Nonparametric Procedures and Chi square Test Order Instructions: Application: SPSS Exercises
Complete the following exercises in your course text Using SPSS for Windows and Macintosh: Analyzing and Understanding Data, by Green and Salkind. Be sure to save your output and export it to your Word document, in which you also must answer the analysis questions and present your results section as indicated:
• Exercises 1–4 on p. 327, Chi-square test (Note: Use “Lesson 40 Exercise File 1.sav”)
• Exercises 1–4 on p. 343, Nonparametric procedures
Exercises 1–5 on p. 354, Nonparametric procedures.
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Nonparametric Procedures and Chi square Test Sample Answer
327, Chi-square test
The hypothesis, in this case, is H0: the two samples are independent, versus, Ha1: the two samples are dependent.
Table 1: Chi-Square Tests |
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Value | df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 96.000a | 4 | .000 |
Likelihood Ratio | 80.352 | 4 | .000 |
Linear-by-Linear Association | .047 | 1 | .828 |
N of Valid Cases | 48 | ||
a. 6 cells (66.7%) have expected count less than 5. The minimum expected count is 1.02. |
The chi-square critical value at (0.05, 4) = 9.488, the null hypothesis is rejected. Thus, the two samples are dependent.
343, Nonparametric procedures
The Mann-Whitney U-test results are as summarized in Table 1 and Table 2. In this case, the null hypothesis to be tested is H0: The two samples come from the same distribution, versus, Ha1: the two sample do not come from the same distribution.
Table 2:
Test Statisticsa |
|
time (seconds) | |
Mann-Whitney U | 112.500 |
Wilcoxon W | 217.500 |
Z | -1.974 |
Asymp. Sig. (2-tailed) | .048 |
Exact Sig. [2*(1-tailed Sig.)] | .048b |
a. Grouping Variable: weight | |
b. Not corrected for ties. |
- The p-value is 0.048 since p-value is smaller than the level of significance the null hypothesis is rejected. Thus, the sample population is
- Z value corrected for ties is -1.974
Table 3: Ranks |
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Weight | N | Mean Rank | Sum of Ranks | |
time (seconds) | Overweight | 26 | 23.17 | 602.50 |
normal weight | 14 | 15.54 | 217.50 | |
Total | 40 |
- Mean rank for normal weighted individuals is 15.54
Table 4:
Independent Samples Test |
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Levene’s Test for Equality of Variances | t-test for Equality of Means | |||||||||
F | Sig. | t | df | Sig. (2-tailed) | Mean Difference | Std. Error Difference | 95% Confidence Interval of the Difference | |||
Lower | Upper | |||||||||
time (seconds) | Equal variances assumed | 1.217 | .277 | 2.156 | 38 | .037 | 12.423 | 5.763 | .757 | 24.089 |
Equal variances not assumed | 2.023 | 22.400 | .055 | 12.423 | 6.140 | -.297 | 25.143 |
The p-value of t-test (assuming the data has equal variance) is 0.37, whereas that of Mann-Whitney U test is 0.048. In t-test, the hypothesis to be tested is H0: The two samples have an equal mean, versus, the Ha1: the two sample population have different mean.
354, Nonparametric procedures
Table 5:
Table Ranks |
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Hair Color | N | Mean Rank | |
Social Extroversion | Blond | 6 | 12.75 |
Brunet | 6 | 10.25 | |
Redhead | 6 | 5.50 | |
Total | 18 |
Table 6:
Test Statisticsa,b |
|
Social Extroversion | |
Chi-Square | 5.963 |
df | 2 |
Asymp. Sig. | .051 |
a. Kruskal-Wallis Test | |
b. Grouping Variable: Hair Color |
- No need to run another test (Post Hoc test), for the difference is insignificant.
Descriptive Statistics | |||||
N | Mean | Std. Deviation | Minimum | Maximum | |
Social Extroversion | 18 | 3.72 | 2.109 | 1 | 10 |
Hair Color | 18 | 2.00 | .840 | 1 | 3 |
d =, where s pooled =
s pooled =
=
=
d =
Thus, the effect size is
- One way ANOVA
Table 7:
ANOVA |
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Hair Color | |||||
Sum of Squares | df | Mean Square | F | Sig. | |
Between Groups | 5.300 | 6 | .883 | 1.450 | .280 |
Within Groups | 6.700 | 11 | .609 | ||
Total | 12.000 | 17 |
At 95% level of significant, these results are not statistically significance, since p-value 0.280 is greater than α = 0.05
- The results of Kruskal-Wallis and One-way ANOVA results indicate that there is no statistically significant difference between the mean of the variables.