Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 142 with 61% successes

Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 142 with 61% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Answer 1

Hi Anni,

Formula for this is:

CI= p^ +/- z*SE

p^= probability estimate of success = 0.61

z*=z-critical value

SE= standard error= sqrt(p^q^/n)

q^= 1-p^ = 0.39

n=sample size=142

SE=sqrt [(0.61*0.39)/142]

SE=0.041

Now, recall some common z-critical values

z_90% = 1.645

z_95% = 1.96

z_99% = 2.576

Those are worth committing to memory, as they are generally the three most common confidence interval percentages. so you now have all you need to compute the confidence interval:

p^ = 0.61

SE = 0.041

z* = 2.576

Now you can compute your confidence interval using CI= p^ +/- (z*SE) I hope this helps.