This SPSS homework focuses on performing non-parametric tests and interpreting the results. It assesses your ability to work with nominal data and evaluate statistical relationships. You will work on three different scenarios, each involving a specific statistical analysis.
Problem Statement 1:
In the recent COVID-19 pandemic, a local precinct observed that of the 75 people who voted, 25 identified as republican, 30 were democrat, and 20 were “other”. Is this similar to pre-pandemic voting based on political party affiliations? At that same precinct, pre-pandemic voting based on political party affiliation was: 35% republican, 44% democrats, and 21% “other”. Enter in the 75 data points and expected values into SPSS to conduct the appropriate statistical test.
Name the the variable of interest in the scenario. How many levels does it have, and what are they?
ANSWER
Voters- Republican, Democrat and others
Calculate the expected frequencies for each of the levels of your variable. Clearly label each group and show all work involving your calculations.
ANSWER
Republican = 35/100*75 = 26
Democrat = 44/100*75 = 33
Others = 21/100*75 = 16
Paste all relevant statistical output in the space provided below:
ANSWER
Table 1.1
Pandemic Voters
Frequency | Percent | Valid Percent | Cumulative Percent | ||
---|---|---|---|---|---|
Valid | Republican | 25 | 33.3 | 33.3 | 33.3 |
Democrat | 30 | 40.0 | 40.0 | 73.3 | |
Others | 20 | 26.7 | 26.7 | 100.0 | |
Total | 75 | 100.0 | 100.0 |
Table 1.2
Pre-pandemic Voters
Frequency | Percent | Valid Percent | Cumulative Percent | ||
---|---|---|---|---|---|
Valid | Republican | 26 | 34.7 | 34.7 | 34.7 |
Democrat | 33 | 44.0 | 44.0 | 78.7 | |
Others | 16 | 21.3 | 21.3 | 100.0 | |
Total | 75 | 100.0 | 100.0 |
Present the results using APA format. This includes a full write-up to include a complete statistical notation as shown in the weekly presentations. Make sure to describe what the conclusions mean in general terms. Additional examples of APA results sections are also available in the “Helpful Hints” document.
ANSWER
Table 1.3
Case Processing Summary
Cases | ||||||
---|---|---|---|---|---|---|
Valid | Missing | Total | ||||
N | Percent | N | Percent | N | Percent | |
Pandemic Voters * Pre-pandemic Voters | 75 | 100.0% | 0 | 0.0% | 75 | 100.0% |
Table 1.4
Pandemic Voters * Pre-pandemic Voters Crosstabulation
Count
Pre-pandemic Voters | Total | ||||
---|---|---|---|---|---|
Republican | Democrat | Others | |||
Pandemic Voters | Republican | 25 | 0 | 0 | 25 |
Democrat | 1 | 29 | 0 | 30 | |
Others | 0 | 4 | 16 | 20 | |
Total | 26 | 33 | 16 | 75 |
Table 1.5
Chi-Square Tests
Value | df | Asymp. Sig. (2-sided) | |
---|---|---|---|
Pearson Chi-Square | 122.742a | 4 | .000 |
Likelihood Ratio | 129.925 | 4 | .000 |
Linear-by-Linear Association | 66.273 | 1 | .000 |
N of Valid Cases | 75 |
1 cells (11.1%) have expected count less than 5. The minimum expected count is 4.27.
The table 1.1 and 1.2 above shows the frequency distribution of the voters before and during pandemic, the result shows that before pandemic 35% of the voters voted for the republican, 44% democrat and 21% other parties, likewise, during pandemic, 33% voted republican, 40% democrat and 27% others. It was noticed that the general percentage of vote for the republican and democrat decreased with a corresponding increase in the other parties vote.
The chi-square test of significance was conducted to determine if there was a significant relationship between the voters interest before and during the pandemic. The result shows that there is a statiscally significant relationship between the voters interest before and during pandemic with a p-value of 0.000.
Problem Statement 2:
Is there a relationship between one’s gender and whether one owns a dog, cat, or reptile? Use the data provided in the table below to answer the following questions.
Dog | Cat | Reptile | Row Totals | |
---|---|---|---|---|
Male | 20 | 17 | 11 | 48 |
Female | 25 | 23 | 5 | 53 |
Column totals | 45 | 40 | 16 | 101 |
Name the two variables of interest and the number of levels in each. Then, list the levels for each variable.
ANSWER
Gender – Male and Female
Pet – Dog, Cat and Reptile
Paste all relevant statistical output in the space provided below.
ANSWER
Table 2.1
Pet Owned * Gender Crosstabulation
Count
Gender | Total | |||
---|---|---|---|---|
Male | Female | |||
Pet Owned | Dog | 20 | 25 | 45 |
Cat | 17 | 23 | 40 | |
Reptile | 11 | 5 | 16 | |
Total | 48 | 53 | 101 |
Calculate the effect size. Show the formula and your calculations in the space provided below:
ANSWER
Table 2.2
Symmetric Measures
Value | Approx. Sig. | ||
---|---|---|---|
Nominal by Nominal | Phi | .185 | .177 |
Cramer's V | .185 | .177 | |
N of Valid Cases | 101 |
- Not assuming the null hypothesis.
- Using the asymptotic standard error assuming the null hypothesis.
