One-Sample t Test
Research Scenario
Question
Janet is interested in determining if male adolescents who do not engage in sports are more or less depressed than the average male adolescent. To test this hypothesis, she obtains Kansas University Depression Inventory (KUDI) score from 30 adolescent boys who indicated that they do not engage in sports outside of school gym classes.
Instrument & Scoring
The KUDI is a depression measure that was standardized to have a mean of 50 for male adolescent which means that scores lower than 50 imply participants are less depressed than a typical male adolescent while scores greater than 50 imply participants are more depressed than a typical male adolescent.
Null Hypothesis
Since Janet is interested in only determining whether a difference in depression scores exist between teen boys who play sports or do not play sports (adolescent males that do not play sports are more or less depressed than a typical male adolescent), the null hypothesis for this test would be that there is no difference among groups. Additionally, since the research question is non-directional, a two-tailed p value is used.
Assumptions
- The test variable is normally distributed in the population.
- The cases represent a random sample from the population, and the scores on the test variable are independent of each other.
Check
Of the two major assumptions for the one-sample t test, only the first can be evaluated using data alone. The second assumption is something that best addressed through implementing proper research/sampling practices–simply put, if the second assumption is found to be violated in the data analysis phase, there is little the researcher can do to address the issue.
Scores are Normally Distributed
To evaluate the first assumption of normality, develop a histogram of the observed scores. Furthermore, and if desired, conduct skewness and kurtosis analyses.
- skewness: -0.23
- kurtosis: 0.27
Conclusion
In reviewing the histogram along with the skewness and kurtosis values above, the data appears to approximate a normal distribution. The first assumption is confirmed.
One-Sample t Test Overview
The one-sample t test determines whether a sample mean is statistically different from a known or hypothesized population mean.
The One Sample t Test is commonly used to test the following:
- Difference between a sample mean and a known or hypothesized value of the mean in the population.
- The difference between the sample mean and the sample midpoint of the test variable.
- The difference between the sample mean of the test variable and chance.
- This approach involves first calculating the chance level on the test variable. The chance level is then used as the test value against which the sample mean of the test variable is compared.
- This approach involves first calculating the chance level on the test variable. The chance level is then used as the test value against which the sample mean of the test variable is compared.
- The difference between a change score and zero.
- This approach involves creating a change score from two variables, and then comparing the mean change score to zero, which will indicate whether any change occurred between the two time points for the original measures. If the mean change score is not significantly different from zero, no significant change occurred.
One-Sample t Test Overall Equation
\[ t = \frac{\overline{X} - \mu}{\left({\displaystyle\frac{s}{\sqrt{n}}}\right)} \]
Simplified Equation
\[t = \frac{\text{Sample Mean - Population Mean}}{\left({\displaystyle\frac{\text{Sample Standard Deviation}}{\sqrt{n}}}\right)} \]
Data Needed to Complete the One-Sample t Test
To conduct a one-sample t test five pieces of information are required; however, only the population mean is provided for Janet’s study.
- population mean (\(\mu\))) = 50
- sample size (n) = ?
- sample mean (\(\overline{X}\)) = ?
- sample standard deviation (s) = ?
- degrees of freedom = ?
To get the other pieces of information, Janet will need to conduct additional descriptives.
| descriptive | value |
|---|---|
| n | 30.00 |
| mean | 54.63 |
| sd | 10.33 |
| df | 29.00 |
Using the table above, Janet has gathered the outstanding pieces of information needed to conduct the one-sample t test:
- population mean (\(\mu\)) = 50
- sample size (n) = 30
- sample mean (\(\overline{X}\)) = 54.63
- sample standard deviation (s) = 10.33
- degrees of freedom = 29
Step 1: Input All of the Data
\[t = \frac{54.63 - 50}{\left({\displaystyle\frac{10.33}{\sqrt{30}}}\right)} \]
Step 2: Simplify the Numerator and Denominator
\[t = \frac{4.63}{{\displaystyle1.89}} \] Step 3: Round the Resulting t Value
\[ t = 2.46 \]
Step 4: Determine Significance
To determine if the computed t value is significant and therefore reject the null hypothesis, consult a table of critical t values. A critical value is the value that a test statistic (in Janet’s case the t statistic) must exceed in order for the null hypothesis to be rejected. In this instance, using degrees of freedom equal to 29 with a p value of 0.05, the critical value for this t test is 2.042.
One-Sample t Test Conclusion
The observed difference appears to be significant since the computed t value of 2.46 is larger than the critical t value of 2.042.
Effect Size
An effect size is a way of quantifying the size of the difference between two groups; or in this case, a measure of magnitude of difference between Janet’s group of non-sport playing male adolescents and typical male adolescents using KUDI scores. To evaluate effect size, use the Cohen’s d effect size.
Cohen’s d Overall Equation
\[ d = \frac{t}{\sqrt{n}} \]
Input t and sample size
\[d = \frac{2.46}{\sqrt{5.48}} \]
Cohen’s d Outcome
\[d = 0.45 \]
The resulting Cohen’s d is considered to be small.
A work by Alex Aguilar
aaguilar@thechicagoschool.edu