Suppose we wish to construct a confidence interval for a population proportion p. If we sample without replacement from a relatively small population of size N, the margin of error E is modified to include the finite population correction factor as follows: $$E = z _ { \alpha } / 2 \sqrt { \frac { \hat { p q } } { n } } \sqrt { \frac { N - n } { N - 1 } }$$
Construct a $90 \%$ confidence interval for the proportion of students at a school who are left handed. The number of students at the school is $N = 410$. In a random sample of 89 students, selected without replacement, there are 10 left handers.
A) $0.057

Question

Suppose we wish to construct a confidence interval for a population proportion p. If we sample without replacement from a relatively small population of size N, the margin of error E is modified to include the finite population correction factor as follows: $$E = z _ { \alpha } / 2 \sqrt { \frac { \hat { p q } } { n } } \sqrt { \frac { N - n } { N - 1 } }$$
Construct a $90 \%$ confidence interval for the proportion of students at a school who are left handed. The number of students at the school is $N = 410$. In a random sample of 89 students, selected without replacement, there are 10 left handers.
A) $0.057 < p < 0.167$
B) $0.070 < \mathrm { p } < 0.155$
C) $0.064 < p < 0.161$
D) $0.074 < p < 0.150$

Quizplus · Accepted Answer

The sample proportion $\hat{p}$ is 10/89. The z-value for a 90% confidence interval is approximately 1.645. Using the finite population correction factor and the formula for the margin of error, the confidence interval is calculated to be approximately $0.064 < p < 0.161$.

Suppose We Wish to Construct a Confidence Interval for a Population