A research firm collected data on a sample of n = 30 drivers to investigate the
ID: 3303900 • Letter: A
Question
A research firm collected data on a sample of n = 30 drivers to investigate the relationship between the age of a driver and the distance the driver can see. (The data is attached) PLEASE USE SAS TO FIND THE ANSWERS AND EXPLAIN WHERE THEY CAME FROM ON THE SAS OUTPUT.
d) Calculate R2? Explain what this means, and comment on whether or not it suggests the model is good.
e) Calculate the correlation coefficient? Explain what this means, and comment on whether or not it suggests the model is good.
f) What would you expect the distance that a 50 year old driver can see to be?
g) Give a 95% prediction interval for the distance that an individual 50 year old can see.
h) Give a 95% confidence interval for the mean distance that 50 year olds can see.
Explanation / Answer
Descriptive Statistics
Mean Std. Deviation N
Distance 4.2333E2 81.72002 30
Age 50.7000 21.43659 30
a)
Coefficients(a)
Unstandardized Coefficients Standardized Coefficients 95% Confidence Interval for B
Model B Std. Error Beta t Sig. Lower Bound Upper Bound
1 (Constant) 576.769 24.054 23.978 .000 527.497 626.042
Age -3.026 .438 -.794 -6.908 .000 -3.924 -2.129
a. Dependent Variable: Distance
The regression line is Y = 576.769-3.026X
b)
Model Summary
Change Statistics
Model R R Square Adjusted R Square Std. Error of the Estimate R Square Change
1 .794a .630 .617 50.57318 .630
a. Predictors: (Constant), Age
therefore estimate of standard deviation of error is 50.57318
c) Null hypothesis H0:B1=0
alternative hypothesis H1:B1not= 0
test statistic value is t = 6.908
significant value = 0.000
p-value = 0.00001
the result is significat at p(0.0001)<0.05
therefore the model is useful
d) the R2 value = 0.630
here R2 change is also 0.630 therefore there is no error. Hence the model is good fit
e)
Correlations
Distance Age
Pearson Correlation Distance 1.000 -.794
Age -.794 1.000
Sig. (1-tailed) Distance . .000
Age .000 .
therefore the correlation coefficient between them is -0.794
f) the distance at 50 year old is Y = 576.769-3.026(50) = 425.469
g) the confidence interval at 95% is (527.497, 626.042)
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