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EPS 525 – INTRODUCTION TO STATISTICS
QUIZ NO. 3 – INDEPENDENT- AND DEPENDENT-SAMPLES t TEST
You may work together on this quiz – however – each individual must turn in a copy of
their own quiz and all SPSS output that assisted with the quiz.
Use the Quiz 3 – Independent t Test data from the Web site and the following scenario:
Dr. Kureous, a teacher at George Junior High, wants to determine if there is a
significant difference between the 6th grade boys and girls in his class on their spelling
exam. He does not have a prediction as to whether the boys or the girls will be better –
he simply wants to see if they differ significantly on the exam.
1. (1 point) State (using symbols or words) the null hypothesis and the alternative hypothesis
for his independent-samples t test.
H0:
1 = 2
or
1 – 2 = 0
There is no (statistically significant) difference between the boys’ spelling
exam mean (1) and the girls’ spelling exam mean (2).

Ha:
Could also make reference to the hypothesized mean difference = 0.
1  2
1 – 2  0
or
or
(1 < 2 and 1 > 2)
There is a (statistically significant) difference between the boys’ spelling exam
mean (1) and the girls’ spelling exam mean (2).

Could also make reference to the hypothesized mean difference  0.
2. Using an alpha () level of .05.
2.a.
(½ point) Has Dr. Kureous met the assumption of homogeneity of variance for the
Spelling Exam t test?
YES
2.b.
or
NO
(circle your answer selection)
(1 point) Justify/explain your answer – that is, how did you come to your conclusion?
(Hint: look at the SPSS printout).
The Levene’s Test for Equality of Variances showed: F = .807, p = .377
Compared to  = .05, p >  (.377 > .05) – retain the null hypothesis of no
difference. Therefore it is not significant and the assumption is met – that is,
equal variances are assumed.
3. (1 point) After computing the test statistic (t-test) – complete the following tables for the
Spelling Exam Independent-Samples t-Test (Do not round – use the values from SPSS):
Gender
N
Mean
Standard
Deviation
Standard Error
of the Mean
Boys
15
74.87
6.791
1.754
Girls
15
81.40
9.485
2.449
t
df
Sig.
(2-tailed)
Mean
Difference
-2.169
28
.039
-6.53
95% CI of the
Difference
Lower
Upper
-12.703
-.363
4. Using an alpha () level of .05, interpret the results from the Spelling Exam t test:
4.a.
(½ point) Did you reject the null hypothesis in favor of the alternative hypothesis
(indicated in question 1)?
YES
4.b.
or
NO
(circle your answer selection)
(1 point) Justify your answer, that is, how did you come to your conclusion?
DO NOT make reference to the t critical value or the confidence intervals. If
applicable, calculate the Effect Size – show your work.
t (28) = -2.169, p = .039
Compared to  = .05, p <  (.039 < .05) – which is significant, therefore the
null hypothesis of no difference is rejected.
d t
N1  N 2
15  15
30
 2.169
 2.169
 2.169 .133333  2.169(.365148)
N1 N 2
(15)(15)
225
d = .7920068 = .79
4.c.
(1 point) Briefly discuss your findings (i.e., what do the results indicate or mean). Be
sure to make reference to the group means.
The girls (M = 81.40) performed significantly better (higher) than the boys (M
= 74.87) on the spelling test for this sample at the .05 alpha level, with an
effect size of nearly one (d = .79) standard deviation.
EPS 525 Quiz No. 3
Page – 2
Using the Quiz 3 – Dependent t Test data and given the following scenario:
Dr. Stats would like to see if there is a difference between Exam_1 and Exam_2 for her
entire class. She does not have a prediction as to whether the performance will be
higher or lower for the two sets of exam comparisons – she simply wants to determine if
the class performed significantly different on the two test occasions.
5. (1 point) State (using symbols or words) the null hypothesis and the alternative hypothesis
for her dependent-samples t test.
H0:
1 = 2
or
1 – 2 = 0
There is no (statistically significant) difference between the Exam_1 mean
(1) and the Exam_2 mean (2).

Ha:
Could also make reference to the hypothesized mean difference = 0.
1  2
1 – 2  0
or
or
(1 < 2 and 1 > 2)
There is a (statistically significant) difference between the Exam_1 mean (1)
and the Exam_2 mean (2).

Could also make reference to the hypothesized mean difference  0.
6. (½ point) Using the Table of Critical Values of Student’s t Distribution – what would the t
critical value (tcrit) be for the Exam_1 / Exam_2 Dependent-Samples t-Test:
With df = 44 – for a two-tailed test using  = .05, tCV = +2.021
We used df = 40 since there was not a df = 44 – to error on the side of being
more conservative (i.e., produce a greater critical value to compare against).
7. Using an alpha () level of .05, interpret the results from the Exam_1 / Exam_2 t test:
7.a.
(½ point) Did you reject the null hypothesis in favor of the alternative hypothesis?
YES
or
NO
(circle your answer selection)
EPS 525 Quiz No. 3
Page – 3
7.b. (1 point) Justify your answer, that is, how did you come to your conclusion? You may
make reference to any of the three methods of determination. Be sure to include all
applicable values for the method that you choose. If applicable, calculate the Effect
Size – show your work.
1)
Comparing tobt = |-2.026| to the tcv = |-2.021|.

2)
Comparing the Sig. (probability) = .049 to the a priori alpha level,  = .05

3)
tobt > tcv – therefore, we reject the null hypothesis of no difference.
p <  – therefore, we reject the null hypothesis of no difference.
Examining the confidence interval (Lower = -11.52, Upper = -.03) to see if
zero (0) is contained within it’s limits.

The confidence interval CI95 does not contain zero – therefore, we
reject the null hypothesis of no difference.
7.c.
Using:
d
Using:
d
Mean  5.78

 .3021906 = .30
SD
19.127
t
N

 2.026
45

 2.026
 .3011238
6.7082039
= .30
(1 point) Briefly discuss your findings (i.e., what do the results indicate or mean). Be
sure to make reference to the group means.
The mean for Exam_1 (M = 75.56) was significantly lower than the mean for
the Exam_2 (M = 81.33). The mean difference (-5.78) was significantly
different from zero (0) at the .05 alpha level for this sample, with an effect size
of nearly one-third (d = .30) of a standard deviation.
EPS 525 Quiz No. 3
Page – 4