Example 8.7 Let denote the true average tread life of a certain type of tire. Co
ID: 3364875 • Letter: E
Question
Example 8.7 Let denote the true average tread life of a certain type of tire. Consider testing H0: > 30,000 based on a sample of size n = 16 from a normal population distribution with = 1500, A test with = 0.01 requires z-Z0.01-2.33. The probability of making a type II error when -31,000 is 30,000 versus Ha: ) - 2.33 . 30,000 31,000 1500/16 0) (rounded to two decimal places) = 0.3669 Since zo.1 1.28, the requirement that the level 0.01 test also have (31,000-0.1 necessitates 150012.33 +128)12-(-5.42)2 = 30,000- 31,000 rounded to two decimal places) n= The sample size must be an integer, so n - should be used. (rounded up to the next whole number) tiresExplanation / Answer
8.7
Given that,
Standard deviation, =1500
Sample Mean, X =31000
Null, H0: <30000
Alternate, H1: >30000
Level of significance, = 0.01
From Standard normal table, Z /2 =2.3263
Since our test is right-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-30000)/1500/(n) < -2.3263 OR if (x-30000)/1500/(n) > 2.3263
Reject Ho if x < 30000-3489.45/(n) OR if x > 30000-3489.45/(n)
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Suppose the size of the sample is n = 16 then the critical region
becomes,
Reject Ho if x < 30000-3489.45/(16) OR if x > 30000+3489.45/(16)
Reject Ho if x < 29127.64 OR if x > 30872.36
Implies, don't reject Ho if 29127.64 x 30872.36
Suppose the true mean is 31000
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(29127.64 x 30872.36 | 1 = 31000)
= P(29127.64-31000/1500/(16) x - / /n 30872.36-31000/1500/(16)
= P(-4.99 Z -0.34 )
= P( Z -0.34) - P( Z -4.99)
= 0.3669 - 0 [ Using Z Table ]
= 0.37
For n =16 the probability of Type II error is 0.37
since Z0.1 = 1.28,the requirement of level 0.01 test also have beta(31000)=0.1
sample size (n)= (1500(2.33+1.28)/(30000-31000))^2 = (-5.42)^2 =29.3764 = 29.37
sample size n= 30(approximately)
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