We want to know whether the percentage of people who paid lower taxes was different based on whether they sought tax advice or not.
H
0:
padvice - pno advice = 0,
H
A:
padvice - pno advice > 0
Conditions:
* Independence: The people who filed tax reports were randomly selected and do not influence each other.
* Random Condition: The people who filed tax reports were randomly selected.
* 10% Condition: 72 is less than 10% of people who didn't get tax advice, and 105 is less than 10% of people who did get tax advice.
* Success/Failure: All observed counts (48, 19, 24, and 86) are at least 10.
Because all conditions have been satisfied, we can model the sampling distribution of the difference in proportions with a Normal model. We can perform a two-proportion
z-test.
Let 'Advice' group be the people who sought tax advice and the 'No advice' group be the people who did not seek tax advice.
We know:
nadvice = 105,
nno advice = 72,
advice = 0.819,
no advice = 0.333.
pooled =
= 0.621
SEpooled (
advice -
no advice) =
= 0.074
The observed difference in sample proportions is
advice -
no advice = 0.819 - 0.333 = 0.486
z =
=
= 6.62
P =
P(
z > 6.62) < 0.0001
The P-value is small, so we reject the null hypothesis. There is strong evidence of a difference in the tax percentages paid between the group who has tax advice and the group who did not have tax advice. It appears that if you have tax advice you are more likely to pay a lower percentage of taxes to the government.