Question

A committee at the College Board has been asked to study the SAT math scores for students in Pennsylvania and Ohio. A sample of 45 students from Pennsylvania had an average score of 580, whereas a sample of 38 students had an average score of 530. The sample standard deviations for Pennsylvania and Ohio are 105 and 114 respectively. Does the study suggest that the SAT math score for students in Pennsylvania and Ohio differ

Answer 1

Answer:

Step-by-step explanation:

From the given information:

The null hypothesis and the alternative hypothesis can be computed as:

$H_0 :\mu_1 -\mu_2 = 0$   (i.e. there is no difference between the SAT score for students in both locations)

$H_1 :\mu_1 -\mu_2 \geq0$ (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:

$t = \dfrac{\overline x_1 -\overline x_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2} } }$

$t = \dfrac{580 -530}{\sqrt{\dfrac{105^2}{45}+\dfrac{114^2}{38} } }$

$t = \dfrac{50}{\sqrt{\dfrac{11025}{45}+\dfrac{12996}{38} } }$

$t = \dfrac{50}{\sqrt{245+342 } }$

$t = \dfrac{50}{\sqrt{587} }$

$t = \dfrac{50}{24.228}$

t = 2.06

degree of freedom = ($n_1 + n_2$ ) -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 $H_o$  if the p-value is less than the significance level

Conclusion: We reject the $H_o$ , 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.