Asked by Jodene
A survey of 90 families showed that 40 owned at least one gun. Find the 95%
confidence interval of the true proportion of families who own at least one gun.
Example
confidence interval of the true proportion of families who own at least one gun.
Example
Answers
Answered by
MathGuru
Example of a proportional confidence interval formula:
CI95 = p + or - (1.96)[√(pq/n)]
...where p = x/n, q = 1 - p, and n = sample size.
Note: + or - 1.96 represents 95% confidence interval.
For p in your problem: 40/90 = 0.44
For q: 1 - p = 1 - 0.44 = 0.56
n = 90
I let you take it from here to calculate the interval.
CI95 = p + or - (1.96)[√(pq/n)]
...where p = x/n, q = 1 - p, and n = sample size.
Note: + or - 1.96 represents 95% confidence interval.
For p in your problem: 40/90 = 0.44
For q: 1 - p = 1 - 0.44 = 0.56
n = 90
I let you take it from here to calculate the interval.
Answered by
Anonymous
89.98 to 90.02