ x1 + x2, data = mydata)
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
> anova(lmfit)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x1 1 7245.5 7245.5 12.116 0.006930 **
x2 1 10571.2 10571.2 17.677 0.002292 **
Residuals 9 5382.2 598.0
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> confint(lmfit)
2.5 % 97.5 %
(Intercept) 228.445660 601.780336
x1 -17.585176 4.399619
x2 -6.926686 -2.080438
> newdata = data.frame(x1=18.9,x2=43)
> predict(lmfit, newdata, interval="confidence")
fit lwr upr
1 96.85631 78.30315 115.4095
> predict(lmfit, newdata, interval="prediction")
fit lwr upr
1 96.85631 38.50805 155.2046
>
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