MATHEMATICS COLLEGE

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

Answers

Answer 1

Answer:

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:

$t = \frac{\overline x_1 -\overline x_2}{\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2) } }$

$t = \frac{580 -530}{\sqrt{(105^2)/(45)+(114^2)/(38) } }$

$t = \frac{50}{\sqrt{(11025)/(45)+(12996)/(38) } }$

$t = (50)/(√(245+342 ) )$

$t = (50)/(√(587) )$

$t = (50)/(24.228)$

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.

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Law School According to the Law School Admission Council, in the fall of 2007, 66% of law school applicants wereaccepted to some law schooL4 The training program LSATisfaction claims that 163 of the 240 students trained in 2006were admitted to law school. You can safely consider these trainees to be representative of the population of law schoolapplicants. Has LSAﬁsfaction demonstrated a real improvement over the national average?a) What are the hypotheses?b) Check the conditions and ﬁnd the P-value.c) Would you recommend this program based on what you see here? Explain.

Answers

Answer:

a) $H_(0): p = 0.66\nH_A: p > 0.66$

b) P-value = 0.2650

c) No, this programme will not be recommended as there is no real improvement over the national average.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 240

p = 66% = 0.66

Alpha, α = 0.05

Number of students admitted to law school , x = 163

a) First, we design the null and the alternate hypothesis

$H_(0): p = 0.66\nH_A: p > 0.66$

This is a one-tailed(right) test.

Formula:

$\hat{p} = (x)/(n) = (163)/(240) = 0.6792$

$z = \frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}$

Putting the values, we get,

$z = \displaystyle\frac{0.6792-0.66}{\sqrt{(0.66(1-0.66))/(240)}} = 0.6279$

b) Now, we calculate the p-value from the table.

P-value = 0.2650

c) Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is no real improvement over the national average.

No, this programme will not be recommended as there is no real improvement over the national average.

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Answers

Answer:

Step-by-step explanation:

Summarizing what we know:

Casey has n nickels, and Meghan 4 times as many: 4n

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Answers

Answer:

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Answers

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Answers

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Answers

Answer: B

Step-by-step explanation:

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