91Ó°ÊÓ

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

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

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

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

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

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.