Question

We are interested in whether math score (math – a continuous variable) is a significant predictor of science score (science – a continuous variable) using the High School and Beyond (hsb2) data.

State the null and alternative hypotheses and the level of significance you intend to use.

Ho:β=0

H1:β≠0

Alph:0.05

Write the equation for the appropriate test statistic.

t =b/SE(b)

What is your decision rule? Be sure to include the degrees of freedom.

If our t value is greater than the critical value of 1.96 we reject the null hypothesis.

FD= n-2=200-2= 198=1.96

Using SAS, estimate the means, variances and covariances for math and science scores. Copy and paste the relevant SAS output below.

variable	label	DF	Peramieter Estimate	Standered Error	tvalue	Pr>\t\	95% CI
intercept	intercept	1	21.7	2.75	7.88	<0.001	16.26,27.13
science	science score	1	0.596	0.052	11.44	<0.001	0.49,0.69

Using the output from (d), calculate by hand the slope. Be sure to show your work.

Using the output from (d), calculate by hand the intercept. Be sure to show your work

Answer 1

In the given SAS output above the value of slope b1 is 0.596 and value of intercept b0 is 21.7

( To calculate slope and intercept by hand , we need value of mean , variances and covariances for math(y) and science scores(x))

Test for slope :

Ho:β=0

H1:β≠0

: 0.05

t =b₁/SE(b₁)

b₁ is slope value = 0.596 and SE(b₁) is standard error of slope = 0.052 ...(From the SAS output )

t = 11.46

Now we need to find critical value , we have degrees of freedom (d.f) = 200-2 = 198

and : 0.05 .

So we have to use excel function =TINV( , d.f ) to find the critical value.

=TINV( 0.05,198) = 1.97

Therefore critical value = 1.97

Decicion rule : Reject H0 if | t | > critical value or

Fail to H0 , if |t| < critical value

As t ( 11.46 ) is greater than (>) critical value (1.97 ) , we reject the null hypothesis H0.

Conclusion : We have significant evidence that slope β is different that 0 at 0.05 level of significance.