Chandra wants to estimate the percentage of people who smoke. She surveys 100 individuals and finds that 15 of them smoke. Find the margin of error for the confidence interval for the population proportion with a 98% confidence level.

M.E = 0.083

the margin of error for the confidence interval for the population proportion with a 98% confidence level is 0.083

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)

p+/-M.E

Given that;

M.E = margin of error

Proportion p = 15/100 = 0.15

Number of samples n = 100

Confidence interval = 98%

z(at 98% confidence) = 2.33

Substituting the values we have;

0.15 +/- 2.33√(0.15(1-0.15)/100)

0.15 +/- 2.33√0.001275

0.15 +/- 2.33(0.035707142142)

0.15 +/- 0.083197641192

0.15 +/- 0.083

So;

p = 0.15

M.E = 0.083

the margin of error for the confidence interval for the population proportion with a 98% confidence level is 0.083

i pretty sure it is