Sanitarium administration of malaria cases. One of the most sedate health challenges in India is malaria. Accordingly, the Indian sanitarium director's must-have—the coffers to treat the high volume of admitted malaria cases. A study published in the National Journal of Community Medicine (Vol. 1, 2010) delved into whether the malaria admission rate is more advanced in months than in others. In a sample of 192 sanitarium cases admitted in January, 32 were treated for malaria.

In an independent sample of 403 cases admitted in May (4 months latterly), 34 were treated for malaria.

a. Describe the two populations of stake in this study.

b. Give a point estimate of the contrast in the malaria admission rates in January and May.

c. Find a 90% confidence interval for the contrast in the malaria admission rates in January and May.

d. Based on the interval, part c, can you conclude that contrast exists in the authentic malaria admission rates in January and May? Simplify.

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,

$$\begin{gathered} \underset{-}{P_1}=\frac{x_1}{n_1} \\ =\frac{32}{92} \\ =0.167 \end{gathered}$$

= 0.167

Consider x₂ = 34 and n₂ = 403.

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

$$\begin{gathered} \underset{-}{P_2}=\frac{x_1}{n_1} \\ =\frac{34}{403} \\ =0.084 \end{gathered}$$

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

= 0.05

From Appendix D, Table II, the critical value for the two-tailed test with α = 0.10 is =±1.645 Hence, the rejection region is $\left|z\right|$> 1.645.

The 90% confidence interval is obtained below:

( $\frac{\left(p_1-p_2\right)\pm z_{0.05}}{{\displaystyle \frac{p_1\left(1-p_1\right)}{n_1}}+\frac{p_2\left(1-p_2\right)}{n_2}}$) = 0.083±1.645$\sqrt{\frac{0.167\left(1-0.167\right)}{192}+\frac{0.084\left(1-0.084\right)}{403}}$

= 0.083±1.645(0.0303)

=0.083±0.050

= (0.033,0.133)

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.