Question

Suppose Jonathan works for the local transportation authority and knows that for the last few years, only 62% of buses have arrived at their destination on time. However, the city recently completed improvements to the main roads, and Jonathan thinks that a higher proportion of buses now arrive on time because of these improvements. Jonathan collects a random sample of bus arrival times for 500 buses and finds that 347 buses arrived on time. He conducts a one-sample, right-sided z-test for a proportion, using a significance level of a = 0,01. Jonathan wants to know the power of the test to reject the null hypothesis if the true proportion of on time arrivals is 65% or more. What is the minimum sample proportion Ộ that rejects H, at a = 0.01? Please give your answer as a decimal precise to at least four decimal places.

Answer 1

a) As this is a one sample right tailed z test, for 0.01 level of significance, we have from standard normal tables:
P(Z < 2.326) = 0.99

Therefore P(Z > 2.326) = 1 - 0.99 = 0.01

Therefore the minimum sample proportion value here is computed as:

$p_{crit} = P + 2.326*\sqrt{\frac{P(1-P)}{n}} = 0.62 + 2.326*\sqrt{\frac{0.62*0.38}{500}} = 0.6705$

Therefore 0.6705 is the required minimum sample proportion value here.

b) Given a true proportion of 0.65, the probability to correctly reject the null hypothesis here is computed as:

P( p > 0.6705)

Converting it to a standard normal variable, we have here:

$P(Z > \frac{0.6705 - 0.65}{\sqrt{\frac{0.65*0.35}{500}}})$

P(Z > 0.9606)

Getting it from the standard normal tables, we have here:

Therefore 0.1684 is the required probability here. (which is the power of the test as well )