Sampling Distrubtions

Question

It is known that 20% of people ride public transport without paying for it. We talk to 63 people who have been on public transport in the last month. What is the probability that the sample proportion is more than 23%?

 (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

p_o = 0.20 (population proportion where people don't pay for riding public transit)

n = 63 people surveyed who rode public transit in the last month (sample size)

p-hat = 0.23 (sample proportion)

1.) Given that population proportion is 0.20, we can assume that the sample proportion follows a normal distribution.

Mean: μ = p_o = 0.20

Standard Deviation:  σ = √(p_o * (1 - p_o) / n) = √(0.20)*(1 - 0.20) / 63) = √(0.20)*(0.80) / 63) = 0.0504

2.) We want to use the z-score formula to find the probability. Make sure to have a z-score table either from your textbook or online. In this case, we have

z = p-hat - p_o / σ = (0.23 - 0.20) / 0.0504 = 0.03 / 0.0504 ≈ 0.60

P(z > 0.60) = 1 - P(z ≤ 0.60) = 1 - 0.7257 = 0.2743

The probability that the sample proportion is more than 23% is about 0.2743.