USFEC rents the flatbed trailers on which the organizations participating in the
ID: 3362284 • Letter: U
Question
USFEC rents the flatbed trailers on which the organizations participating in the parade build their floats. The company that owns these trailers determines their rental price based on the mean weight of a parade float. The standard deviation of float weight is widely known to be 125 pounds. The types of floats used in the Martinsville parade typically fall into the second classification on the company's pricing scale. This classification places an upper limit of 850 pounds on the mean. Therefore, USFEC wishes to perform hypothesis test about the mean weight of a parade float using a level of significance of O. 1 . USFEC wishes to detect a true mean of 900 pounds with probability 0.95 What sample size should USFEC use? (a) (b) State the null and alternative hypothesis. What is the critical region for this test? (c) What is the power of this test if the true mean float weight is 885 pounds?Explanation / Answer
Solution:
Here we have
Population SD = 125, Population Mean = 900
Sample Mean = 850, Mean Difference = 900-850=50
(a) Sample size n = (1.645*1.645*125*125)/(50*50) = 16.91265625
The USDEC should use a sample size of 17.
(b) Null Hypothesis (Ho): The population mean is 900 pounds. (µ = 900)
Alternative Hypothesis (Ha): The population mean is different from 900 pounds. (µ900)
Here level of significance = 0.10 and the critical values corresponding to 10% are ±1.645
Critical region is defined as
Reject Ho if Z>Z(upper critical value) or Z<-Z(lower critical value)
(c) Test statistics is
Z = (Sample Mean – Population Mean)/Population SD/sqrt(n)
We reject Ho if Z>1.645 or Z <-1.645
Sample mean is calculated as
x-bar=1.645*125/sqrt(17) = 49.85 50
Power is given as
=2*P(Z>850-885/30.32)
= 2*P(Z>35/30.32)
=2*P(Z>-1.15)
=2*P(0<Z<1.15)
= 2*0.3749
= 0.7498
So the power of this test if the true mean float weight is 885 pounds is 0.7498.
(d) Test statistics is
Z = (Sample Mean – Population Mean)/Population SD/sqrt(n)
= 868-900/30.32
= -32/30.32
= -1.055
We reject Ho if Z >1.645 or Z < -1.645
Since Z test statistics lies between the upper and lower critical value so we will not be able to reject the null hypothesis and conclude that the population mean is 900 pounds.
(e) P-value of the test is calculated as
P(Z>test statistics value) + P(Z<- test statistics value)
= 2*P(0<Z<1.055)
= 2*0.1459 = 0.2912
So the smallest level of significance that would have resulted in rejecting the null hypothesis is 0.7087
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