Answer:
H0 is rejected and Ha is accepted that the students show a higher population mean math score on the SAT if their parents attained a higher level of education.
Explanation:
Part a:
The null and alternate hypothesis can be formulated as
H0 : u1 ≤ u2 the two means ( of students whose parents did or did not attain a higher level of education) are equal .
against the claim
Ha: u1 > u2 the students show a higher population mean math score on the SAT if their parents attained a higher level of education.
Part b:
The point estimate of the difference between the means for the two populations is the difference of sample means
x1`- x2`
525- 487= 38
Student’s Parents
College Grads
x x²
485 487 235,225 237,169
534 533 285,156 284,089
650 526 422,500 267,676
554 410 306,916 168,100
550 515 302,500 265,225
572 578 327,184 334,084
497 448 247,009 200,704
592 469 350,464 219,961
∑xi = 8400 ∑xi ²= 4,462,962
x1`= ∑ xi/n1= 8400/16= 525
Using statistic calculator Using formula : σ(n-1)
s1= 59.4205
High School Grads
x x²
442 492 195,364 242,064
580 478 336,400 228,484
479 425 229,441 180,625
486 485 236,196 235,225
528 390 278,784 152,100
524 535 274,576 286,225
∑xi = 5844 ∑xi ²= 2,875,484
x2`= ∑ xi/n2= 5844/12= 487
Using statistic calculator Using formula : σ(n-1)
s2= 51.7476
x1`- x2`= 525- 487= 38
The test statistic is
t= (x1`- x2`) / √ s1²/n1+ s2²/n2
t= 38/ √(59.4205)²/16 + (51.7476)²/12
t=38 / √3530.7958/16 + 2677.8141/12
t= 1.804
and the degrees of freedom is
υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1
= [3530.7958/16 + (2677.8141/12) ]²/ (3530.7958/16)²/15 +(2677.8141/12)²/11
≈ 25
The degrees of freedom is always rounded in this calculation
From the table t∝ = 1.708
Hence critical value is t ≥ t∝
Reject H0:
Part C.
The p-value is 0.041647.
The result is less than 0.05.
Result:
H0 is rejected and Ha is accepted that the students show a higher population mean math score on the SAT if their parents attained a higher level of education.