Question

5. (This problem is inspired by one of my best students, who went on to get his PhD in materials science at MIT) If you look at car tires, they tend to average about 30000 miles before being replaced. In general, tire lifetimes are known to be normally distributed with SD 4500 miles. Now, my student claims to have a new manufacturing process that raises this average (he is right!). Let μ be the true average lifespan of tires with the new mianufacturing process, when you draw a sample of size n, the power of this HT is 0.2 when μ-32000, and the power is 0.85176 when -35000. Find a and n for this HT

Answer 1

lifetime is normally distributed with a standard deviation $\sigma =4500$

Now the claim is that the manufacturing process has raised the average lifetime from 30000 miles

Hence the hypothesis formed are

$H_0:\mu =30000$

$H_a:\mu >30000$

. This is a right-tailed test

We will fail to reject the null (commit a Type II error) if we get a Z statistic less than A.

This 'A' Z-critical value corresponds to some X critical value ( X critical), such that

$\frac{X_{critical}-30000}{\frac{4500}{\sqrt{n}}}=A$

$X_{critical}=30000+\frac{4500A}{\sqrt{n}}$.......................(1)

So I will incorrectly fail to reject the null as long as a draw a sample mean that less than $X_{critical}$

. To complete the problem what I now need to do is compute the probability of drawing a sample mean less than $X_{critical}$given $\mu$

for $\mu =32000$ Power of the Hypothesis test is 0.2

The probability of type II error is 1-0.2=0.8

$P(Z<\frac{X_{critical}-32000}{\frac{4500}{\sqrt{n}}})=P(Z<0.8416)=0.8$

$\frac{X_{critical}-32000}{\frac{4500}{\sqrt{n}}}=0.8416$.....................(2)

Evaluating (1) and (2)

$2.25(A-0.8416)-\sqrt{n}=0$ ..........................................(3)

for $\mu =35000$ Power of the Hypothesis test is 0.85176

The probability of type II error is 1-0.85176=0.14824

$P(Z<\frac{X_{critical}-35000}{\frac{4500}{\sqrt{n}}})=P(Z<-1.0440)=0.14824$

$\frac{X_{critical}-35000}{\frac{4500}{\sqrt{n}}}=-1.0440$.....................(4)

Evaluating (1) and (4)

$0.9(A&plus;1.0440)-\sqrt{n}=0$ ..........................................(5)

Now solving (3) and (5)

$A=2.0987$

$n=8$

The sample size n = 8 and the significance level is $P(Z>2.0987)=P(Z<-2.0987)=0.0179$