Question

The College Board provided comparisons of Scholastic Aptitude Test (SAT) scores based on the highest level of education attained by the test taker's parents. A research hypothesis was that students whose parents had attained a higher level of education would on average score higher on the SAT. The overall mean SAT math score was 514.† SAT math scores for independent samples of students follow. The first sample shows the SAT math test scores for students whose parents are college graduates with a bachelor's degree. The second sample shows the SAT math test scores for students whose parents are high school graduates but do not have a college degree.

College Grads

469	487
518	533
666	526
538	426
566	499
572	578
513	464
592	469

High School Grads

442	492
580	478
479	425
486	485
528	390
524	535

(a)

Formulate the hypotheses that can be used to determine whether the sample data support the hypothesis that students show a higher population mean math score on the SAT if their parents attained a higher level of education. (Let μ₁ = population mean verbal score of students whose parents are college graduates with a bachelor's degree and μ₂ = population mean verbal score of students whose parents are high school graduates but do not have a college degree.)

H₀: μ₁ − μ₂ ≤ 0

H_a: μ₁ − μ₂ > 0

H₀: μ₁ − μ₂ ≠ 0

H_a: μ₁ − μ₂ = 0

    

H₀: μ₁ − μ₂ = 0

H_a: μ₁ − μ₂ ≠ 0

H₀: μ₁ − μ₂ ≥ 0

H_a: μ₁ − μ₂ < 0

H₀: μ₁ − μ₂ < 0

H_a: μ₁ − μ₂ = 0

(b)

What is the point estimate of the difference between the means for the two populations?

(c)

Find the value of the test statistic. (Round your answer to three decimal places.)

Compute the p-value for the hypothesis test. (Round your answer to four decimal places.)

p-value =

Answer 1

college	HS
469	442
518	580
666	479
538	486
566	528
572	524
513	492
592	478
487	425
533	485
526	390
426	535
499
578
464
469

this is 2- sample independent t-test

t-Test: Two-Sample Assuming Equal Variances
	college	HS
Mean	526	487
Variance	3559.6	2677.818182
Observations	16	12
Pooled Variance	3186.538462
Hypothesized Mean Difference	0
df	26
t Stat	1.809158526
P(T<=t) one-tail	0.04100213
t Critical one-tail	1.70561792
P(T<=t) two-tail	0.08200426
t Critical two-tail	2.055529439

a)
H0: μ1 − μ2 ≤ 0

Ha: μ1 − μ2 > 0

b)

point estimate = Xbar1 - Xbar2 = 526 -487 = 39

c) test statistic = 1.809

p-value = 0.041