Drug hypothesis testing

Question

A drug claims to cure patients in $80$% of all cases. A sample of $10$ patients was taken, $5$ got cured, $5$ not. I need to decide if the manufacturer should change their claim. Assume $5$% to be the significance level.

• $H_0$: p = $0.8$
• $H_A$: p < $0.8$

So, my first approach:

1. sample size = $10$, p = $0.5$, std dev = $0.25$
2. calculate the z-value for the sample: $z = \frac{(mean - sample \ mean)}{std \ error} $

3. here, I am not sure how to calculate $std \ error$. I am considering two formulas:

   • $std \ error_1 = {expected \ std \ dev} = \sqrt{0.8 \times 0.2} = \sqrt{1.6} = 1.264$
   • $std \ error_2 = {actual \ std \ dev} = \sqrt{\frac{p \times (1-p)}{sample \ size}} = \sqrt{\frac{0.5 \times (1-0.5)}{10}} = \sqrt{2.5} = 1.58$

4. check the z-value to be less than $-1.65$, if so, reject $H_0$

My second approach:

1. calculate the margin of error for p = 0.8:
2. mean = 8 patients $(0.8 \times 10)$

3. margin of error = $\pm z \ value \times std \ dev = \pm 1.96 \times \sqrt{10 \times 0.8 \times 0.2} = \pm 2.48$

4. see if the number of cured patients, 5, is within the margin from mean 8: [8-2.48, 8+2.48]

5. no, so reject $H_0$

Which approach makes sense?

Answer 1

Hy, the first approach of testing proportions makes sense. However since you already have $P$ values you don't need to use estimate and should find variance directly using $P = .8$ not $.5$