PROBLEM 3 (39 pts) The Allied Express overnight delivery company claims an on-ti

Question

PROBLEM 3 (39 pts) The Allied Express overnight delivery company claims an on-time delivery rate of 95%. A consumer group is interested in testing this claim to determine if the company is overstating the proportion of on-time deliveries. a)(3 pts) Assume that the company’s claim is defined as the null hypothesis. State the null and alternative hypotheses in terms of the population parameter, the proportion of late deliveries. b)(4 pts) Verbally state the type I and II errors in the context of this problem. c)(4 pts) A sample of 25 deliveries is conducted. The consumer group decides to reject the firm’s claim if at least three late deliveries are found in the sample. State the decision rule using the appropriate random variable. Be sure to define this random variable and state its distribution. d)(9 pts) Using the information of parts a and c, calculate the probability of a type I error. e)(5 pts) Suppose the consumer group conducts a sample of 150 deliveries (out of a very large number of deliveries made by Allied Express). Twelve of the sampled deliveries were late. What is the sampling distribution of the proportion of late deliveries in this sample, assuming that Allied’s claim is true? f)(10 pts) Calculate the probability of obtaining a sample proportion of late deliveries at least as large as the one reported in part e, assuming that Allied’s claim is true. g)(4 pts) Based on your anwer to part f, what conclusion can be drawn regarding the validity of Allied Express’s on time delivery claim of 95%? Does the result provide support for the null or alternative hypothesis?

We are given that p0 = 0.95

a) The null hypothesis will be,

against the alternative hypothesis,

For the efficiency of the test we set alpha as low as possible here.

c) The sample proportion for the hypothesis test will be,

I only need help with questions 3E-3G, please show work if possible.

Explanation / Answer

Part-E

Sampling distribution will be normal with mean =0.05 and standard deviation s=sqrt(0.05*(1-0.05)/150)=0.0178

Part-F

P(p>0.05)=P((p-phat)./s>(0.05-12/150)/0.0178)=P(Z>-1.685)=0.954

Part-G

As this probability is quite large we conclude that late deliveries are higher than 0.05 and so on time deliveris are less than 95%

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