Question

A tire company claims that the lifetimes of its tires average 49750 miles. The standard deviation of tire lifetimes is known to be 4750 miles. You sample 75 tires and will test the hypothesis that the mean tire lifetime is at least 49750 miles against the alternative that it is less. Assume, in fact, that the true mean lifetime is 49250 miles.

It is decided to reject H₀ if the sample mean is less than 49150. Find the level and power of this test. Round the answers to four decimal places.

At what level should the test be conducted so that the power is 0.75?

You are given the opportunity to sample more tires. How many tires should be sampled in total so that the power is 0.75 if the test is made at the 5% level? Round the answer to the next largest integer.

Answer 1

Given that,
population mean(u)=49750
standard deviation, σ =4750
sample mean, x =49150
number (n)=75
null, Ho: μ=49750
alternate, H1: μ<49750
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 49150-49750/(4750/sqrt(75)
zo = -1.09
| zo | = 1.09
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.09 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.09 ) = 0.14
hence value of p0.05 < 0.14, here we do not reject Ho
ANSWERS
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null, Ho: μ=49750
alternate, H1: μ<49750
test statistic: -1.09
critical value: -1.645
decision: do not reject Ho
p-value: 0.14
we do not have enough evidence to support the claim that A tire company claims that the lifetimes of its tires average is less than 49750 miles.
b.
Given that,
Standard deviation, σ =4750
Sample Mean, X =49150
Null, H0: μ=49750
Alternate, H1: μ<49750
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-49750)/4750/√(n) < -1.6449 OR if (x-49750)/4750/√(n) > 1.6449
Reject Ho if x < 49750-7813.275/√(n) OR if x > 49750-7813.275/√(n)
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Suppose the size of the sample is n = 75 then the critical region
becomes,
Reject Ho if x < 49750-7813.275/√(75) OR if x > 49750+7813.275/√(75)
Reject Ho if x < 48847.8007 OR if x > 50652.1993
Implies, don't reject Ho if 48847.8007≤ x ≤ 50652.1993
Suppose the true mean is 49250
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(48847.8007 ≤ x ≤ 50652.1993 | μ1 = 49250)
= P(48847.8007-49250/4750/√(75) ≤ x - μ / σ/√n ≤ 50652.1993-49250/4750/√(75)
= P(-0.7333 ≤ Z ≤2.5565 )
= P( Z ≤2.5565) - P( Z ≤-0.7333)
= 0.9947 - 0.2317 [ Using Z Table ]
= 0.763
For n =75 the probability of Type II error is 0.763
power of the test = 1- type 2 error
power of the test = 1-0.763 =0.237
c.
power = 0.75
type 2 error = beta = 1-0.75 =0.25
sample size =(( standard deviation^2)*(Zalpha +Z beta )^2/(population mean - true mean )^2)
sample size = ((4750^2)*(Z0.05+Z0.25)^2 / (49750-49250)^2)
sample size = ((4750^2)*(1.645 +0.674)^2/(49750-49250)^2) = 485.342
sample size = 486