Question

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The article "The Influence of Temperature and Sunshine on the Alpha-Acid Contents of Hops"t reports the following data 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: 21.3 X1 16.7 17.4 18.4 16.8 18.9 17.1 17.3 18.2 21.2 20.7 18.5 30 47 43 48 50 56 60 X2 42 47 41 44 43 210 110 103 103 91 76 73 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(y~x1+x2) summary(mod) (a)According to the output, what is the least squares regression equation y = bo+61X1+b2x2: (Round each value to 3 decimal places.) + X1+ X2 (b) What is the estimate for o? s = (Hint: This is referred to as the residual standard error in R output) (c) According to the model what is the predicted value for y when x1 = 17.3 and x2 = 48 and what is the corresponding residual? (Round your answers to four decimal places.) Residual: y - y = (d) Test Ho: Bı = P2 = 0 versus Ha: either ß1 or ß2 # 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.) f = p-value = State the conclusion in the problem context. significant predictor of the response There is no suggestive evidence at least one of the explanatory variables is a There is slightly 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 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 = 17.3 and x2 = 48 is sy = 10.13. Use this to obtain the 95% CI for /uy 17.3, 48. (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 = 17.3 and x2 = 48. (Round your answers to two decimal places.) (g) Given that x2 is in the model, would you retain X1? 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 You may use the f table at: http://www.stat.purdue.edu/~jtroisi/STAT350Spring2015/tables/FTable.pdf

Answer 1

(a)

On running the given code, the summary output is,

> 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)

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

The estimated regression equation is,

$\hat{y}$ = 415.113 + -6.593 x1 + -4.504 x2

(b)

Estimate of $\sigma$ is,

24.45

(c)

For x1 = 17.3, x2 = 48, the predicted y is,

$\hat{y}$ =  415.113  - 6.593 * 17.3 - 4.504 * 48 = 84.8621

y = 73

Residual = y - $\hat{y}$ = 73 - 84.8621 = -11.8621

(d)

From the summary output,

f = 14.9

P-value = 0.0014

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and

There is convincing evidence that at least one of the explanatory variables is a significant predictor of the response.

(e)

The degree of freedom of residual standard error is 9

Critical t value for 95% confidence interval and df = 9 is 2.262

Margin of error = t * sd = 2.262 * 10.13 = 22.91406

Estimated $\hat{y}$ = 84.8621

95% CI is,

(84.8621 - 22.91406,  84.8621 + 22.91406)

(61.95, 107.78)

(f)

Residual standard error $\sigma$ = 24.45

Margin of error for 95% PI = t * $\sigma$
11/n V
= 2.262 * 24.45 *
11/12 V
= 57.56421

Estimated $\hat{y}$ = 84.8621

95% PI is,

(84.8621 - 57.56421,  84.8621 + 57.56421)

(27.30, 142.43)

*Note that, the answers can differ because of rounding issue.

(g)

Since, p-value for x1 (0.207913) is greater than 0.05 significance level, x1 is not significant in the model. Thus,

No, there isn't evidence this factor is significant. It should be dropped from the model.