Question

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Test the null hypothesis that the slope is zero versus the two-sided alternative in each of the following settings using the  α = 0.05  significance level.

(a)n = 20, ŷ = 24.5 + 1.6x, and SE_b1 = 0.75

(b)n = 30, ŷ = 30.1 + 2.7x, and SE_b1 = 1.35

(c)n = 100, ŷ = 29.4 + 2.7x, and SE_b1 = 1.35

Answer 1

a )

n = 20 , SE_b1 = 0.75

and regression equation is ,

, ŷ = 24.5 + 1.6x

Where , 24.5 is y intercept and 1.6 is estimate of slope ( b1 )

Hypothesis :

Two tailed test.

{ Where , is slope }

Test statistic :

Test statistic t follows student t distribution with df = n - 2

P-value :

P-value for this two tailed test is ,

Using Excel function , =TDIST( t , df , tails )

df = n - 2 = 20 - 2 = 18 , tails = 2

So, P-value = TDIST (2.1333, 18 , 2 ) = 0.0469

P-value = 0.0469

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level .

Given , = 0.05

It is observed that p-value ( 0.0469 ) is less than 0.05

So reject null hypothesis.

Conclusion :

There is no sufficient evidence to conclude that slope is zero.

b)

n = 30 , SE_b1 = 1.35

and regression equation is ,

ŷ = 30.1 + 2.7x

Where , 30.1 is y intercept and 2.7 is estimate of slope ( b1 )

Hypothesis :

Two tailed test.

{ Where , is slope }

Test statistic :

Test statistic t follows student t distribution with df = n - 2

P-value :

P-value for this two tailed test is ,

Using Excel function , =TDIST( t , df , tails )

df = n - 2 = 30 - 2 = 28 , tails = 2

So, P-value = TDIST (2, 28 , 2 ) = 0.0553

P-value = 0.0553

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level .

Given , = 0.05

It is observed that p-value ( 0.0553) is greater than 0.05

So, fail reject null hypothesis.

Conclusion :

There is sufficient evidence to conclude that slope is zero.

c)

n = 100 , SE_b1 = 1.35

and regression equation is ,

ŷ = 29.4 + 2.7x

Where , 29.4 is y intercept and 2.7 is estimate of slope ( b1 )

Hypothesis :

Two tailed test.

{ Where , is slope }

Test statistic :

Test statistic t follows student t distribution with df = n - 2

P-value :

P-value for this two tailed test is ,

Using Excel function , =TDIST( t , df , tails )

df = n - 2 = 100 - 2 = 98 , tails = 2

So, P-value = TDIST (2, 98 , 2 ) =

P-value = 0.0483

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level .

Given , = 0.05

It is observed that p-value ( 0.0483) is less than 0.05

So, reject null hypothesis.

Conclusion :

There is no sufficient evidence to conclude that slope is zero.