The manufacturer of a new compact car claims that the car will average at least
ID: 3315701 • Letter: T
Question
The manufacturer of a new compact car claims that the car will average at least 35 miles per gallon in general highway driving. For 100 test runs, the car averaged 35.5 miles per gallon with a standard deviation of 4 miles per gallon. Suppose that the manufacturer tests the following hypotheses: H subscript 0 : mu equals 35 H subscript 1 : mu greater than 35 Consider the 10% significance level. What is the power of the test if the actual average fuel efficiency is 36 miles per gallon? a. 95.254% b. 99.086% c. 80.511% d. 56.749% e. 88.877%
Explanation / Answer
Given that,
Standard deviation, =4
Sample Mean, X =35.5
Null, H0: =35
Alternate, H1: >35
Level of significance, = 0.1
From Standard normal table, Z /2 =1.2816
Since our test is right-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-35)/4/(n) < -1.2816 OR if (x-35)/4/(n) > 1.2816
Reject Ho if x < 35-5.1264/(n) OR if x > 35-5.1264/(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 35-5.1264/(100) OR if x > 35+5.1264/(100)
Reject Ho if x < 34.48736 OR if x > 35.51264
Implies, don't reject Ho if 34.48736 x 35.51264
Suppose the true mean is 36
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(34.48736 x 35.51264 | 1 = 36)
= P(34.48736-36/4/(100) x - / /n 35.51264-36/4/(100)
= P(-3.7816 Z -1.2184 )
= P( Z -1.2184) - P( Z -3.7816)
= 0.1115 - 0.0001 [ Using Z Table ]
= 0.1114
For n =100 the probability of Type II error is 0.1114
power of the test = 1- beta
= 1-0.1114
= 0.8886
= 88.86%
= 88.877%
option :E
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