5. A courier service advertises that its average delivery time is 45 minutes for

Question

5. A courier service advertises that its average delivery time is 45 minutes for deliveries within the city limit. The distribution of delivery times is known to be normal, with standard deviation 12 min. A random sample of 15 deliveries yielded a sample mean 51 minutes delivery time of hours. Consider a test of the null hypothesis that the population mean is 45 minutes. a. What is the test statistic? b. What is the P-value of the test? c. By interpreting the P-value in (b) can the null hypothesis be rejected at 5% significant level? d. Can the null hypothesis be rejected at 10% significant level?

Explanation / Answer

Sol:

Null hypothesisL

H0:mu=45

Alternative Hypothesis:

Ha:mu not =45

two tail t test

T=sample mean-popmean/samplesd/sqrt(n)

=51-45/12/sqrt(!5)

=1.936

t=1.936

Solutionb:

df=n-1=15-1=14

p=0.073323

Decsion rule:p<alpha,Reject Null hypothesis

P>alpha fail to reject null hypothesis.

SOlutionc:

p=0.073323

alpha=0.05=5%

p>alpha

Fail to reject null hypothesis

Accept NullHypothesis.

SOlutiond:

p=0.073323

alpha=10%=0.1

p<alpha

Reject Null hypothesis

Acept Alternative Hypothesis

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