Use the following information to answer the next three questions: A rankoan sanu

Question

Use the following information to answer the next three questions: A rankoan sanuple of vight drivers insured with a coupany and having similar auto insurance policies w selected. The folhowing table lists their driving experiences (in years)(x) and monthly auto insurasice premiuas () 5 2 12 9 15 6 25 16 Driving Experience (sears) (X) Monthly Auto Insurance Premiun(inS) (Y) 64 87 50 71 44 56 42 60 It is desired to undertake a study to relate Mouthly Auto Insurance Premium (Y) to Driving Experience (sears)(X). To do this. regression of Y on X is used and a partial Excel output is MS F Signilicance F Regression 1 918.493 918.493 8.624 0.0261 Residual Total 63.007 106.501 7 1557.500 Coefficients Standard Frror t Stat P-value Intercept76.660 1.347 6.961 0,527 11.012 3.33384E-05 2.937 0.026058804 6. The value and the the interpretation of R2 are: (a)R2 = 58.97%, about 58.97% of the variation in Driving Experience can be explained by Monthly Auto Insurance Premium Ra 58.97%. about 58.97% of the variation in Monthly Auto Insurance Pre nium can be explained by Driving Experience. ( R2-58.97%. More than 50% on the variability in in Driving Experience can be explained by Monthly Auto Insurance Premium. (d) None of the above. 7. The least squares line is given ly (a) Monthly Auto Insurance Premium =-1.547 + 76.600Driving Experience Monthly Auto Insurance Premium = 76660 1.547Driving Experience (c) Driving Experience =-1.547 + 76.66°Monthly Auto Insurance Premium (d) Driving Experience 76.660 -1.347Mouthly Auto Insurance Premium 8. (5pt) The cotrelation coefficieut in this case is a) r= 0.5897 b) r 0.7081 r-0.7681 d) r=-5807

Explanation / Answer

Sol:

Executed in R:code:

Driving_exp <- c(5,2,12,9,15,6,25,16)
MonthlyAutoins <- c(64,87,50,71,44,56,42,60)
model1 <- lm(MonthlyAutoins~Driving_exp)
summary(model1)
cor(MonthlyAutoins,Driving_exp)

Output:

model1 <- lm(MonthlyAutoins~Driving_exp)

> summary(model1)

Call:

lm(formula = MonthlyAutoins ~ Driving_exp)

Residuals:

Min 1Q Median 3Q Max

-11.3748 -8.4286 -0.4465 8.1428 13.4348

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 76.660 6.961 11.012 3.33e-05

Driving_exp -1.548 0.527 -2.937 0.0261

(Intercept) ***

Driving_exp *  

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Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.32 on 6 degrees of freedom

Multiple R-squared: 0.5897, Adjusted R-squared: 0.5213

F-statistic: 8.624 on 1 and 6 DF, p-value: 0.02606

-0.7679342

Answers:

SOLUTION6:

R sq=Multiple R-squared: 0.589

58.,9% variation in monthly insurance is covered by model

OPTIONB

SOLUTION7:

FROM COEFFICIENTS TABLE

LEAST SQUARES LINE IS

monthly auto insurance premium=76.660-1.547(drivers exp)

OPTIONB

SOLUTION8

r=-0.767931

there exists a negative relationship between monthly insurance and drivers exp.

OPTIONC

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