Question

a random sample of 300 people, 60 own a bicycle.

95% confidence interval, what is the proportion of people wjo own a bicyle?

Answer 1

Proportion, p = 60 / 300 = 0.2 (Proportion of people who own a bicycle)

Significance level = 1 - 0.95 = 0.05

Critical z-vale at 0.05 alpha= Z$\alpha /2$= 1.96 (from the table)

Standard Error for p = SE =

Vp* (1 - p)/n

SE =

2 *.8/300

SE = 0.02

Let Q = Z$\alpha /2$ *

Vp* (1 - p)/n

Q = 1.96*0.02

Q = 0.04

Lower Limit of Confidence Interval = p - Q = 0.2 - 0.04 = 0.16

Upper Limit of Confidence Interval = p + Q = 0.2 + 0.04 = 0.24

So, 95% confidence Interval: 0.16 < p < 0.24

**if you liked the answer, then please upvote. Would be motivating for me. Thanks.