Question

A transportation company is suspicious of the claim that the average useful life of certain tires is at least 28,000 miles. To verify that, 40 tires are placed in trucks and an average useful life of 27,463 is obtained with a standard deviation of 13,48 miles. Test this hypothesis with a level of significance of 1%.

Fail to reject the null hypothesis

Reject the null hypothesis

If the population mean under the alternative hypothesis is 27,230, calculate the probability of Type II Error.

A. 5%

B. 85%

C. 5%

D. 99%

Answer 1

Given that,
population mean(u)=28000
standard deviation, σ =1348
sample mean, x =27463
number (n)=40
null, Ho: μ=28000
alternate, H1: μ!=28000
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 27463-28000/(1348/sqrt(40)
zo = -2.52
| zo | = 2.52
critical value
the value of |z α| at los 1% is 2.576
we got |zo| =2.52 & | z α | = 2.576
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.52 ) = 0.012
hence value of p0.01 < 0.012, here we do not reject Ho
ANSWERS
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null, Ho: μ=28000
alternate, H1: μ!=28000
test statistic: -2.52
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.012
we do not have enough evidence to support the claim that the average useful life of certain tires is at least 28,000 miles.
b.
Given that,
Standard deviation, σ =1348
Sample Mean, X =27463
Null, H0: μ=28000
Alternate, H1: μ<28000
Level of significance, α = 0.01
From Standard normal table, Z α/2 =2.3263
Since our test is left-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-28000)/1348/√(n) < -2.3263 OR if (x-28000)/1348/√(n) > 2.3263
Reject Ho if x < 28000-3135.8524/√(n) OR if x > 28000-3135.8524/√(n)
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Suppose the size of the sample is n = 40 then the critical region
becomes,
Reject Ho if x < 28000-3135.8524/√(40) OR if x > 28000+3135.8524/√(40)
Reject Ho if x < 27504.1782 OR if x > 28495.8218
Implies, don't reject Ho if 27504.1782≤ x ≤ 28495.8218
Suppose the true mean is 27230
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(27504.1782 ≤ x ≤ 28495.8218 | μ1 = 27230)
= P(27504.1782-27230/1348/√(40) ≤ x - μ / σ/√n ≤ 28495.8218-27230/1348/√(40)
= P(1.2864 ≤ Z ≤5.939 )
= P( Z ≤5.939) - P( Z ≤1.2864)
= 1 - 0.9008 [ Using Z Table ]
= 0.0992
For n =40 the probability of Type II error is 0.0992
the probability of Type II error is 9.92%