Question

(Use Excel) A recent report criticizes SAT-test-preparation providers for promising big score gains without any hard data to back up such claims (The Wall Street Journal, May 20, 2009). Suppose eight college-bound students take a mock SAT, complete a three-month test-prep course, and then take the real SAT. Let the difference be defined as Score on Mock SAT minus Score on Real SAT. Use Table 2.

Student	Score on Mock SAT	Score on Real SAT
1	1,895	1,841
2	1,709	1,702
3	1,961	2,004
4	2,268	2,070
5	1,605	1,614
6	1,849	1,980
7	1,923	1,890
8	1,691	1,665

Let the difference be defined as scores on Mock SAT – Real SAT.

a.	Specify the competing hypotheses that determine whether completion of the test-prep course increases a student’s score on the real SAT.
	A) H0: μD = 0; HA: μD ≠ 0 B) H0: μD ≥ 0; HA: μD < 0 C) H0: μD ≤ 0; HA: μD > 0

b.

Assuming that the SAT scores difference is normally distributed, calculate the value of the test statistic and its associated p-value. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Compute the p-value using your unrounded test statistic value. Round "Test statistic" value to 2 decimal places and "p-value" to 4 decimal places.)

Test statistic _____
p-value ______

c.	At the 5% significance level, do the sample data support the test-prep providers’ claims?
	A) Reject H0 since the p-value is less than α. B) Reject H0 since the p-value is more than α. C) Do not reject H0 since the p-value is less than α. D) Do not reject H0 since the p-value is more than α.

Answer 1

The statistical software output for this problem is:

Paired T hypothesis test:
μ_D = μ₁ - μ₂ : Mean of the difference between Score on Mock SAT and Score on Real SAT
H₀ : μ_D = 0
H_A : μ_D > 0
Hypothesis test results:

Difference	Mean	Std. Err.	DF	T-Stat	P-value
Score on Mock SAT - Score on Real SAT	16.875	32.93578	7	0.51236072	0.3121

Hence,

a) Option C is correct.

b) Test statistic = 0.51

P - value = 0.3121

c) Option D is correct.