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An article 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: 30 42 47 47 43 41 48 44 43 50 56 60
y: 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)

Required:
According to the output, what is the least squares regression equation ลท = bo + b1x1 + b2x2:

Respuesta :

MrRoyal

Answer:

[tex]y = 415.113 -6.593 x_1 -4.504 x_2[/tex]

Step-by-step explanation:

Given

[tex]x_1: 16.7\ 17.4\ 18.4\ 16.8\ 18.9\ 17.1\ 17.3\ 18.2\ 21.3\ 21.2\ 20.7\ 18.5[/tex]

[tex]x_2: 30\ 42\ 47\ 47\ 43\ 41\ 48\ 44\ 43\ 50\ 56\ 60[/tex]

[tex]y: 210\ 110\ 103\ 103\ 91\ 76\ 73\ 70\ 68\ 53\ 45\ 31[/tex]

Required

Use R code to determine the regression equation

[tex]y = b_o + b_1x_1 + b_2x_2[/tex]

First, write the following code in a R program

[tex]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)[/tex]

[tex]x2 <- c(30,42,47,47,43,41,48,44,43,50,56,60)[/tex]

[tex]y <- c(210,110,103,103,91,76,73,70,68,53,45,31)[/tex]

[tex]mod <- lm(y[/tex]~[tex]x1+x2)\\[/tex]

[tex]summary(mod)[/tex]

Next, run the program

See attachment for program and output

From the output, go to coefficients:

Check Estimate Std. column, you have the following:

[tex](Intercept) = 415.113[/tex]

[tex]x_1 = -6.593[/tex]

[tex]x_2 = -4.504[/tex]

[tex]y = b_o + b_1x_1 + b_2x_2[/tex] implies that:

[tex]b_o \to[/tex] Intercept

[tex]b_1 \to x_1[/tex]

[tex]b_2 \to x_2[/tex]

Hence, the least square regression equation is:

[tex]y = 415.113 -6.593 x_1 -4.504 x_2[/tex]

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