Question

Edwin conducted a survey to find the percentage of people in an area who smoked regularly. He defined the label "smoking regularly" for males smoking 30 or more cigarettes in a day and for females smoking 20 or more. Out of 635 persons who took part in the survey, 71 are labeled as people who smoke regularly. What is the 90% confidence interval for this population proportion? Answer choices are rounded to the hundredths place.

Answer 1

Answer:

The confidence interval for this proportion = (0.09, 0.13)

Step-by-step explanation:

Confidence Interval for Proportion =

p ± z ×√p(1 - p/n)

p = proportion = x /n

x = 71

n = 635

p = 71/635

p = 0.1118110236

p ≈ 0.1118

z score for 90% confidence interval = 1.645

Confidence Interval =

0.1118 ± 1.645 × √0.1118 (1 - 0.1118)/635

0.1118 ± 1.645 × √0.09930076/635

0.1118 ± 1.645 × √0.0001563791

0.1118 ± 1.645 × 0.0125051629

0.1118 ± 0.020570993

Confidence Interval =

0.1118 - 0.020570993

= 0.091229007

Approximately to the nearest hundredth ≈ 0.09

0.1118 + 0.020570993

= 0.132370993

Approximately to the nearest hundredth = 0.13

Therefore, the confidence interval for this proportion = (0.09, 0.13)