Question

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 having a common attribute. The second sample consists of 1900 people with 1337 of them having the same common attribute. Compare the results from a hypothesis test of p 1 equals 2 ​(with a 0.01 significance​ level) and a 99​% confidence interval estimate of p 1 minus 2.

Answer 1

Answer:

• There is no significant evidence that p1 is different than p2 at 0.01 significance level.

• 99% confidence interval for p1-p2 is  -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$: p1-p2=0

$H_{a}$: p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

• p1 is the sample proportion of the common attribute in population1 ($\frac{16}{30} =0.533$)

• p2 is the sample proportion of the common attribute in population2 ($\frac{1337}{1900} =0.704$)

• p is the pool proportion of p1 and p2 ($\frac{16+1337}{30+1900}=0.701$)

• n1 is the sample size of the people from population1 (30)

• n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is  0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2±$z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

• z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704±$2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)