Download Ind t - Practice Exercise - KEY

EPS 525 – INTRODUCTION TO STATISTICS
INDEPENDENT-SAMPLES t TEST – PRACTICE EXERCISE – KEY
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. State (using symbols and words) the null hypothesis and the alternative hypothesis for this
independent-samples t test.
H0:
µ1 = µ2
or
µ1 – µ2 = 0
There is no (statistically significant) difference between the boys’ spelling
exam mean (µ1) and the girls’ spelling exam mean (µ2).
•
Ha:
Could also make reference to the hypothesized mean difference = 0.
µ1 ≠ µ2
or
µ1 – µ2 ≠ 0
There is a (statistically significant) difference between the boys’ spelling exam
mean (µ1) and the girls’ spelling exam mean (µ2).
•
Could also make reference to the hypothesized mean difference ≠ 0.
2. Has the assumption of independence been met for this data?
YES
or
NO
(circle your answer selection)
Justify/explain your answer – that is, how did you come to your conclusion?
The two groups (boys and girls) are independent of each other.
3. Has the assumption of normality been met for this data? Use an alpha (α) level of .001
looking at the Shapiro-Wilks test to justify your answer.
YES
or
NO
(circle your answer selection)
Justify/explain your answer – that is, how did you come to your conclusion? (Hint: look at
the SPSS printout). Be sure to show all applicable values and symbols.
Looking at the p (Sig.) values on the Shapiro-Wilks test, we find that neither
group/level was significant, thereby indicating that each of the levels of the
independent variable (gender) are statistically normal (normally distributed).
Boys: p (.818) > α (.001)
Girls: p (.464) > α (.001)
4. Has the assumption of homogeneity of variance been met for this data? Use an alpha (α)
level of .05 looking at the Levene’s Test for Equality of Variances.
YES
or
NO
(circle your answer selection)
Justify/explain your answer – that is, how did you come to your conclusion? (Hint: look at
the SPSS printout). Be sure to show all applicable values and symbols.
The Levene’s Test for Equality of Variances showed: F = .807, p = .377
Compared to α = .05, p (.377) > α (.05) – retain the null hypothesis of no difference.
Therefore it is not significant and the assumption is met – that is, equal variances
are assumed.
5. After computing the test statistic (t-test) – complete the following tables for the independentsamples 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
6. Using an alpha (α) level of .05, interpret the results from the independent-samples t test:
6.a.
Did you reject the null hypothesis for the group means in favor of the alternative
hypothesis (indicated in question 1)?
YES
6.b.
or
NO
(circle your answer selection)
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. Be sure to include the
applicable values and symbols.
t (28) = -2.169, p = .039
Compared to α = .05, p (.039) < α (.05) – which is significant, therefore the
null hypothesis of no difference is rejected.
INDEPENDENT-SAMPLES t TEST – KEY
PAGE – 2
If applicable (or indicate otherwise, and why), calculate the Effect Size – be sure to
show your work.
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 = .79σ
σ
5.c.
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 girls (M = 81.40, SD = 9.49) performed significantly better (higher) than
the boys (M = 74.87, SD = 6.79) on the spelling test for this sample at the .05
alpha level, t(28) = 2.17, p < .05, d = .79. The mean difference (6.53) was
significantly different from zero (0), with an effect size of just over threefourths (d = .79) of a standard deviation.
Note: there are several ways to interpret the results, the key is to indicate that
there was a significant difference between the boys and girls on the spelling
test at the .05 alpha level – and include, at a minimum, reference to the group
means and effect size (if applicable).
t(28) = 2.17, p < .05, d = .79
t
Indicates that we are using a t-Test
(28)
Indicates the degrees of freedom associated with this t-Test
2.17
Indicates the obtained t statistic value (tobt)
p < .05
Indicates the probability of obtaining the given t value by chance alone
d = .79
Indicates the effect size for the significant effect (measured in standard
deviation units)
INDEPENDENT-SAMPLES t TEST – KEY
PAGE – 3
EPS 525 – INTRODUCTION TO STATISTICS
INDEPENDENT-SAMPLES t TEST – PRACTICE EXERCISE – KEY
Tests of Normality
a
Spelling
Gender
Boys
Girls
Kolmogorov-Smirnov
Statistic
df
Sig.
.130
15
.200*
.183
15
.187
Statistic
.967
.946
Shapiro-Wilk
df
15
15
Sig.
.818
.464
*. This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Independent T-Test
Group Statistics
SPELLING
GENDER
1 Boys
2 Girls
N
Mean
74.87
81.40
15
15
Std. Deviation
6.791
9.485
Std. Error
Mean
1.754
2.449
Independent Samples Test
Levene'
s Test for
Equality of Variances
SPELLING
Equal variances
assumed
Equal variances
not assumed
F
.807
Sig.
.377
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-2.169
28
.039
-6.53
3.012
-12.703
-.363
-2.169
25.367
.040
-6.53
3.012
-12.732
-.334