Question

"Achievement test scores are declining all around us," brooded Professor Probity. "Here are my final exam scores for last year and this year on the same exam." What did Professor Probity discover? (18 pts).

This Year | Last Year (this year on the left, last year in the right column)

29    27

27    28

26    25

22    24

19    22

15    20

14    18

10    16

10    12

Indicate:

a. a statement of your hypotheses in words (2 pts)

b. the t-value you compute (5 pts)

c. your decision regarding H₀ and why, and (2 pts)

d. Results of homogeneity of variance test and effect size, if appropriate

e. an APA conclusion based on your results (4 pts)

Answer 1

a)

Null hypothesis: The null hypothesis is defined as the mean difference exam score is zero, i.e.

Alternative hypothesis: The alternative hypothesis test the claim that the mean difference of exam score is not zero, i.e.

where Difference, D = Exam score (this year) - Exam score (last year)

b)

Test-statistic

The t-statistic is obtained using the formula,

From the data values,

	This year	Last year	Difference
	29	27	2
	27	28	-1
	26	25	1
	22	24	-2
	19	22	-3
	15	20	-5
	14	18	-4
	10	16	-6
	10	12	-2
Average	19.1111	21.3333	-2.2222
StdDev	7.2877	5.3151	2.6352
n	9	9	9

P-value

The P-value is obtained from the t distribution table for t = -2.53 and degree of freedom = n - 1 = 9 - 1 = 8 for two-tailed hypothesis.

P-value = 0.0353

c)

Conclusion:

Since the p-value = 0.0353 < 0.05 at a 5% significance level, the null hypothesis is rejected. Hence we can conclude that this exam score is not the same as last year.

d)

An F test is used to test the equality of two population variances, and  .

Hypotheses

The Null and Alternative Hypotheses are defined as,

Test-statistic

The F statistic is computed using the formula,

From the data values,

P-value

The P-value for the F statistic is obtained using the F distribution table for F = 1.88 and numerator degree of freedom = 9-1=8 and denominator degree of freedom = 9 - 1 = 8,

P-value = 0.1804

Conclusion:

Since the p-value = 0.1804 > 0.05 at a 5% significance level, the null hypothesis is not rejected. hence there is no difference in the two population proportions. Hence the assumption of homogeneity of variances is met.

Effect size

The eta square value is obtained using the following formula,

e)

There is a significant difference in the exam score between this year and last year.; t(8) = -2.53, p-value < 0.05.