Question

Treating AIDS The drug AZT was the first drug that seemed effective in delaying

the onset of AIDS. Evidence for AZT’s effectiveness came from a large randomized

comparative experiment. The subjects were $870$volunteers who were infected with HIV,

the virus that causes AIDS, but did not yet have AIDS. The study assigned $435$of the

subjects at random to take $500$milligrams of AZT each day and another $435$to take a

placebo. At the end of the study, $38$of the placebo subjects and $17$of the AZT subjects

had developed AIDS.

a. Do the data provide convincing evidence at the $\alpha =0.05$level that taking AZT lowers the proportion of infected people like the ones in this study

who will develop AIDS in a given period of time?

b. Describe a Type I error and a Type II error in this setting and give a consequence of

each error.

Short answer

(a)Yes,the data provide convincing evidence at the $\alpha =0.05$level that taking AZT lowers the proportion of infected people.

(b)Type I error : This error occurs when our null hypotheses is true and we reject our null hypotheses.

consequence of Type I error :Our decision is wrong.

Type II error : This error occurs when our null hypotheses is false and we accept our null hypotheses.

consequence of Type II error :Our decision is wrong.

Step-by-step solution

Part (a) Step 1:Given Information

We have been given that,

Number of subjects assigned to take AZT each day(n₁)=$435$

Number of AZT subjects who had developed AIDS (r₁)=$17$

Number of subjects assigned to take placebo each day (n₂) =$435$

Number of placebo subjects who had developed AIDS (r₂)=$38$

Part (a) Step 2:Explanation

P₁=Observed proportion in first sample

P₁= $\frac{r_1}{n_1}=\frac{17}{436}=0.03$

P₂=Observed proportion in second sample

P₂=$\frac{r_2}{n_2}=\frac{38}{435}=0.08$

p₁=proportion in first population

p₂=proportion in second population

Observed value of difference in proportions=P₁-P₂

Expected value of difference in proportions=p₁-p₂=$0$

p=(r₁+r₂)/n₁+n₂

p=$\frac{17+38}{435+435}=0.06$

q=$1-p=1-0.06=0.94$

Standard Deviation=$\sqrt{p\times q\times (\frac{1}{n_1}+\frac{1}{n_2}})=\sqrt{0.06\times 0.94\times (\frac{1}{435}+\frac{1}{435})}=0.016$

Null hypotheses: AZT do not lowers the proportion of infected people

Alternative hypotheses: AZT lowers the proportion of infected people

localid="1654269083875" $\left|Z\right|=\frac{P1-P2-0}{\sqrt{p\times q\times ({\displaystyle \frac{1}{n1}}+{\displaystyle \frac{1}{n2}}})}$

localid="1654269118341" $\left|z\right|=\frac{0.03-0.08-0}{0.016}$

localid="1654269502461" $\left|z\right|=-3.125$

Tabulated value at $\alpha =0.05$level of significance =$\pm 1.96$

Since z_cal>z_tab

Null hypotheses is rejected.

Conclusion:AZT lowers the proportion of infected people

Part (b) Step 1:Given Information

We have to describe a Type I error and a Type II error

Part (b) Step 2:Explanation

Type I error : This error occurs when our null hypotheses is true and we reject our null hypotheses.

consequence of Type I error :Our decision is wrong.

Type II error : This error occurs when our null hypotheses is false and we accept our null hypotheses.

consequence of Type II error :Our decision is wrong.