# Test the hypothesis using the P-value approach.

Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test.
versus ${H}_{1}:p<0.52$
$N=150,X=72,\alpha =0.1$
Is $N{P}_{0}\left(1-{P}_{0}\right)$ greater than or equal to 10?
Use technology to find the P-Value.

You can still ask an expert for help

## Want to know more about Sampling distributions?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

pattererX

Null hypothesis, ${H}_{0}:P=0.52$
Alternative hypothesis, ${H}_{1}:P<0.52$
Under ${H}_{0}$, the population proportion is ${P}_{0}=0.52$
Now consider,
$N{P}_{0}\left(1-{P}_{0}\right)=150×0.48×\left(1-0.48\right)$
$=37.44$
$>10$
Since $N{P}_{0}\left(1-{P}_{0}\right)$  is greater than or equals to 10, use normal approximation to test the true proportion.
Given information:
Number of trials, $N=150$
Number of sucesses in 150 cases, $X=72$
Sample proportion,
$p=\frac{X}{N}$
$=\frac{72}{150}$
$=0.48$
Compute test statistic.
$z=\frac{p-{P}_{0}}{\sqrt{\frac{{P}_{0}\left(1-{P}_{0}\right)}{N}}}$
$=\frac{0.48-0.52}{\sqrt{\frac{\left(0.52\right)\left(1-0.52\right)}{N}}}$
$=-0.98058$
$=-0.98$
Alternative hypothesis has symbol "<", so the test is based on left-tailed.
Using Excel function "=NORMSTD(z)"
$P-value=P\left(Z\le z\right)$
$=P\left(Z\le -0.98\right)$
$=NORMSDIST\left(-0.98\right)$
$=0.1635$
When testing the hypothesis ${H}_{0}:P=0.52$ versus ${H}_{1}:P<0.52$, using the sample information $N=150$ and $X=72$, the P-value of test is 0.1635