Question

In the past, 44% of those taking a public accounting qualifying exam have passed the exam on their first try. Latterly, the availability of exam preparation books and tutoring sessions may have improved the likelihood of an individual’s passing on his first try. In a sample of 250 recent applicants, 130 passed on their first attempt. At the 0.05 level of significance, what is the calculated value of test statistic? (Specify your answer to the 2nd decimal.)

Answer 1

The calculated value of test statistic is z=2.48.

This has a P-value of P=0.00657.

If we state the null hypothesis
`H_0: \pi\leq0.44` at a significance level of
`\alpha=0.05`, we would reject this null hypothesis as
`P-value<\alpha`.

Explanation:

We have in this problem, a hypothesis test of proportions.

The test statistic for this is the z-value, and is calculated like that:

`z=(p-\pi-0.5/N)/(\sigma)`

Where the term 0.5/N is the correction for continuity and is negative in the cases that p>π.

p: proportion of the sample; π: proportion of the population; σ: standard deviation of the population.

The standard deviation of the population has to be calculated as:

`\sigma=\sqrt{(\pi(1-\pi))/(N) } =\sqrt{(0.44(1-0.44))/(250) }=√(0.0009856)=0.0314`

The proportion of the sample (p) is
`p=130/250=0.52`.

Then, the test statistic z is

`z=(p-\pi-0.5/N)/(\sigma)=(0.52-0.44-0.5/250)/(0.0314) =(0.078)/(0.0314) =2.48`

The P-value of this statistic is P(z>2.48)=0.00657

If we state the null hypothesis
`H_0: \pi\leq0.44` at a significance level of
`\alpha=0.05`, we would reject this null hypothesis as
`P-value<\alpha`.