In: Statistics and Probability
In a simple random sample of size 56, taken from a population, 22 of the individuals met a specified criteria.
a) What is the margin of error for a 90% confidence interval for p, the population proportion? Round your response to at least 4 decimal places.
b) What is the margin of error for a 95% confidence interval for p? Round your response to at least 4 decimal places. NOTE: These margin of errors are greater than .10 or 10%.
c) How big of a sample is needed to be certain that we have a margin of error less than .10 (or 10%) at 90% confidence?
d) How big of a sample is needed to be certain that we have a margin of error less than .10 (or 10%) at 95% confidence?
Solution :
Given that,
n = 56
x = 22
Point estimate = sample proportion = = x / n = 22/56 = 0.393
a)
At 90% confidence level the z is ,
Margin of error = E = Z / 2 * √ ((
* (1 -
)) / n)
= 1.645 * (√((0.393*0.607) / 56)
= 0.1074
b)
At 95% confidence level the z is ,
Margin of error = E = Z / 2 * √ ((
* (1 -
)) / n)
= 1.96 * (√((0.393*0.607) / 56)
= 0.1279
c)
margin of error = E = 0.10
At 90% confidence level the z is ,
sample size = n = (Z / 2 / E )2 *
* (1 -
)
= (1.645/0.10)2 * 0.393*0.607
= 64.55
sample size = 65
d)
margin of error = E = 0.10
At 95% confidence level the z is ,
sample size = n = (Z / 2 / E )2 *
* (1 -
)
= (1.96/0.10)2 * 0.393*0.607
= 91.64
sample size = 92