Answer:
Step-by-step explanation:
From the given information:
The null hypothesis and the alternative hypothesis can be computed as:
[tex]H_0 :\mu_1 -\mu_2 = 0[/tex] (i.e. there is no difference between the SAT score for students in both locations)
[tex]H_1 :\mu_1 -\mu_2 \geq0[/tex] (i.e. there is a difference between the SAT score for students in both locations)
The test statistics using the students' t-test for the two-samples; we have:
[tex]t = \dfrac{\overline x_1 -\overline x_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2} } }[/tex]
[tex]t = \dfrac{580 -530}{\sqrt{\dfrac{105^2}{45}+\dfrac{114^2}{38} } }[/tex]
[tex]t = \dfrac{50}{\sqrt{\dfrac{11025}{45}+\dfrac{12996}{38} } }[/tex]
[tex]t = \dfrac{50}{\sqrt{245+342 } }[/tex]
[tex]t = \dfrac{50}{\sqrt{587} }[/tex]
[tex]t = \dfrac{50}{24.228}[/tex]
t = 2.06
degree of freedom = ([tex]n_1 + n_2[/tex] ) -2
degree of freedom = (45+38) -2
degree of freedom = 81
Using the level of significance of 0.05
Since the test is two-tailed at the degree of freedom 81 and t = 2.06
The p-value = 0.0426
Decision rule: To reject [tex]H_o[/tex] if the p-value is less than the significance level
Conclusion: We reject the [tex]H_o[/tex] , thus, there is no sufficient evidence to conclude that there is a significant difference between the SAT math score for students in Pennsylvania and Ohio.
