Answer:
c. an observed difference is 0.12 and P < 0.0001
Step-by-step explanation:
Let p1 be the proportion of the students who received musical instruction received a passing grade
Let p2 be the proportion of the students who didn't receive musical instruction received a passing grade
Null and Alternative hypotheses are:
: p1-p2=0
: p1-p2≠0
Test statistic can be found using the equation:
where
- p1 is the sample proportion of the students who received musical instruction received a passing grade (
)
- p2 is the sample proportion of the students who didn't receive musical instruction received a passing grade (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the students who received musical instruction (3239)
- n2 is the sample size of the students who didn't receive musical instruction (2787)
Thus, ≈ 11.84 gives p-value < 0.0001
Observed difference is: 0.87-0.75=0.12