Answer to Question #315769 in Statistics and Probability for Ashianna

Question

A certified public accountant claims that more than 30% of all accountants advertise. A sample of 400 accountants in Rizal province showd that 112 had used some form of advertising. At a ?0.01, is there enough evidence to support the claim? (Use p-value method)

Answer 1

$H_0:p=0.3$

$H_1:p>0.3$

Test statistic for proportion can be found the following way

$T={\frac {{\frac w n} -p} {\sqrt{{\frac {p*(1-p)} n}}}}$ , where w - number of satisfiable observations, n - sample size, p - assumed proportion

So, in the given case we have

$T={\frac {{\frac {112} {400}} -0.3} {\sqrt{{\frac {0.3*(1-0.3)} {400}}}}}={\frac {-0.02} {0.0105}}=-1.9$

Since sample size is big, then we can use Z-score. According to the form of the alternative hypothesis, right-tailed test is appropriate. So, p-value for obtained T is

$p=P(Z>T)=P(Z>-1.9)=0.97128>0.01$

So, based on the given sample, there is no enough evidence to conclude that population propotion is greater than 0.3