Null hypothesis, \(H_0:P=0.52\)
Alternative hypothesis, \(H_1:P<0.52\)
Under \(H_0\), the population proportion is \(P_0=0.52\)
Since \(NP_0(1-P_0)\) is greater than or equals to 10, use normal approximation to test the true proportion.
Number of trials, \(N=150\)
Number of sucesses in 150 cases, \(X=72\)
Compute test statistic.
Alternative hypothesis has symbol "<", so the test is based on left-tailed.
Using Excel function "=NORMSTD(z)"
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