91Ó°ÊÓ

In the study, the first population of interest is all patients admitted in January, and the second population of interest is all patients admitted in May.

Consider x₁ = 32 and n₁= 192.

The point estimate for the malaria admission rates in January is,

$\begin{array}{l}{\underline{\underset{\_}{P}}}_1=\frac{x_1}{n_1}\\ =\frac{32}{92}\end{array}$

= 0.167

Consider x₂ = 34 and n₂ = 403.

The point estimate for the malaria admission rates in May is,

$\begin{array}{l}{\underline{\underset{\_}{P}}}_2=\frac{x_2}{n_2}\\ =\frac{34}{403}\\ =0.084\end{array}$

$$\begin{gathered} \underline{P_1}-\underline{P_2}=0.167-.084 \\ =0.083 \end{gathered}$$

Thus, the point estimate for the contrast in the malaria admission rates in January and May is 0.083.

The critical value for a two-tailed test is obtained below:

Here, the test is two-tailed, and the significance level is α=0.10.

The rejection region for the two-tailed test is$\left|z\right|>z_{\frac{a}{2}}$.

The confidence coefficient is 0.90.

So,

(1-α) = 0.90

α =0.10

$\frac{\mathrm{Î\pm }}{2}=0.05$

From Appendix D, Table II, the critical value for the two-tailed test with α = 0.10 is $z_{\frac{a}{2}(−0.05)}$=±1.645 Hence, the rejection region is $\left|z\right|$> 1.645.

The 90% confidence interval is obtained below:

$\left(\frac{({\underset{\_}{P}}_1−{\underset{\_}{p}}_2)Â\pm z_{0.05}}{\frac{{\underset{\_}{P}}_1(1−{\underset{\_}{P}}_1)}{n_1}+\frac{{\underset{\_}{P}}_2(1−p_2)}{n_2}}\right)=0.083Â\pm 1.645\sqrt{\frac{0.167(1−0.167)}{192}+\frac{0.084(1−0.084)}{403}}$

$$\begin{gathered} =0.083Â\pm 1.645(0.0303) \\ =0.083Â\pm 0.050 \\ =(0.033,0.133) \end{gathered}$$

90% confidence interval for the contrast in the malaria admission rates in January and May is (0.033.0.133).

Yes, it can be concluded that the contrast exists in the authentic malaria admission rates in January and May.

Explanation

The 9 z- confidence interval for (p1 – p2) is (0.033.0.133), which doesn't contain the hypothecated value 0.

That is, the hypothecated value p0 = 0 falsehoods outside the interval (0.033,0.133)

So, by the condition, if the hypothecated value (p0) lies outside the corresponding 100 (1-α) Z- confidence interval for (P1 – P2), also reject the null hypothesis.

Therefore, it can be concluded that reject the null thesis H0 at α = -0.05.

Hence, the contrast exists in the authentic malaria admission rates in January and May.