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 346 buses arrived on time. He conducts a one-sample, right‑sided ?-test for a proportion, using a significance level of ?=0.10. 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 ?̂ that rejects ?0H0 at ?α = 0.10? Please give your answer as a decimal precise to at least four decimal places.

?̂ =

Determine the power of the test to correctly reject the null hypothesis if the population proportion is actually 0.65 That is, compute the probability that if the population proportion is 0.65, the sample proportion would be at least as great as the value of ?̂ you found in the first step. Please give your answer as a decimal precise to at least four decimal places.

power =

If the lighting company decides to return the shipment to the manufacturer, what is the probability that a type I error has been made? Give your answer as a decimal precise to two decimal places.

Answer 1

-Ans a ) we are testing here whether the population Proportion is more than 0.62; therefore the test here is statistic Computed as P-P VP (1-P)/n As this is a one tailed test, we have from standard normal tables here. PLZ>2-576) = 0.005 .. The critical Proportion value here is computed as. Porit -P = 2.576 P(1-P) n Perit 0.62 = 2.576 0.62 (1-0.62) SOO Perit = 0.6759 b) for a true proportion of 0.65 the power of the test hare is computed as the probability to reject. the nult hypothesis, which is computed here as. P (P >0.6759) Converting it to a standard normal variable, we have here.

0.6759-0065 e Z 0.65 0.35 500 P (2>1.215) Getting it from the standard normal tables we have here. P(Z > 1,215) = 0.1122. : 0.1122 org 11-221. is the required power of the test.