Question
You will analyze the following variables in the grades.sav data set:
SPSS Variables and Definitions
SPSS Variable Definition
Gender female = 1; male = 2
GPA previous grade point average
Solution
t-Test Application and Interpretation
Variable used in this t-Test analysis is the “gender variable” and it is a Nominal variable since the categories cannot be ordered from high to low. The gender variable containing the male and female entities. The value of the female gender corresponds to the value of the male gender, for each case in the data set. The research question will determine the relationship between the gender and student’s results, total and final. The data is sorting and presenting information of teacher’s recording on student demographic and performance on quizzes and a final exam across the course.
Research question
The research question entails conduction of t-Test on a nominal variable data provided – homogeneity of variance and SPSS output showing Levene’s test for equality variances.
Null hypothesis
The null hypothesis proves / states that there is no statistical relationship between two variables.
Alternate hypothesis
The alternate hypothesis defines that there is a statistically important relationship between two variables.
Assumptions of t-tests—homogeneity of variance.
Set P-value is .01
P-value of the variances = .250.
Independent Samples Test | |||||||||||||||
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 | ||||||||||||||
total | Equal variances assumed | 2.132 | .147 | 1.157 | 103 | .250 | 3.102 | 2.682 | -2.217 | 8.420 | |||||
Equal variances not assumed | 1.111 | 74.084 | .270 | 3.102 | 2.792 | -2.461 | 8.665 | ||||||||
Non-significant result greater than .01 indicates that the assumption of homogeneity of variance is statistically significant.
Levene’s test for equality variances
Independent Samples Test | ||||||||||
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 | |||||||||
total | Equal variances assumed | 2.132 | .147 | 1.157 | 103 | .250 | 3.102 | 2.682 | -2.217 | 8.420 |
Equal variances not assumed | 1.111 | 74.084 | .270 | 3.102 | 2.792 | -2.461 | 8.665 |
T= 1.157, with a degree of freedom of 103 and a significant value of .250. Due to the higher p value of .250 as compared to the p value of .01, the null hypothesis is no difference between the means and the state that in all probability, the difference in total value between the male gender and the female gender is statistically significant / approximately equal.
t-Test
Group Statistics |
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gender | N | Mean | Std. Deviation | Std. Error Mean | |
total | Female | 64 | 101.30 | 12.379 | 1.547 |
Male | 41 | 98.20 | 14.880 | 2.324 |
Independent Samples Test | ||||||||||
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 | |||||||||
total | Equal variances assumed | 2.132 | .147 | 1.157 | 103 | .250 | 3.102 | 2.682 | -2.217 | 8.420 |
Equal variances not assumed | 1.111 | 74.084 | .270 | 3.102 | 2.792 | -2.461 | 8.665 |
Means and standard deviations for each group
Descriptive Statistics | ||||||||||
N | Mean | Std. Deviation | ||||||||
gender | 105 | 1.39 | .490 | |||||||
Valid N (listwise) | 105 | |||||||||
Case Processing Summary | ||||||||||
Cases | ||||||||||
Included | Excluded | Total | ||||||||
N | Percent | N | Percent | N | Percent | |||||
gender | 105 | 100.0% | 0 | 0.0% | 105 | 100.0% | ||||
Report | |
gender | |
Mean | Std. Deviation |
1.39 | .490 |
Means
the mean of the female gender is 101.30.
the mean of the male gender is 98.20.
Standard deviations
The standard deviation of the female gender is 12.379.
The standard deviation for the male gender is 14.880.
Results of the t-test using the “Assume equal variances” row
Since the significant variances are not found to be equal
i.e., equal variances assumed = .250
equal variances not assumed = .270,
then, the F statistic will indicate that,
the null hypothesis shows that the variances of the two groups are approximately equal, hence the distribution of competency scores for the male gender is similar in shape to the distribution of competency scores in the female gender.
The alternate hypothesis states that the two distributions are significantly different in shape
Level of significance = .01
Interpretation of the statistical results against the null hypothesis and stating whether they are accepted or rejected.
mean difference = female mean – male mean
= 101.30 – 98.30
= 3.1.
Standard error associated with the mean difference is 2.682.
Confidence interval = 95%; this states that its 95% confident that the actual difference of the main competency for the male and the female gender is between -2 and +8.
Hence concluding that sometime the male gender will have a smaller competency score than the female gender and there is a chance that the male gender could have a higher competency score (+8), which is why there is no significant difference because the chances can go either way.
The null hypothesis is accepted since the p-value is lower than the significant value.
Statistical Conclusions
Given the nominal variable (Gender) on the basis of conducting an SPSS data analysis, the summary is given as follows;
Under the gender variable, with a total number of 105, the mean is 1.39 and the standard deviation of the variable is .490.
The female gender carries a total number of 64 entities and the male gender carries a total of 41 entities.
The null hypothesis carried out indicates that the error variance of the dependent variable is equal across the following groups;
- Dependant variable; Totals
- Design; intercept + gender
Using a P-Value of .01, the significant value after the analysis is .250, stating a higher value than the p-value.
This proves that the null hypothesis is accepted.
The confident level is 95%. This shows that the actual difference between the gender competency is between -2 and +8.
This rate declares that the male gender has a smaller competency than the female gender, though, it has a higher chance of having a higher competency against the female gander (+8).
Application.
A t-Test is any statistical hypothesis test in which the test statistic follows a student’s t-distribution under the null hypothesis. It is most commonly applied when the test statistic would follow a normal distribution if the value of a scaling term in the test statistic were known (typically, the scaling term is unknown and therefore a nuisance parameter. When the scaling term is estimated based on the data, the test statistic—under certain conditions—follows a Student’s t distribution.
The t-test’s most common application is to test whether the means of two populations are different.t-Test analysis could be used in job set to;
- compare two means of two groups
- hypothesis testing to determine whether a process or treatment actually has an effect on population of interest, or whether two groups are different from one another.
The outcome Prediction: Understanding the relationships between different variables of the data set, one may be able to use t-Test analysis to check if the results are statistically significant or not.
i.e. one may be able to predict a student’s performance based on the confidence rate given.
t-Test Trends: t-Test analysis can help in identification for trends in data such as whether there is a high rate of competency or low in the data set.
Decision making: By understanding the relationship between different variables, t-Test analysis can be used to make up a decision.
i.e. Results of mean data and standard deviation can help in deciding which variable is good in competence or performance.
Substantially: t-Test analysis is a useful tool for understanding how significant the difference between the means of two groups while taking their variance into account.
References
- O’Mahony, Michael (1986). Sensory Evaluation of Food: Statistical Methods and Procedures. CRC Press. p. 487. ISBN0-82477337-3.
- Press, William H.; Teukolsky, Saul A.; Vetterling, William T.; Flannery, Brian P. (1992). Numerical Recipes in C: The Art of Scientific Computing. Cambridge University Press. p. 616. ISBN0-521-43108-5.