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.

μ1    College Grads		μ2    High School Grads
469		442
550		580
650		479
570		486
534		528
572		524
497		492
592		478
487		425
517		485
526		390
426		535
515
562
464
453

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

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

H₀: μ₁ − μ₂ ≤ 0 & H_a: μ₁ − μ₂ > 0

	College grads	High school Grads
sample mean x =	524.000	487.000
std deviation s=	58.105	51.748
sample size n=	16	12
std error se=s/√n=	14.526	14.938
degree freedom=(se12+se22)2/(se12/(n1-1)+se22/(n2-1))=			25

Point estimate =x1-x2=		37.000
standard error of difference Se=√(S21/n1+S22/n2)=			20.8365
test statistic t =(x1-x2-Δo)/Se=		1.776
p value : =	0.0440		from excel: tdist(1.7757,25,1)