To estimate the total number of successes in a large finite population of size N,using a sample of size n,the confidence interval estimator is:
A) $$\tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde
{ p } ( 1 - \tilde { p } ) / n }$$
B) $$N \left[ \tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$
C) $$( N + n ) \left[ \tilde { p } \pm z _ { a / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$
D) $$( N - n ) \left[ \tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$

Question

To estimate the total number of successes in a large finite population of size N,using a sample of size n,the confidence interval estimator is:
A) $$\tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde
 { p } ( 1 - \tilde { p } ) / n }$$
B) $$N \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde  { p } ( 1 - \tilde { p } ) / n } \right]$$
C) $$( N + n ) \left[ \tilde { p } \pm z _ { a / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$
D) $$( N - n ) \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$

Quizplus · Accepted Answer

The confidence interval estimator for the total number of successes in a large finite population is given by multiplying the sample proportion by the population size, which gives the estimated total number of successes. Then we use the same formula as in the case of estimating a proportion, except that the denominator of the square root term is adjusted for the finite population:

$$N \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde  { p } ( 1 - \tilde { p } ) / n } \right]$$

Option A is the confidence interval estimator for a proportion, not a total number of successes.

Option C is the confidence interval estimator for the population mean, not a total number of successes.

Option D includes (N-n) in the formula which gives the wrong estimation for what we are required in this problem.

To Estimate the Total Number of Successes in a Large p~±zα/2p~(1−p~)/n\tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n }p~​±zα/2​p~​(1−p~​)/n​

To Estimate the Total Number of Successes in a Large $\tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n }$