Answer:
There is a sufficient evidence that the coefficient b1 is not zero
There is a sufficient evidence that the coefficient b2 is not zero
Step-by-step explanation:
Given
[tex]n = 30[/tex]
[tex]b_1 = 2.815[/tex]
[tex]Sb_1 = 0.75[/tex]
[tex]b_2 = -1.249[/tex]
[tex]Sb_2 = 0.41[/tex]
[tex]\alpha = 0.05[/tex]
Claim: Coefficient is zero
The null and alternate hypotheses are:
[tex]H_0: \beta_1 =0\\\\H_1: \beta_1 \ne 0[/tex]
The test is two tailed because the alternate hypothesis contains [tex]\ne[/tex]
Calculate the rejection region
[tex]P(t < -t_0) = P(t > t_0) = \frac{\alpha}{2}[/tex]
[tex]P(t < -t_0) = P(t > t_0) = \frac{0.05}{2}[/tex]
[tex]P(t < -t_0) = P(t > t_0) = 0.025[/tex]
Calculate the degrees of freedom
[tex]df = n -2[/tex]
[tex]df = 30 -2[/tex]
[tex]df = 28[/tex]
On the student's T distribution table, the t value at 28 and column with [tex]\alpha = 0.05[/tex] (two tailed) is:
[tex]t\ value = 2.048[/tex]
The rejection value will contain all value lesser than -2.048 and all values greater than 2.048.
So: We reject [tex]H_0[/tex] when [tex]t < -2.048[/tex] and [tex]t >2.048[/tex]
Testing the first independent variable
Calculate test statistic
[tex]t = \frac{b_1 - 0}{Sb_1}[/tex]
[tex]t = \frac{2.815 - 0}{0.75}[/tex]
[tex]t = \frac{2.815 }{0.75}[/tex]
[tex]t = 3.753[/tex]
[tex]t = 3.753 > 2.048[/tex]
This implies that, we reject [tex]H_o[/tex] and accept [tex]H_1[/tex]
There is a sufficient evidence that the coefficient b2 is not zero
Testing the second independent variable
Calculate test statistic
[tex]t = \frac{b_2 - 0}{Sb_2}[/tex]
[tex]t = \frac{-1.249 - 0}{0.41}[/tex]
[tex]t = \frac{-1.249 }{0.41}[/tex]
[tex]t = -3.0463[/tex]
[tex]t = -3.0463 < -2.048[/tex]
This implies that, we reject [tex]H_o[/tex] and accept [tex]H_1[/tex]
There is a sufficient evidence that the coefficient b2 is not zero
