Random samples of size $n_1=400$ and$n_2=500$ were drawn from populations 1 and 2, respectively. The samples yielded${\text{x}}_1=105$ and $x_2=140$. Test$H_0:\left(p_1-p_2\right)⩾0$ against$H_a:\left(p_1-p_2\right)<0$at the 1% level of significance.

We have

The size of the samples is$n_1=400$and$n_2=500$

And

The number of successes is and

The sample proportion of successes is$x_1=105$and$x_2=140$

$\begin{array}{rcl}{\hat{p}}_1& =& \frac{105}{400}\\ & =& 0.2625\end{array}$

And

$\begin{array}{rcl}{\hat{p}}_2& =& \frac{140}{500}\\ & =& 0.28\end{array}$

For the large sample size (at least 30), the distribution of the sample proportion is approximately normal as per the central limit theorem.

The sample proportion can be viewed as means of the number of successes per trial in the respective samples, so the Central Limit Theorem applies when the sample sizes are large.

The test statistic for testing these hypotheses is

$$\begin{gathered} Z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}} \\ =\frac{0.2625-0.28}{\sqrt{\frac{0.2625\left(1-0.2625\right)}{400}+\frac{0.28\left(1-0.28\right)}{500}}} \\ =\frac{-0.0175}{\sqrt{0.000484+0.0004032}} \\ =\frac{-0.0175}{0.02979} \\ =-0.59 \end{gathered}$$

Here

$\alpha =.01$

Using the standard normal table, the critical value at the 1% significance level and left-tailed test (alternative hypothesis is left-tailed) is -2.326

That is

$\begin{array}{rcl}{Z}_{\alpha }& =& Z_{0.01}\\ & =& -2.326\end{array}$

We can see that

$Z>Z_{\alpha }$

That is,$-0.59>-2.326$

Hence, we failed to reject the null hypothesis.

At a 1% significance level, we do not have sufficient evidence to conclude that the difference between the two population proportions is less than 0.