Question

In the study of urinary tract infection, the sample and proportion in the group of

fosfomycin / trometamol was 25, 0.92 and trimethoprim / sulfamethoxazole was

16, 0.61, respectively.

a) Find the difference in proportions. (0.5)

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b) Standard error of difference in proportions. (0.5)

c) Compute a 95% confidence interval of difference in proportions

Answer 1

Given,

population 1

$n_1=25,\\\hat p_1 =0.92$

population 2

$n_2= 16,\\\hat p_2= 0.61,$

$a)$

The difference in proportion is $\hat p_1-\hat p_2=0.92-0.61=0.31$

$b)$

The standard error of difference in proportions is given as,

$SE(\hat p_1-\hat p_2)=\sqrt{{\hat p_1\times (1-\hat p_1)\over n_1}+{\hat p_2\times (1-\hat p_2)\over n_2}}=\sqrt{{0.92\times0.08\over 25}+{0.61\times0.39\over16}}=\sqrt{0.002944+0.01486875}=0.13346441$

Therefore, the standard error of difference in proportions is 0.13346441

$c)$

A 95% confidence interval of difference in proportions is given as,

$CI=(\hat p_1-\hat p_2)\pm Z_{\alpha\over2}SE(\hat p_1-\hat p_2)$ where $Z_{\alpha\over2}$ is the standard normal table value at $\alpha=0.05$. Thus $Z_{0.05\over2}=Z_{0.025}=1.96$

Hence,

$CI=0.31\pm(1.96\times 0.13346441)=0.31\pm0.26159024$

Therefore, a 95% confidence interval for the difference in proportions is, $(0.0484,0.5716)$