I am Ann Brenda Mathews. Currently associated with statisticsassignmenthelp.com as a statistics assignment help Expert. After completing my Master’s in Statistics, at the University of Liverpool, I was in search of an opportunity that would expand my area of knowledge hence I decided to help students with their assignments. I have written several Hypothesis Assignments to help students overcome numerous difficulties.

1. For More Visit:
https://www.statisticsassignmenthelp.com/
support@statisticsassignmenthelp.com
+1 (315) 557-6473
Hypothesis Assignment
Help

2. Q1: Is there a statistically significant difference among last semester gpa
(GPA_last) based on race (student ethnic origin)?
Null hypothesis: There is no significant difference among last semester GPA based on
race.
Alternative hypothesis: There is a significant difference among last semester GPA
based on race.
Assumption testing:
Independence assumption: The assumption states that the sample drawn should be
independent of one another. The assumption has been met by the data as the students
are independent of one another.
Normality assumption: The normality assumption states that the distribution of last
semester GPA for each category of race should be normally distributed. The results
for the Shapiro Wilk test (under the null hypothesis which states that the distribution
is normal) show that the p-value for the test is close to zero for all the three categories
of race. At 5% alpha, since the p-value is less than 0.05, we have enough evidence to
reject the null hypothesis. Thus, we conclude that the distribution of the variable is
not normal. However, the central limit theorem, which states that for the large sample
size the distribution can be considered to be normal, applies here as the sample size
for each of the category of race is large. Thus, the assumption is met by the data.
www.statisticsassignmenthelp.com

3. Tests of Normality
Race Kolmogorov-Smirnova Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
GPA LAST
TERM
Black .141 673 .000 .953 673 .000
White .141 889 .000 .930 889 .000
Other .124 158 .000 .951 158 .000
a. Lilliefors Significance Correction
Constant variance: The homoskedasticity assumption states that the variance of the three
categories of race should be equal. The results for the Levene statistic show that the p-
value for the test is 0.519. At 5% alpha, since the p-value is more than 0.05, we do not
have enough evidence to reject the null hypothesis (which states that the variances of the
groups are equal). Thus, we conclude that the variances of the three groups are equal.
Test of Homogeneity of Variances
GPA LAST TERM
Levene Statistic df1 df2 Sig.
.657 2 1717 .519
Hypothesis results: The results for the One-way ANOVA show that the p-value for the test is
close to zero. At 5% alpha, since the p-value is less than 0.05, we have enough evidence to
reject the null hypothesis.
www.statisticsassignmenthelp.com

4. Thus, we conclude that there is a significant difference among last semester GPA
based on race.
ANOVA
GPA LAST TERM
Sum of
Squares
df Mean Square F Sig.
Between Groups 42.390 2 21.195 24.159 .000
Within Groups 1506.320 1717 .877
Total 1548.711 1719
Multiple Comparisons
Dependent Variable: GPA LAST TERM
(I)
Race
(J) Race Mean
Difference
(I-J)
Std.
Error
Sig. 95% Confidence
Interval
Lower
Bound
Upper
Bound
Tukey
HSD
Black
White -.32067* .04786 .000 -.4329 -.2084
Other -.03784 .08280 .891 -.2321 .1564
White
Black .32067* .04786 .000 .2084 .4329
Other .28283* .08087 .001 .0931 .4725
Other
Black .03784 .08280 .891 -.1564 .2321
White -.28283* .08087 .001 -.4725 -.0931
*. The mean difference is significant at the 0.05 level.
www.statisticsassignmenthelp.com

5. Discussion: The results show that the GPA in the last term differs on the basis of the race
categories. The GPA in the last term is significantly higher for the White people (M=2.84,
SD=0.93) than Black people (M=2.52, SD=0.94) and “Other” race people (M=2.56,
SD=0.98). But there is no significant difference found in the last term GPA for Black and
“Other” race people.
Q2: Is there a significant difference between last semester GPA (GPA_last) and the
current semester GPA (GPA_curr) for white students only?
Null hypothesis: There is no significant difference in last semester GPA and the current
semester GPAfor white students only.
Alternative hypothesis: There is a significant difference in last semester GPA and the
current semester GPAfor white students only.
Assumption testing:
Dependent variable must be continuous: Both the variables used in the analysis namely
last semester GPA and current semester GPA are continuous variables and thus, the
assumption is met by the data.
Independence assumption: The assumption states that the sample drawn should be
independent of one another. The assumption has been met by the data as the students are
independent of one another.
www.statisticsassignmenthelp.com

