Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 133 with 61.7% successes. Enter your answer as an open-interval (i. e., parentheses) using decimals (not percents) accurate to three decimal places.

The 99% confidence interval is (0.508, 0.726).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

99% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence interval is (0.508, 0.726).

answer: i don't understand it

step-by-step explanation: sorry

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.