Answer to Question #190723 in Statistics and Probability for jennifer

Question

A survey of CPAs across the US found that the average net income for sole proprietor CPAs is not more than $74,924. Because this survey is more than 15 years old, an accounting researcher who wanted to test this finding took a sample of 112 sole proprietor accountants and found the net income to be $78,695 with a population standard deviation of $14,530. If you want to test at the 5% significance level if the survey findings still hold true, what are your null and alternative hyposthesis?

Ho : μ ≤ 74,924; Ha : μ > 74,924

Ho : μ > 74,924; Ha : μ < 74,924

Ho : x̄ < 74,924; Ha : x̄ > 74,924    

Ho : μ = 74,924; Ha : μ≠74,924

Answer 1

Sample size n=112

sample means x=78695

Population standard deviation=14530

Null Hypothesis$(H_o):\mu\le 74924$ ( The average net income for sole proprietor CPAs is not more than $74,924. 

Alternate hypothesis ($H_a):\mu> 74924$ ( the average net income for sole proprietor CPAs is more than $74,924.

Level of significance=0.05

Critical value at 5%=$\pm 1.96$

Test statics-

$z=\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}=\dfrac{(78695-74924)\sqrt{112}}{14530}=2.3539$

As calculated value of z is bigger than 1.96, so Null hypothesis is rejected i.e. the average net income for sole proprietor CPAs is more than $74,924.