Question

A health officer is trying to study the malaria situation of Ethiopia. From the records of

seasonal blood survey (SBS) results he came to understand that the proportion of people

having malaria in Ethiopia was 3.8% in 1978 (Eth. Cal). The size of the sample

considered was 15000. He also realised that during the year that followed (1979), blood

samples were taken from 10,000 randomly selected persons. The result of the 1979

seasonal blood survey showed that 200 persons were positive for malaria. Help the

health officer in testing the hypothesis that the malaria situation of 1979 did not show

any significant difference from that of 1978 (take the level of significance, α =.01).

Answer 1

$$\hat{p}_1=0.038, N_1=15000, X_1=0.038(15000)=570$$

$$\hat{p}_2=\dfrac{X_2}{N_2}=\dfrac{200}{10000}=0.02, N_2=10000, X_2=200$$

The value of the pooled proportion is computed as 

$$\bar{p}=\dfrac{X_1+X_2}{N_1+N_2}=\dfrac{570+200}{15000+10000}=0.0308$$

The following null and alternative hypotheses for the population proportion needs to be tested:

​$H_0:p_1=p_2$

$H_1:p_1\not=p_2$

This corresponds to a two-tailed test, and a z-test for two population proportions will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\} .$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(\dfrac{1}{N_1}+\dfrac{1}{N_2})}}$$

$$=\dfrac{0.038-0.02}{\sqrt{0.0308(1-0.0308)(\dfrac{1}{15000}+\dfrac{1}{10000})}}$$

$$=8.0699$$

Since it is observed that $|z| = 8.0699 > 2.5758=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z>8.0699)\approx0,$ and since $p = 0 < 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p_1$

is different than $p_2,$ at the $\alpha = 0.01$ significance level.