Using the degrees of freedom provided by your SPSS output and an alpha value of .05, find the critical value in the appropriate table in the Appendix of your Jackson e-book. Do not round – present all three decimal places. Clearly identify the critical value from your e-book and the obtained value from your SPSS output. Based on this information, would you reject or fail to reject the null hypothesis? Does this mean there is a significant difference or no significant difference?
ANSWER
Alternative Hypothesis: There is a significant relationship between the pets owned and gender.
Null Hypothesis: There is no significant relationship between the pets owned and gender.
The chi-square test of significance between the gender and pets owned is shown in table 2.4 below. The findings indicate that, with a p-value of 0.177, which is higher than the alpha value, there was no statistically significant relationship between gender and the number of owned pets at 2 degrees of freedom. Since the alternative hypothesis is rejected, we accept the null hypothesis.
Present the results using APA format. This includes a full write-up to include a complete statistical notation as shown in the weekly presentations. Make sure to describe what the conclusions mean in general terms. Additional examples of APA results sections are also available in the “Helpful Hints” document.
ANSWER
Table 2.3
Case Processing Summary
Cases | ||||||
---|---|---|---|---|---|---|
Valid | Missing | Total | ||||
N | Percent | N | Percent | N | Percent | |
Pet Owned * Gender | 101 | 99.0% | 1 | 1.0% | 102 | 100.0% |
Table 2.4
Chi-Square Tests
Value | df | Asymp. Sig. (2-sided) | |
---|---|---|---|
Pearson Chi-Square | 3.467a | 2 | .177 |
Likelihood Ratio | 3.518 | 2 | .172 |
Linear-by-Linear Association | 1.724 | 1 | .189 |
N of Valid Cases | 101 |
0 cells (0.0%) have expected count less than 5. The minimum expected count is 7.60.
Problem Statement 3:
A student researcher was surprised to learn that the 2017 NCAA Student-Athlete Substance Use Survey supported that college athletes make healthier decisions in many areas than their peers in the general student body. He collected data of his own, focusing exclusively on male student-athletes to see if such habits vary based on one’s sport. He asked 93 male student-athletes whether they had engaged in binge-drinking in the last month (> 5 drinks in a single sitting). Data are provided in the table below.
Lacrosse | Hockey | Swimming | Row Totals | |
---|---|---|---|---|
Yes – Binge | 20 | 17 | 15 | 52 |
No – did not binge | 16 | 15 | 10 | 41 |
Column totals | 36 | 32 | 25 | 93 |
Solution
Name the two variables of interest. Identify all levels associated with each variable
ANSWER
Student Athlete Sport – Lacrosse, Hockey and Swimming
Binge Drinking – Yes and No
Paste all relevant statistical output in the space provided below:
ANSWER
Table 3.1
Binge Drinking * Student Sport Crosstabulation
Count
Student Sport | Total | ||||
---|---|---|---|---|---|
Lacrosse | Hockey | Swimming | |||
Binge Drinking | Yes-Binge | 20 | 17 | 15 | 52 |
No-did not binge | 16 | 15 | 10 | 41 | |
Total | 36 | 32 | 25 | 93 |
Calculate the effect size. Show the formula and your calculations in the space provided below:
ANSWER
Table 3.2
Symmetric Measures
Value | Approx. Sig. | ||
---|---|---|---|
Nominal by Nominal | Phi | .054 | .873 |
Cramer's V | .054 | .873 | |
N of Valid Cases | 93 |
- Not assuming the null hypothesis.
- Using the asymptotic standard error assuming the null hypothesis.
Using the degrees of freedom provided by your SPSS output and an alpha value of .05, find the critical value in the appropriate table in the Appendix of your Jackson e-book. Do not round – present all three decimal places. Clearly identify the critical value from your e-book and the obtained value from your SPSS output. Based on this information, would you reject or fail to reject the null hypothesis? Does this mean there is a significant difference or no significant difference?
ANSWER
Alternative Hypothesis: There is a significant relationship between student athlete sport and binge drinking.
Null Hypothesis: There is no significant relationship between student athlete sport and binge drinking.
In other to test this hypothesis the chi-square test of significance was use to check for significant relationship as seen in the table 3.4 below. The result of the analysis shows that there is no statistically significant association between the student athelete sport and binge drinking at 2 degrees of freedom with p-value>0.05. therefore, we accept the null hypothesis and reject the alternative hypothesis.
Present the results using APA format. This includes a full write-up to include a complete statistical notation as shown in the weekly presentations. Make sure to describe what the conclusions mean in general terms. Additional examples of APA results sections are also available in the “Helpful Hints” document.
ANSWER
Table 3.3
Case Processing Summary
Cases | ||||||
---|---|---|---|---|---|---|
Valid | Missing | Total | ||||
N | Percent | N | Percent | N | Percent | |
Binge Drinking * Student Sport | 93 | 100.0% | 0 | 0.0% | 93 | 100.0% |
Table 3.4
Chi-Square Tests
Value | df | Asymp. Sig. (2-sided) | |
---|---|---|---|
Pearson Chi-Square | .272a | 2 | .873 |
Likelihood Ratio | .273 | 2 | .872 |
Linear-by-Linear Association | .089 | 1 | .765 |
N of Valid Cases | 93 |
0 cells (0.0%) have expected count less than 5. The minimum expected count is 11.02.
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