Side effects A drug manufacturer claims that less than $10\%$of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of $300$out of $5000$Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, $25$of the subjects experience nausea.

a. Describe a Type I error and a Type II error in this setting, and give a possible

consequence of each.

b. Do these data provide convincing evidence for the drug manufacturer’s claim?

It is given that $\alpha =0.05$

$n=300$

$x=25$

Claim is less than$10\%$

According to given data:

Null hypothesis: $H_0:p=10\%=0.10$

Alternate hypothesis: $H_1:p<0.10$

Type I error: There is proof that percentage of patients suffering from nausea $<10\%$where the percentage of patients experience nausea is actually $10\%$

Consequence can be that doctors underestimate patients experiencing nausea are side effects are worse.

Type II error: There is sufficient convincing proof that percentage of patients suffering from nausea $<10\%$where the percentage of patients experience nausea is actually $10\%$

Consequence is doctors overestimate patients experiencing nausea and side effects are better than expected.

The condition of normality is: $n{p}_0=300(0.10)=30$and $\mathrm{n}\left(1-p_0\right)=300(1-0.10)=300(0.90)=270$

Both are greater than $10$

We can use hypothesis test.

Sample proportion: $\hat{p}=\frac{x}{n}=\frac{25}{300}=\frac{1}{12}=0.0833$

Test static: $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.0833-0.10}{\sqrt{\frac{0.10(1-0.10)}{300}}}=-0.96$

$P$value is $P=P(Z<-0.96)=0.1685$

Now, $P>0.05\Rightarrow \text{Fail to reject}{H}_0$

There is not convincing evidence that manufacturer's claim is false.