Question

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. Use these data to perform a test of the drug manufacturer's claim at the $\alpha =0.05$significance level.

Short answer

The data does not provide convincing evidence to support the manufacturers claim.

Step-by-step solution

Introduction

In measurable speculation testing, an outcome has factual importance when having happened given the invalid hypothesis is far-fetched.

Explanation

The population proportion is p is $37\%=0.37$

$p_0=1-p=1-0.37=0.63$

The number of patients is n = $300$

Patients that experienced nausea x = $25$

Standard proportion $\stackrel{-}{p}=\frac{x}{n}=\frac{25}{300}=0.0834$

Calculating the null and alternative hypotheses,

localid="1662024456534" $\begin{array}{r}{H}_0:p=0.10\\ H_1:p<0.10\end{array}$

Using,

localid="1662024901851" $z=\frac{\stackrel{-}{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

$z=\frac{0.0834-0.63}{\sqrt{\frac{{\displaystyle 0}{\displaystyle .}{\displaystyle 63}\left(1-0.63\right)}{300}}}=0.168$

Test statistic. 

The critical value is defined as below

$Z_{0.05}=1.64$its from the Z- table such that $P(Z>1.64)=0.05$

But test is left-tailed, therefore the critical value is $-1.64$

From the above we may define Reject $H_o$if$Z<-1.64$

Now, define the test statistic as

$$\begin{gathered} z=\frac{\stackrel{-}{p}-p_0}{\sqrt{\frac{pq}{n}}} \\ z=\frac{0.0833-0.1}{\sqrt{\frac{0.1\times 0.9}{300}}} \\ z=-0.96 \end{gathered}$$

This is our test statistic.

It is observed that $z>-1.64,$ so we do not reject H₀.

Hence, we conclude that the data does not provide convincing evidence to support the manufacturers claim.