10.63
In the past, 44% of those taking a public accounting qualifying exam
have passed the exam on their first try. Lately, the availability of
exam preparation books and tutoring session may have improved the
likelihood of an individual’s passing on his or her first try. In a
sample of 250 recent applicants, 130 passed on their first attempt. At
the 0.05 level of significance, can we conclude that the proportion
passing on the first try has increased? Determine and interpret the
p-value for the test. (Hypothesis Testing)

Hi, I need clear step by step work answers. However, not long
explanations or professional answers. Thank you very much.

Hello wta2k:

First, let's list what we know:

This is a population proportion question where:
* pi = .44
* p = 130/250 = .52
* n = 250
* alpha = 0.05

Our null hypothesis is 

H_0: pi = .44

Our alternate hypothesis is

H_a: pi > .44

Therefore, a one-tail test is required. 

Since n > 30, the z statistic can be used as the test statistic. 

Since alpha=0.05, the critical value of the test statistic is z=1.645.

Reject H_0 and accept H_a if the absolute value of the sample test
statistic z > 1.645

sigma_p = sqrt(pi*(1-pi)/n) 
        = sqrt(.44*(.56)/250)
        = sqrt(.0009865)
        = .03139

sample test statistic: z = (p-pi)/sigma_p
                         = (.52-.44)/.03139
                         = 2.54

Since the absolute value of the sample test statistic z > the critical
value, reject H_0 and accept H_a.

Therefore, we can conclude that the proportion passing on the first
try has increased.

As for the p-value, we know from the z-table that the area between mu
and z = 2.54 is .5000 - .4945 = .0055. Therefore, the p-value is
0.0055 or 0.55%.


I hope this helps. 

websearcher-ga


Search Strategy: none, but you could try:

"hypothesis testing"
"z-values"
"p-values"

Thanks a lot!! Clear and Straight!!