Question

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.

Answer 1

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