There are 32 students in the class. They are separated into groups of five and one group of seven. How many, g groups of five we
2 answers:
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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
: p1-p2=0
: p1-p2≠0
Test statistic can be found using the equation:
where
- p1 is the sample proportion of the common attribute in population1 (
)
- p2 is the sample proportion of the common attribute in population2 (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the people from population1 (30)
- n2 is the sample size of the people from population2 (1900)
Then ≈ 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± where
- z is the z-statistic for the 99% confidence (2.58)
Thus 99% confidence interval is
0.533-0.704± ≈ -0.171 ±0.237 that is (−0.408, 0.066)