Question

The article "The Influence of Temperature and Sunshine on the Alpha-Acid Contents of Hops"t 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: 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 X2 48 30 42 47 47 43 41 44 43 50 56 60 76 73 210 110 103 103 91 70 68 53 45 31 y 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 Im(yx1+x2) summary(mod) the least squares regression equation y bo+b1x1+b22: (Round each value to 3 decimal places.) (a)According to the output, what X1+ X2 (b) What is the estimate for o? s (Hint: This is referred to as the residual standard error in output) (c) According to the model what is the predicted value for ý when X1 = 21.3 and x2 43 and what is the corresponding residual? (Round your answers to four decimal places.) Residual: y V = (d) Test Ho: B1 = P2 = 0 versus Ha: either B1 or B2 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.) p-value= State the conclusion in the problem context There is moderately suggestive evidence at least one of the explanatory variables is a significant predictor of the response. OThere is no 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 significant predictor of the response. OThere 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 = 21.3 and x2 43 is sy 16.58. Use this to obtain the 95% CI for uy. 21.3, 43- (Round your answers to two decimal places.) (f) Use the information in parts (b) and (e to obtain a 95% PI for yield in a future experiment when x1 = 21.3 and x2 = 43. (Round your answers to two decimal places.) (g) Given that x is in the model, would you retain x1? OYes, there is evidence this factor is significant. It should remain in the model. ONo, there isn't evidence this factor is significant. It should be dropped from the model. You may use the f table at: http://www.stat.purdue.edu/jtroisi/STAT350Spring2015/tables/FTable.pdf

Answer 1

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)

predict(mod,data.frame(x1=21.3,x2=43),interval="confidence")
predict(mod,data.frame(x1=21.3,x2=43),interval="prediction")

> summary(mod)

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

> 
> predict(mod,data.frame(x1=21.3,x2=43),interval="confidence")
       fit      lwr      upr
1 81.03364 43.52379 118.5435
> predict(mod,data.frame(x1=21.3,x2=43),interval="prediction")
       fit      lwr      upr
1 81.03364 14.19586 147.8714

a)

y^ = 415.113 -6.593 x1 -4.504 x2
b)
24.45

c)
y^ = 81.0336

residual = yi - y^
= 68 - 81.0336
= -13.0336

d)
F = 14.9
p-value = 0.0014
conclusion
option D) convincing evidence

e)
(43.52,118.54)

f)
(14.2,147.87)