Equivalence of Hypothesis Test and Confidence Interval Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 10 having a common attribute. The second sample consists of 2000 people with 1404 of them having the same common attribute. Compare the results from a hypothesis test of $p_1=p_2$(with a 0.05 significance level) and a 95% confidence interval estimate of$p_1-p_2$.

In a sample consisting of 20 people, 10 possess a given attribute. In another sample of 2000 people, 1404 people possess the attribute.

It is claimed that the proportion of people having the attribute corresponding to the first population is equal to the proportion of people having the attribute corresponding to the second population.

The following hypotheses are set up:

Null Hypothesis:The proportion of people having the attribute corresponding to the first population is equal to the proportion of people having the attribute corresponding to the second population.

$H_0:p_1=p_2$

Alternative Hypothesis:The proportion of people having the attribute corresponding to the first population is not equal to the proportion of people having the attribute corresponding to the second population.

$H_1:p_1\ne p_2$

The test is two-tailed.

Here, $n_1$is the sample size for the first sample and $n_2$is the sample size for the second sample.

Thus, $n_1$is equal to 20 and $n_2$is equal to 2000.

Let ${\hat{p}}_1$denote the sampleproportion of people having the attribute in the first sample.

$$\begin{gathered} {\hat{p}}_1=\frac{10}{20} \\ =0.5 \end{gathered}$$

Let ${\hat{p}}_2$denote the sampleproportionof people having the attribute in the second sample.

$$\begin{gathered} {\hat{p}}_2=\frac{1404}{2000} \\ =0.702 \end{gathered}$$

The value of the pooled proportion is calculated as follows:

$$\begin{gathered} \overline{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)} \\ =\frac{\left(10+1404\right)}{\left(20+2000\right)} \\ =0.7 \end{gathered}$$

$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.7 \\ =0.3 \end{gathered}$$

The value of the test statistic is computed as shown below:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}\right)}} \\ =\frac{\left(0.5-0.702\right)-0}{\sqrt{\left(\frac{0.7\times 0.3}{20}+\frac{0.7\times 0.3}{2000}\right)}} \\ =-1.962 \end{gathered}$$

Thus, the value of the test statistic is -1.962.

Referring to the standard normal distribution table, the critical values of z corresponding to $\alpha =0.05$for a two-tailed test are -1.96 and 1.96.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0498.

Since the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to reject the claim that the two population proportions are equal.

The general formula for confidence interval estimate of the difference in the two proportions is written below:

$\text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) ...\left(1\right)$

The margin of error (E) has the following expression:

$E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)}$

For computing the confidence interval, first find the critical value $z_{\frac{\alpha }{2}}$.

The confidence level is 95%; thus, the value of the level of significance for the confidence interval becomes $\alpha =0.05$.

Hence,

$$\begin{gathered} \frac{\alpha }{2}=\frac{0.05}{2} \\ =0.025 \end{gathered}$$

The value of $z_{\frac{\alpha }{2}}$from the standard normal table is equal to 1.96.

Now, the margin of error (E) is equal to:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)} \\ =1.96\times \sqrt{\left(\frac{0.5\times 0.5}{20}+\frac{0.704\times 0.298}{2000}\right)} \\ =0.2200 \end{gathered}$$

Substitute the value of E in equation (1) as follows:

$$\begin{gathered} \text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) \\ =\left(\left(0.5-0.702\right)-0.2200 , \left(0.5-0.702\right)+0.2200\right) \\ =\left(-0.4220 ,\hspace{0.17em} 0.0180\right) \end{gathered}$$

Thus, the 95% confidence interval for the difference between the two proportions is.

The above interval contains the value 0. This implies that the difference in the two proportions can be equal to 0. In other words, the two population proportions have a possibility to be equal.

Thus, there is not enough evidence to reject the claim that the two population proportions are equal.

It can be observed that the conclusion using the p-value method is different from the conclusion obtained using the confidence interval method.

Thus, it can be said that the hypothesis test method and the confidence interval method are not always equivalent when testing the difference between the two population proportions.