Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 315 with 37% 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 = 0.37 +/- 0.070

= (0.300, 0.440)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 37% = 0.37

Number of samples n = 315

Confidence interval = 99%

z value(at 99% confidence) = 2.58

Substituting the values we have;

0.37 +/- 2.58√(0.37(1-0.37)/315)

0.37 +/- 2.58√(0.00074)

0.37 +/- 2.58(0.027202941017)

0.37 +/- 0.070183587825

0.37 +/- 0.070

= (0.300, 0.440)

The 99% confidence interval is = 0.37 +/- 0.070

= (0.300, 0.440)

21/58.

hope to be marked brainliest

stops on an even number.