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