Download Dep t - Practice Exercise - KEY

EPS 525 – INTRODUCTION TO STATISTICS
DEPENDENT-SAMPLES t TEST – PRACTICE EXERCISE – KEY
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.
1. State (using symbols and words) the null hypothesis and the alternative hypothesis for her
dependent-samples t test.
µ1 = µ2
H0:
µ1 – µ2 = 0
or
There is no (statistically significant) difference between the Exam_1 mean
(µ1) and the Exam_2 mean (µ2).
•
Could also make reference to the hypothesized mean difference = 0.
µ1 ≠ µ2
Ha:
or
µ1 – µ2 ≠ 0
There is a (statistically significant) difference between the Exam_1 mean (µ1)
and the Exam_2 mean (µ2).
•
Could also make reference to the hypothesized mean difference ≠ 0.
2. After computing the test statistic (t-test) – complete the following tables for the dependentsamples t -test (Do not round – use the values from SPSS):
Test
Occasion
N
Mean
Standard
Deviation
Standard Error
of the Mean
Exam 1
45
75.56
15.308
2.282
Exam 2
45
81.33
13.915
2.074
t
df
Sig.
(2-tailed)
Paired Mean
Difference
Paired
Standard
Deviation
2.026
44
.049
5.78
19.127
95% CI of the
Difference
Lower
Upper
-11.52
-.03
3. Using the Table of Critical Values of Student’s t Distribution – what would the t critical
value (tcrit) be for dependent-samples t -test:
With df = 44 – for a two-tailed test using α = .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).
4. Using an alpha (α) level of .05, interpret the results from the dependent-samples t test:
4.a.
Did you reject the null hypothesis in favor of the alternative hypothesis?
YES
NO
or
(circle your answer selection)
4.b. 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 the Sig. (probability) = .049 to the a priori alpha level, α = .05
•
2)
Comparing tobt = |-2.026| to the tcv = |-2.021|.
•
3)
p < α – therefore, we reject the null hypothesis of no difference.
tobt > tcv – 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 its limits.
•
The confidence interval CI95 does not contain zero – therefore, we
reject the null hypothesis of no difference.
4.c.
Using:
d=
Using:
d=
Mean − 5.78
=
= .3021906 = .30σ
σ
SD
19.127
t
N
=
− 2.026
45
=
− 2.026
= .3011238
6.7082039
= .30σ
σ
Briefly discuss your findings (i.e., what do the results indicate or mean). Be sure to
make reference to the group means and effect size (if applicable). Be sure to show all
applicable values and symbols including the statistical strand for the results.
The mean for Exam_2 (M = 81.33) was significantly higher than the mean for
the Exam_1 (M = 75.56). 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.
t(44) = 2.03, p < .05, d = .30
DEPENDENT-SAMPLES T TEST – KEY
PAGE – 2
Dependent T-Test
Paired Samples Statistics
Pair
1
EXAM_1
EXAM_2
Mean
75.56
81.33
N
45
45
Std. Deviation
15.308
13.915
Std. Error
Mean
2.282
2.074
Paired Samples Correlations
Pair 1
EXAM_1 & EXAM_2
N
45
Correlation
.146
Sig.
.339
Paired Samples Test
Paired Differences
Pair 1
EXAM_1 - EXAM_2
Mean
-5.78
Std. Deviation
19.127
Std. Error
Mean
2.851
95% Confidence
Interval of the
Difference
Lower
Upper
-11.52
-.03
t
-2.026
df
44
Sig. (2-tailed)
.049
DEPENDENT-SAMPLES T TEST – KEY
PAGE – 3