Grades in a statistics course and an operations research course taken simultaneously were as follows for a group of students.
Statistics Grade/ Operation Research Grade | A | B | C | Other |
---|---|---|---|---|
A | 25 | 6 | 17 | 13 |
B | 17 | 16 | 15 | 6 |
C | 18 | 4 | 18 | 10 |
Other | 10 | 8 | 11 | 20 |
Are the grades in statistics and operations research related? Use $\alpha = 0.01$ in reaching your conclusion. What is the P-value for this test?
Solution
The observed data is
OR A | OR B | OR C | OR Other | Sum | |
---|---|---|---|---|---|
ST A | 25 | 6 | 17 | 13 | 61 |
ST B | 17 | 16 | 15 | 6 | 54 |
ST C | 18 | 4 | 18 | 10 | 50 |
ST Other | 10 | 8 | 11 | 20 | 49 |
Sum | 70 | 34 | 61 | 49 | 214 |
Number of rows $r=4$, number of coulmns $c=4$.
Step 1 The null and alternative hypothesis are as follows:
$H_0:$ The grades in Statistics and Operation Research are not related.
$H_1:$ The grades in Statistics and Operation Research are related.
Step 2 Test statistic:
The test statistic for testing above hypothesis is
$$ \begin{equation*} \chi^2= \sum \sum \frac{(O_{ij} -E_{ij})^2}{E_{ij}} \sim \chi^2_{(r-1)(c-1)}\\ \end{equation*} $$
Step 3 Level of Significance
The level of significance is $\alpha =0.01$.
Step 4 Critical value of $\chi^2$
The level of significance is $\alpha =0.01$. Degrees of freedom $df=(r-1)(c-1)=(4-1)(4-1) =9$.
The critical value of $\chi^2$ for $df=9$ and $\alpha=0.01$ level of significance is $\chi^2_{0.01,9} =21.666$.
Step 5 Computation of test Statistic
The expected frequency for $(i,j)^{th}$ cell is given by
$$ \begin{equation*} E_{ij} =\frac{i^{th}\text{ row total }\times j^{th}\text{ column total}}{N} \end{equation*} $$
For example, $E_{11}$ is given by
$$ \begin{eqnarray*} E_{11} & = &\frac{1^{st}\text{ row total }\times 1^{st}\text{ column total}}{N}\\ &=& \frac{61*70}{214}\\ &=&19.95. \end{eqnarray*} $$
Table of expected frequencies
OR A | OR B | OR C | OR Other | Sum | |
---|---|---|---|---|---|
ST A | 19.95 | 9.69 | 17.39 | 13.97 | 61.00 |
ST B | 17.66 | 8.58 | 15.39 | 12.36 | 53.99 |
ST C | 16.36 | 7.94 | 14.25 | 11.45 | 50.00 |
ST Other | 16.03 | 7.79 | 13.97 | 11.22 | 49.01 |
Sum | 70.00 | 34.00 | 61.00 | 49.00 | 214.00 |
The test statistic is
$$ \begin{eqnarray*} \chi^2&=& \sum \sum \frac{(O_{ij} -E_{ij})^2}{E_{ij}} \sim \chi^2_{(r-1)(c-1)}\\ &=&\frac{(25-19.95)^2}{19.95}+\cdots + \frac{(20-11.22)^2}{11.22}\\ &=& 25.5545. \end{eqnarray*} $$
Step 6 Decision (Traditional approach)
The test statistic is $\chi^2 =25.5545$ which falls $inside$ the critical region bonded by the critical value $\chi^2_{0.01,9}=21.666$, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision ($p$-value approach)
The p-value is $P(\chi^2_{9}>25.5545) =0.00242$.
As the p-value $0.0024$ is $\textit{less than}$ the significance level of $\alpha = 0.01$, we $\textit{reject}$ the null hypothesis.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators