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# population proportion?

construct the 95% and the 99% confidence intervals for the population proportion p using the sample statistics below

p=0.733

q=0.267

n=4408

the 95% confidence interval for the population proportion p is ___________ _____________

the 99% confidence interval for the population proportion p is ___________ _____________

### 2 Answers

- MLv 71 decade agoFavorite Answer
ANSWER: 95% Resulting Confidence Interval for 'true mean': = [0.72, 0.746]

Why???

POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION

n = Number of samples 4408

Number of Successes 3231.064

p = POPULATION PROPORTION 0.733

significant digits 3

Confidence Level 95

"Look-up" Table 'z-critical value' 1.960

from Cumulative Normal Distribution Table "Look-up" of 'z-critical value'.

Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical.

95% Resulting Confidence Interval for 'true mean':

p +/- (z critical value) * SQRT[p * (1 - p)/n] = 0.733 +/- 1.96 * SQRT[0.733 * (1 - 0.733)/4408] = [0.72, 0.746]

ANSWER: 99% Resulting Confidence Interval for 'true mean': = [0.716, 0.75]

Why???

POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION

n = Number of samples 4408

Number of Successes 3231.064

p = POPULATION PROPORTION 0.733

significant digits 3

Confidence Level 99

"Look-up" Table 'z-critical value' 2.576

from Cumulative Normal Distribution Table "Look-up" of 'z-critical value'.

Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical.

99% Resulting Confidence Interval for 'true mean':

p +/- (z critical value) * SQRT[p * (1 - p)/n] = 0.733 +/- 2.576 * SQRT[0.733 * (1 - 0.733)/4408] = [0.716, 0.75]

- gainsLv 44 years ago
because you recognize the classic deviation, a z-try will suffice. on the ninety% aspect of self assurance, your z-score is z = a million.645. enable s be the classic deviation and n the pattern length. Then the margin of mistakes is defined as: E = zs / n^(a million/2) = (a million.645)(21.9) / 22^(a million/2) = 16.375.