Question

Consider the following data to be used in a regression. xi yi 1 0 2 10 3 25 4 30 5 35 (a) Find the values of b0 and b1. (b) Find the Coefficient of Determination. (c) Find the estimated standard deviation of b1 and the corresponding t-statistic. At the 1% level of significance, can you reject the null hypothesis? Make sure you state the null and alternative hypotheses. (d) Find the F-statistic. Is the equation significant at the 1% level? Make sure you state the null and alternative hypotheses. Use the p-value approach.

Answer 1

following data to be used in a regression.

we will use R-software to obtaine regression output

{ note if any Software output is not required only mannual output is required , then you can tell it in comment box }

Now regresssion model is

y = b0 + b1 x

where estimate of b0 and b1 are given by

= cov(x,y) / var(x)

= - *

where mean of yi and is mean of xi

From R

First we will import data into R

> x=c(1,2,3,4,5)
> y=c(0,10,25,30,35)
> data.frame(x,y)
x    y
1 1 0
2 2 10
3 3 25
4 4 30
5 5 35

# to fit regreesion model in R we use command " lm() "

> fit=lm(y~x)
> summary(fit)

Call:
lm(formula = y ~ x)

Residuals:
1 2 3 4 5
-2 -1 5 1 -3

Coefficients:
             Estimate Std. Error   t value Pr(>|t|)
(Intercept)   -7.000     3.830 -1.828    0.1650
x               9.000     1.155    7.794    0.0044 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.651 on 3 degrees of freedom
Multiple R-squared : 0.9529,    Adjusted R-squared: 0.9373
F-statistic: 60.75 on 1 and 3 DF, p-value: 0.004395

(a) Find the values of b0 and b1. (b)

Thus estimate of b0 is   ,   = -7.0

           estimate of b1 is   ,   = 9.0

(b) Find the Coefficient of Determination.

The coefficient of determination, R2, is used to analyze how differences in one variable can be explained by a difference in a second variable.

From R- output

Multiple R-squared : 0.9529

Thus Coefficient of Determination is 95.29% ( or 0.9529 )

{

General formula to find coefficient of determination R2 is

R2   = ( r )2

where r = cov(x,y) / ( x * y )

here x is standard deviation of xi and y is standard deviation of yi

> r = cov(x,y)/ (sd(x)*sd(y))            # corelation between xi and yi
> r
[1] 0.9761871
> R2=r^2
> R2            # Coefficient of Determination
[1] 0.9529412

}

(c) Find the estimated standard deviation of b1 and the corresponding t-statistic. At the 1% level of significance, can you reject the null hypothesis

From R Output we have obtained this-

Coefficients:
             Estimate Std. Error   t value Pr(>|t|)
(Intercept)   -7.000     3.830 -1.828    0.1650
x               9.000     1.155    7.794    0.0044 **

Now standard error of b1 and the corresponding t-statistic 1.155

Let standard deviation -SD , standard error - SE

Now standard error SE = SD /

here n = 5

Thus SD = SE * = 1.155 * = 2.582659

Thus estimated standard deviation of b1 and the corresponding t-statistic 2.582659

Now the null and alternative hypotheses are

H0 : b1 = 0              ( term b1 do not contributes significantly to the model)

H1 : b1 0            ( term b1 contributes significantly to the model)

Here Test Statistics = = = 7.792208

Thus Test Statistics = 7.792208

We reject null hupothsis if Test Statistics |TS | value is greater than

Here     = =

i.e t-distribution with n-(1+1) = 5-2 = 3 degree of freedom an at 1 % of level of significance

Now t-distributed t-table can be obtaine from statistical book or from any software like R

From R

> qt(1-0.01/2,3)         #t-distribution t-table value
[1] 5.840909

Here Test Statistics |TS | value = 7.792208

and t-table value = 5.840909

Thus   |TS |   > t-table value

So we reject null hypothesis at 1% level of significance

At the 1% level of significance, we reject the null hypothesis, and hence we conclude that term b1 contributes significantly to the model .

(d) Find the F-statistic. Is the equation significant at the 1% level? Make sure you state the null and alternative hypotheses. Use the p-value approach.

Null and Alternative hypothesis are

H0 : bj = 0              ( given model is not significantly )

H1 : bj 0            ( given model is significantly )

We have obtain ANOVA in R-Output which was as follow

> anova(fit)
Analysis of Variance Table

Response: y
         Df    Sum Sq Mean Sq F value    Pr(>F)
         x          1    810    810.00      60.75 0.004395 **
Residuals 3    40          13.33                  
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

F-statistics is given by

F-statistics = MSR / MSRes = ( 810 / 1) / (40 / 3 ) = 810 / 13.33 = 60.75

Thus

F-statistics = 60.75

Using P-value Approch

Also Given P-value is 0.004395376          { Pr(>F) = 0.004395 ** }

We reject null hypothsis is P-value is less than 0.01

{ here 0.01 because we are given 1% level of significance }

Here P-value =0.004395376 < 0.01                   

Thus we reject null hypothesis at 1% of level of significance .

And hecne conclude that model is significant .