Question

Your probability professor has been dedicating part of the time she used to spend commuting to doing longer yoga practices. She took a random sample of her yoga practices since Hofstra has ceased in-person classes and found 3 under ten minutes and 19 over ten minutes. Construct a 99% confidence interval for the proportion of her yoga practices that will be over ten minutes given her current lifestyle.

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 19

n = 3+19 = 22

P = x/n = 19/22 = 0.863636364

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.863636364 ± 2.5758* sqrt(0.863636364*(1 – 0.863636364)/22)

Confidence Interval = 0.863636364 ± 2.5758*0.0732

Confidence Interval = 0.863636364 ± 0.1885

Lower limit = 0.863636364 - 0.1885 = 0.6752

Upper limit = 0.863636364 + 0.1885 = 1.0521

Confidence interval = (0.7, 1.0)