The article \"The Influence of Temperature and Sunshine on the Alpha-Acid Conten
ID: 3240062 • Letter: T
Question
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 (x_1), and mean percentage of sunshine during the same period (x_2) 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) (a) According to the output, what is the least squares regression equation = b_0+b_1+b_2x_2: (Round each value to 3 decimal places.) = + x_2 + x_2 (b) What is the estimate for sigma? S = (c) According to the model what is the predicted value for whenx_1 = 21.3 and x_2 = 43 and what is the corresponding residual (Round your answers to four decimal places.) (d) Test H_0: beta_1 = beta_2 = 0 versus H_a: either B_1 or beta_2 notequalto 0. From the output state the test statistic and the p-value. Round your test stat to one decimal place and your p-value to 4 decimal places.) State the conclusion in the problem context. There is slightly suggestive evidence at least one of the explanatory significant predictor of the response. There is convincing evidence at least one of the explanatory variables significant predictor of the response. There is moderately suggestive evidence at least one of the explanatory significant predictor of the response. There is no suggestive evidence at least one of the explanatory variables is a significant predictor of the response. (e) The estimated standard deviation of when x_1 = 21.3 and x_2 = 43 is s_1 = 16.58. Use this to obtain the 95% CI for mu r 213.40 (Round your answers to two decimal places.) (middot) (f) Use the information in parts (b) and (e) to obtain a 95% PI for yield in a future experiment when x_1 = 21.3 and x_2 = 43. (Round your answers to two decimal places.) (g) Given that x_2 is in the model, would you retain x_1 Yes, there is evidence this factor is significant. It should remain in the model. No, there isn't evidence this factor is significant. It should be dropped from the model.Explanation / Answer
Answer:
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)
R output: Call:
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
a).
y=415.113 +(-6.593)x1 +(-4.504)x2
b). s=24.45
c).
y =415.113 +(-6.593)*21.3 +(-4.504)*43 =81.0101
residual =68-81.0101 = -13.0101
d). F=14.9
P=0.0014
There is convincing evidence at least one of the explanatory variables is significant predictor of the response.
e).
newdata=data.frame(x1=21.3, x2=43)
# confidence interval
predict(mod, newdata, interval="confidence")
#prediction interval
predict(mod, newdata, interval="predict")
confidence interval
> predict(mod, newdata, interval="confidence",level=.95)
fit lwr upr
1 81.03364 43.52379 118.5435
95% CI = (43.52, 118.54)
f).
#prediction interval
> predict(mod, newdata, interval="predict",level=.95)
fit lwr upr
1 81.03364 14.19586 147.8714
95% PI=(14.20, 147.87)
g).
No, there is not evidence this factor is significant. It should be dropped from the model.
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