6. Normality assumption: The normality assumption states that the distribution of last
semester GPA and current semester GPA should be normally distributed. The results
for the Shapiro Wilk test (under the null hypothesis which states that the distribution is
normal) show that the p-value for the test for both the variables is close to zero. At 5%
alpha, since the p-value is less than 0.05, we have enough evidence to reject the null
hypothesis. Thus, we conclude that the distribution of the variable is not normal.
However, the central limit theorem, which states that for the large sample size the
distribution can be considered to be normal, applies here as the sample size for both
the variables is large. Thus, the assumption is met by the data.
Tests of Normality
Kolmogorov-Smirnova Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
GPA LAST
TERM
.141 889 .000 .930 889 .000
Current GPA .069 889 .000 .961 889 .000
a. Lilliefors Significance Correction
No outliers in the data: The assumption states that there are no outliers in the data. The
box plot presented below shows that there are no outliers in the data for GPA (last term)
variable but there are a few outliers in the data for current GPA. Hence, the assumption
is not met by the data for Current GPA.
www.statisticsassignmenthelp.com

7. Hypothesis results: The results for the paired samples t-test show that the p-value
for the test is close to zero. At 5% alpha, since the p-value is less than 0.05, we
have enough evidence to reject the null hypothesis. Thus, we conclude that there is
a significant difference in last semester GPA and current GPA for white students
only.
www.statisticsassignmenthelp.com

8. Paired Differences t df Sig.(2
-
tailed)
Mean Std.
Deviatio
n
Std
Error
Mean
95% Confidence
Interval of the
Difference
Lower Upper
Pair
GPA
LAST
TERM-
Current
GPA
0.7166 .58280 .001955 -.11003 -0.3330 3.66
6 888 .000
Discussion: The results show that for white students, there is a difference in the GPA in
the last term and current term. The mean GPA in the last term for the White people was
2.84 with standard deviation of 0.93) but the same for current GPA for white people is
2.91 with standard deviation of 0.72. The mean of the difference in last term GPA and
current GPA is -0.072 which means that on an average, the students have scored better in
current GPA than last term GPA.
Q3: Was the performance of Black students better or worse than the average
performance of all students during the last semester?
www.statisticsassignmenthelp.com

9. Null hypothesis: There is no significant difference in performance of Black students in
comparison to all students during last semester.
Alternative hypothesis: There is a significant difference in performance of Black students in
comparison to all students during last semester.
Assumption testing:
Dependent variable must be continuous: The dependent variable used in the analysis namely
last semester GPA is continuous variable and thus, the assumption is met by the data.
Independence assumption: The assumption states that the sample drawn should be
independent of one another. The assumption has been met by the data as the students are
independent of one another.
Normality assumption: The normality assumption states that the distribution of last semester
GPA for black students should be normally distributed. The results for the Shapiro Wilk test
(under the null hypothesis which states that the distribution is normal) show that the p-value
for the test for both the variables is close to zero. At 5% alpha, since the p-value is less than
0.05, we have enough evidence to reject the null hypothesis. Thus, we conclude that the
distribution of the variable is not normal. However, the central limit theorem, which states
that for the large sample size the distribution can be considered to be normal, applies here as
the sample size for the variable is large. Thus, the assumption is met by the data.
www.statisticsassignmenthelp.com

10. Tests of Normality
Kolmogorov-Smirnova Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
GPA LAST TERM .141 673 .000 .953 673 .000
a. Lilliefors Significance Correction
No outliers in the data: The assumption states that there are no outliers in the data. The
box plot presented below shows that there are outliers in the data for GPA (last term)
variable. Hence, the assumption is not met by the data.
www.statisticsassignmenthelp.com

11. Hypothesis results: The results for the one-sample t-test show that the p-value for the
test is close to zero. At 5% alpha, since the p-value is less than 0.05, we have enough
evidence to reject the null hypothesis. Thus, we conclude that there is a significant
difference in performance of Black students in comparison to all students during last
semester.
One-Sample Test
Test Value = 2.69
t df Sig. (2-
tailed)
Mean
Difference
95% Confidence
Interval of the
Difference
Lower Upper
GPA LAST
TERM
-4.720 672 .000 -.17082 -.2419 -.0998
Discussion: The results show that for black students, the mean GPA in the last term
was 2.51 with standard deviation of 0.94 and this is lower than the average GPA of
.last semester by all the students at 2.69. Thus, the performance of the Black students
is worse than the average performance of all students during the last semester.
www.statisticsassignmenthelp.com

12. Q4: Whether or not students believe instructors provide helpful comments
(item10) or used examples (item 20).
Helpful Comments
Null hypothesis: The students believe instructors provide helpful comments.
Alternative hypothesis: The students believe instructors do not provide helpful comments.
Assumption testing:
Dependent variable must be continuous: The dependent variable used in the analysis
namely helpful comments is continuous variable and thus, the assumption is met by the
data.
Independence assumption: The assumption states that the sample drawn should be
independent of one another. The assumption has been met by the data as the students are
independent of one another.
Normality assumption: The normality assumption states that the distribution of helpful
comments should be normally distributed. The results for the Shapiro Wilk test (under the
null hypothesis which states that the distribution is normal) show that the p-value for the
test for both the variables is close to zero. At 5% alpha, since the p-value is less than 0.05,
we have enough evidence to reject the null hypothesis. Thus, we conclude that the
distribution of the variable is not normal. However, the central limit theorem, which
states that for the large sample size the distribution can be considered to be normal,
applies here as the sample size for the variable is large. Thus, the assumption is met by
the data.
www.statisticsassignmenthelp.com

13. Tests of Normality
Kolmogorov-Smirnova Shapiro-Wilk
Statisti
c
df Sig. Statisti
c
df Sig.
ITEM 10:HELPFUL
COMMENTS
.185 205 .000 .854 205 .000
a. Lilliefors Significance Correction
No outliers in the data: The assumption states that there are no outliers in the data.
The box plot presented below shows that there are no outliers in the data for helpful
comments variable. Hence, the assumption is met by the data.
www.statisticsassignmenthelp.com

14. Hypothesis results: The results for the one-sample t-test show that the p-value for the test
is close to zero. At 5% alpha, since the p-value is less than 0.05, we have enough evidence
to reject the null hypothesis. Thus, we conclude that students believe instructors do not
provide helpful comments.
One-Sample Test
Test Value = 4
t df Sig. (2-
tailed)
Mean
Difference
95% Confidence
Interval of the
Difference
Lower Upper
ITEM
10:HELPFUL
COMMENTS
-
17.451
204 .000 -1.351 -1.50 -1.20
Discussion: The results show that the mean rating of question related to helpful comments
was 2.65 with standard deviation of 1.11. This means that the students on an average either
“disagree” or “agree” with the question of helpful comments and this is lower than the
rating of helpful comments at 4, suggesting strongly agree. Thus, the students believe
instructors do not provide helpful comments.
www.statisticsassignmenthelp.com

15. Used Examples
Null hypothesis: The students believe instructors provide used examples.
Alternative hypothesis: The students believe instructors do not provide used examples.
Assumption testing:
Dependent variable must be continuous: The dependent variable used in the analysis
namely used example sis continuous variable and thus, the assumption is met by the
data.
Independence assumption: The assumption states that the sample drawn should be
independent of one another. The assumption has been met by the data as the students are
independent of one another.
Normality assumption: The normality assumption states that the distribution of used
examples should be normally distributed. The results for the Shapiro Wilk test (under the
null hypothesis which states that the distribution is normal) show that the p-value for the
test for both the variables is close to zero. At 5% alpha, since the p-value is less than
0.05, we have enough evidence to reject the null hypothesis. Thus, we conclude that the
distribution of the variable is not normal. However, the central limit theorem, which
states that for the large sample size the distribution can be considered to be normal,
applies here as the sample size for the variable is large. Thus, the assumption is met by
the data.
www.statisticsassignmenthelp.com

16. Tests of Normality
Kolmogorov-Smirnova Shapiro-Wilk
Statisti
c
df Sig. Statisti
c
df Sig.
ITEM 20: USED
EXAMPLES
.204 205 .000 .859 205 .000
a. Lilliefors Significance Correction
No outliers in the data: The assumption states that there are no outliers in the data. The
box plot presented below shows that there are no outliers in the data for used examples
variable. Hence, the assumption is met by the data.
www.statisticsassignmenthelp.com

17. Hypothesis results: The results for the one-sample t-test show that the p-value for the
test is close to zero. At 5% alpha, since the p-value is less than 0.05, we have enough
evidence to reject the null hypothesis. Thus, we conclude that students believe
instructors do not provide used examples.
One-Sample Test
Test Value = 4
t df Sig. (2-
tailed)
Mean
Difference
95% Confidence
Interval of the
Difference
Lower Upper
ITEM 20: USED
EXAMPLES
-
17.195
204 .000 -1.263 -1.41 -1.12
Discussion: The results show that the mean rating of question related to “used
examples” was 2.74 with standard deviation of 1.05. This means that the students on
an average either “disagree” or “agree” with the question of used examples and this is
lower than the rating of “used examples” at 4, suggesting strongly agree. Thus, the
students believe instructors do not provide used examples.
www.statisticsassignmenthelp.com