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During a research, the amount of Internet users was measured. Each time three random groups of 10,000 people of the average a

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Answer #1

Solution :-

y = c(1300,1050,140,2900,2400,490,4350,4010,710,6090,5530,1120)

x1 = c(1,1,1,3,3,3,5,5,5,7,7,7)

x2 = c(20,40,60,20,40,60,20,40,60,20,40,60)

mlr = lm(y~x1+x2)

summary(mlr)


## Output 

> summary(mlr)

Call:
lm(formula = y ~ x1 + x2)

Residuals:
    Min      1Q  Median      3Q     Max 
-1566.5  -632.7   158.1   558.9  1321.0 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)  3283.83     888.40   3.696  0.00495 **
x1            567.17     124.40   4.559  0.00137 **
x2            -76.13      17.03  -4.469  0.00156 **
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 963.6 on 9 degrees of freedom
Multiple R-squared:  0.8191,    Adjusted R-squared:  0.7789 
F-statistic: 20.38 on 2 and 9 DF,  p-value: 0.0004552

(a) y hat =  3283.83 + (567.17) * x1 + ( -76.13 ) * x2

(b)  Residual standard error : 963.6 on 9 degrees of freedom

SSRes = 963.6

=> sigma2 hat = SSRes / df = 963.6 / 9 = 107.066666667

So, sigma2 hat = 107

(c) se(intercept) = 888.40

se( beta1 hat ) = 124.40

se( beta2 hat ) = 17.03

(d) Given that : x1 = 4 ; x2 = 40

=> Number of users = 3283.83 + (567.17) * 4  + ( -76.13 ) * 40 = 2507.31

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