Answer:
Explanation:
From the given information:
The number of children that were randomly allocated to each vaccination group; n₁ = 200,000
No of polio cases X₁ = 57
Now, in the vaccine group:
the proportion of polio cases is:
![\hat p_1 = (57)/(200000)](https://img.qammunity.org/2022/formulas/mathematics/college/rmkqrn3lgft6wp9sga43k8oqoer2omjlho.png)
= 0.000285
The number of children that were randomly allocated to the placebo group, n₂ = 200,000
No of polio cases X₂ = 142
In the placebo group
the proportion of polio cases is:
![\hat p_2 = (142)/(200000)](https://img.qammunity.org/2022/formulas/mathematics/college/54ey64r8c17ri5hjzz2ymgbapxbzonteea.png)
Null and alternative hypothesis is computed as follows:
H₀: There is no difference in the proportions of polio cases between both groups.
H₁: There is a difference in the proportions of polio cases between both groups.
Let assume that the level of significance ∝ = 0.05
The test statistic can be computed as:
![Z = \frac{\hat p_1-\hat p_2}{\sqrt{(\hat p_1 \hat q_1)/(n_1)+ (\hat p_2 \hat q_2)/(n_2)}}](https://img.qammunity.org/2022/formulas/mathematics/college/fj3xm1hjd3420tsy5ln9648eo7ihr2f753.png)
![Z = \frac{0.000285-0.000710}{\sqrt{(0.000285(1-0.000285))/(200000)+ (0.000710(1-0.000710))/(200000)}}](https://img.qammunity.org/2022/formulas/mathematics/college/qed29vb8jlz8nekf9tfl4yv9u05z27b2m3.png)
![Z = \frac{-4.25* 10^(-4)}{\sqrt{(0.000285(0.999715))/(200000)+ (0.000710(0.99929))/(200000)}}](https://img.qammunity.org/2022/formulas/mathematics/college/q7p0y2cmkwj8rnb3t0e14hdyjrq7eumsrw.png)
Z = - 6.03
P-value = 2P(Z < -6.03)
From the Z - tables
P-value = 2 × 0.0000
= 0.000
We reject the H₀ provided that P-value is very less.
Therefore, we may conclude that there is a difference in the proportions of polio cases between the vaccine group and placebo group not due to chance.