Answer:
Explanation:
From the given information:
The null hypothesis and the alternative hypothesis can be computed as:
(i.e. there is no difference between the SAT score for students in both locations)
(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:
![t = \frac{\overline x_1 -\overline x_2}{\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2) } }](https://img.qammunity.org/2021/formulas/mathematics/college/p8vcwmkzbjq74wi6hw3ta2ilcuvb4xvth0.png)
![t = \frac{580 -530}{\sqrt{(105^2)/(45)+(114^2)/(38) } }](https://img.qammunity.org/2021/formulas/mathematics/college/w4mmox3q1g258iwdvrwkrzowllqr9rb2ip.png)
![t = \frac{50}{\sqrt{(11025)/(45)+(12996)/(38) } }](https://img.qammunity.org/2021/formulas/mathematics/college/hywga8wp7w8ozkxjpeut5dfyexedgi6lzs.png)
![t = (50)/(√(245+342 ) )](https://img.qammunity.org/2021/formulas/mathematics/college/oi7nfyedgzqan0rx17di7x0hvv4uq5c7ks.png)
![t = (50)/(√(587) )](https://img.qammunity.org/2021/formulas/mathematics/college/ddjttygx5ex3yblum8ektpqmj7jlk10m5a.png)
![t = (50)/(24.228)](https://img.qammunity.org/2021/formulas/mathematics/college/20rfd09xyywm5akvocubp73s0bexl45uqr.png)
t = 2.06
degree of freedom = (
) -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
if the p-value is less than the significance level
Conclusion: We reject the
, 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.