Summary Output
Regression Statistics
Multiple R
0.990069
R Square
0.980237
Adjusted R Square
0.977766

Standard Error
0.096067
Observations
10
ANOVA
df
SS
MS
F
Significance F
Regression
1
3.66198
3.66198

8
0.0000
Residual
8
0.07383
0.45775
Total
9
3.73581
Coefficients
Standard Error
t Stat
P-value
Intercept
?
0.072432
11.77961
0.0000
x
?
0.001442
19.91967

0.00000
I. A small sample data is collected for two variables. Y represents the price of all gas types per
gallon (Bureau of Labor Statistics data), and X represents the average price of crude oil per
barrel (Department of energy data), for a period of ten years.
1. Is the linear model useful for the prediction of price of gas? Test using %u03B1 =0.05. Use
critical value approach.
2. Is the linear model useful for the prediction of price of gas? Test using %u03B1 =0.05. Use p-
value approach.
3. Conduct an F test (Critical value approach) to test the usefulness of the model. Use %u03B1
=0.05.
4. Conduct an F test (p-value approach) to test the usefulness of the model. Use %u03B1 =0.05.
5. Did you come up with the same conclusion in all previous parts? Is it possible to get
conflicting answers? Explain.
6. Develop a 95% confidence interval for the slope of the model. Interpret the meaning of the
interval in the context of the problem.
7. Develop a 90% confidence interval for the slope of the model. Compare it to the answer you
got in the precious part. Which interval is wider? Does it make sense? Explain.
Summary Output

Regression Statistics
Multiple R
0.990069
R Square
0.980237
Adjusted R Square
0.977766
Standard Error
0.096067
Observations
10

ANOVA
df
SS
MS
F
Significance F
Regression
1
3.66198
3.66198
8
0.0000
Residual
8
0.07383
0.45775
Total
9
3.73581

Coefficients
Standard Error
t Stat
P-value
Intercept
?
0.072432
11.77961
0.0000
x
?
0.001442
19.91967
0.00000
Solution
a) yes, beacuse all the coefficients are significant( t value calculates > t table) and F test is also
showing that model is statistically significant( F calculated > F table).So it linear model is useful
b) Since P < 0.05 , so model is significant.So it linear model is useful
3) F calculated = 8 ,F table(1,8) =5.32
and F calculated > F table
therefore it is siginicant and model is useful
4) null hypothseis H0 : model not correct
H1: model is correct
if P < alpha(0.05)

then model is significant
and here p =0
so model is significant
5) yes conclusion is same.In both the cases model is significant
6) (slope value)/ sd = t value of slope coefficient
slope value/0.00142 = 19.9167
slope value = 0.02828
confidence interval ( (0.02828 - 1.96*0.00142/sqrt(n) ,(0.02828 + 1.96*0.00142/sqrt(n)
here n = 9 ( no of observation)
confidence interval(0.019003,0.037557)
if slope value lies in this interval then the value is insignificant otherwise significant
7) at 90% Z critical Z value = 1.65
confidence interval ( (0.02828 - 1.65*0.00142/sqrt(n) ,(0.02828 + 1.65*0.00142/sqrt(n)
here n = 9 ( no of observation)
confidence interval( 0.027499,0.029061)
95 % has wider confidence interval.Because more the significance level the less is the
confodence interval.So initialy it has 5% significance level and later it has 10%significance
level.Soconfidence interval is more in 1st case