Answer....

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 (x7) 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 * 2 30 42 47 47 43 41 48 44 43 50 56 60 210 110 103 103 91 76 73 70 68 53 45 31 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+byx1+b2x2: (Round each value to 3 decimal places.) 9 = 415.113 + 6.593 v x1+ 4.504 * 2 (b) What is the estimate for o? s = 24.45 (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 = 17.1 and x2 = 41 and what is the corresponding residual? (Round your answers to four decimal places.) V = x Residual: y - y = X (d) Test Ho: $1 = 82 = 0 versus Ha: either #1 or 82 = 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. O There is no suggestive evidence at least one of the explanatory variables is a significant predictor of the response. O 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 x] = 17.1 and x2 = 41 is so = 9.76. Use this to obtain the 95% CI for My . 17,1, 41- (Round your answers to two decimal places.) X 1x ) (1) Use the information in parts (b) and (e) to obtain a 95% PI for yield in a future experiment when x1 - 17.1 and x2 - 41. (Round your answers to two decimal places.) X