Answer to Question #282487 in Statistics and Probability for seli

Question

Each of two groups consists of 100 patients who have leukaemia. A new drug is given to

the first group but not to the second (the control group). It is found that in the first

group 60 people have remission for 2 years; but only 40 in the second group. Find 95%

CI.

Answer 1

$$\hat{p}_1=\dfrac{X_1}{N_1}=\dfrac{60}{100}=0.6$$

$$\hat{p}_2=\dfrac{X_2}{N_2}=\dfrac{40}{100}=0.4$$

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\hat{p}_2-\hat{p}_1-z_c\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{N_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{N_2}},$$

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

$$=(0.6-0.4-1.96\sqrt{\dfrac{0.6(1-0.6)}{100}+\dfrac{0.4(1-0.4)}{100}}$$

$$0.6-0.4+1.96\sqrt{\dfrac{0.6(1-0.6)}{100}+\dfrac{0.4(1-0.4)}{100}})$$

$$=(0.0642, 0.3358)$$

Therefore, based on the data provided, the 95% confidence interval for the difference between the population proportions $p_1-p_2$ is $0.0642 < p_1 - p_2 < 0.3358,$ which indicates that we are 95% confident that the true difference between population proportions is contained by the interval $(0.0642, 0.3358).$