Question

# Test the hypothesis using the P-value approach.

Sampling distributions

Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test.
$$H_0:\ P=0.52$$ versus $$H_1:p<0.52$$
$$N=150, X=72,\alpha=0.1$$
Is $$NP_0(1-P_0)$$ greater than or equal to 10?
Use technology to find the P-Value.

2021-05-28

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,
$$NP_0(1-P_0)=150\times0.48\times(1-0.48)$$
$$=37.44$$
$$>10$$
Since $$NP_0(1-P_0)$$  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.
$$z=\frac{p-P_0}{\sqrt{\frac{P_0(1-P_0)}{N}}}$$
$$=\frac{0.48-0.52}{\sqrt{\frac{(0.52)(1-0.52)}{N}}}$$
$$=-0.98058$$
$$=-0.98$$
Alternative hypothesis has symbol "<", so the test is based on left-tailed.
Using Excel function "=NORMSTD(z)"
$$P-value=P(Z\leq z)$$
$$=P(Z\leq-0.98)$$
$$=NORMSDIST(-0.98)$$
$$=0.1635$$
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