In a sample of 200 workers 45% said they missed work because of personal illness. Ten years ago in a sample of 200 workers, 35% said they missed work because of personal illness. At $\alpha = 0.01$ is there a difference in the proportions?
Solution
Given that $n_1 = 200$, $n_2=200$.
The sample proportions are
$\hat{p}_1=0.45$ and $\hat{p}_2=0.35$.
The pooled estimate of sample proportion is
$\hat{p} =\frac{n_1\hat{p}_1+ n_2 \hat{p}_2}{n_1+n_2}=\frac{200*0.45+200*0.35}{200+200} =0.4$
Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : p_1 = p_2$ against $H_1 : p_1 \neq p_2$ ($\textit{two-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned} $$
The test statistic $Z$ follows standard normal distribution $N(0,1)$.
Step 3 Specify the level of significance $\alpha$
The significance level is $\alpha = 0.01$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $Z$ $\text{ are }$ $\text{-2.58 and 2.58}$ (From Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < -2.58 or Z > 2.58}$.
Step 5 Computation
The test statistic under the null hypothesis is
$$ \begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.45-0.35)-0}{\sqrt{\frac{0.4*(1-0.4)}{200}+\frac{0.4*(1-0.4)}{200}}}\\ &= 2.041 \end{aligned} $$
Step 6 Decision
Traditional approach:
The rejection region (i.e. critical region) is $\text{Z < -2.58 or Z > 2.58}$. The test statistic is $Z_{obs} =2.041$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
$p$-value approach:
The test is $\text{two-tailed}$ test, so the p-value is the area to the $\text{extreme}$ of the test statistic ($Z_{obs}=2.041$) is p-value = $0.0412$.
The p-value is $0.0412$ which is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.
There is no sufficient evidence to conclude that the that there is difference in the proportions.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
