Hypothesis Assignment Help

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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.

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Tests of Normality

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 pvalue 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

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.

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
GPA LAST TERM Levene Statistic df1 df2 Sig. .657 2 1717 .519
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Thus, we conclude that there is a significant difference among last semester GPA based on race.

Multiple Comparisons

Dependent Variable: GPA LAST TERM

*. The mean difference is significant at the 0.05 level.

ANOVA
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
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
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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.

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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

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.

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
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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.

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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?

Paired Differences t df Sig. (2tailed) Mean Std. Deviatio n Std Error Mean 95% Confidence Interval of the Difference Lower Upper Pair GPA LAST TERMCurrent GPA 0.7166 .58280 .001955 -.11003 -0.3330 3.66 6 888 .000
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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.

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Tests of Normality

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.

Kolmogorov-Smirnova Shapiro-Wilk Statistic df Sig. Statistic df Sig. GPA LAST TERM .141 673 .000 .953 673 .000
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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 Value = 2.69

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.

Test
t df Sig. (2tailed) Mean Difference 95% Confidence Interval of the Difference Lower Upper GPA LAST TERM -4.720 672 .000 -.17082 -.2419 -.0998
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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.

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Tests of Normality

a.

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.

Kolmogorov-Smirnova Shapiro-Wilk Statisti c df Sig. Statisti c df Sig. ITEM 10:HELPFUL COMMENTS .185 205 .000 .854 205 .000
Lilliefors Significance Correction
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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.

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.

One-Sample Test Test Value = 4 t df Sig. (2tailed) Mean Differenc e 95% Confidence Interval of the Difference Lower Upper ITEM 10:HELPFUL COMMENTS17.451 204 .000 -1.351 -1.50 -1.20
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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.

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Tests of Normality

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.

Kolmogorov-Smirnova Shapiro-Wilk Statisti c df Sig. Statisti c df Sig. ITEM 20: USED EXAMPLES .204 205 .000 .859 205 .000
Lilliefors Significance Correction
a.
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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.

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.

One-Sample Test Test Value = 4 t df Sig. (2tailed) Mean Differenc e 95% Confidence Interval of the Difference Lower Upper ITEM 20: USED EXAMPLES17.195 204 .000 -1.263 -1.41 -1.12
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