Question

A manufacturer produces crankshafts for an automobile engine. the crankshafts wear after 100,000 miles (0.0001 inch) is of interest because it is likely to have an impact on warranty claims. a random sample of n = 15 shafts is tested and x = 2.78. it is known that σ = 0.9 and that wear is normally distributed. (a) test h0 : μ = 3 versus h1: μ ≠ 3 using α = 0.05. (b) what is the power of this test if μ = 3.25? (c) what sample size would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9?

Answer 1

Part A:

Significant level:

α = 0.05

Null and alternative hypothesis:

h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{2.78-3}{0.9/\sqrt{15}} \\ \\ = \frac{-0.22}{0.2324} =-0.9467$

P-value:

P(-0.9467) = 0.1719

Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438

Conclusion:

Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.



Part B:

The power of the test is given by:

$\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\ \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\ \\ =0.8117-0.0012=0.8105$

Therefore, the power of the test if μ = 3.25 is 0.8105.



Part C:

The sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:

$1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\ \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\ \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\ \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898$

$\Rightarrow n=(3.8898)^2=15.13$

Therefore, the sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.