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A bearing used in an automotive application is supposed to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected, and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation σ = 0.01 inch. a. Test the hypothesis

H0:μ=1.5H_0: μ = 1.5

versus

H1:μ1.5H_1: μ ≠ 1.5

using α = 0.01. b. What is the P-value for the test in part (a)? Compute the power of the test if the true mean diameter is 1.495 inches. d. What sample size would be required to detect a true mean diameter as low as 1.495 inches if you wanted the power of the test to be at least 0.9?

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A random sample of n=25n=25 bearings is selected, and the average inside diameter of these bearings is xˉ=1.4975\bar{x}=1.4975 inches and standard deviation σ=0.01\sigma=0.01 inches.

a)

We need to test

H0:μ=1.5H_0: \mu=1.5

versus

H1:μ1.5H_1:\mu\ne 1.5

using α=0.01\alpha=0.01.

Test statistic:

z0=xˉμσn=1.49751.50.0125=1.25.z_0=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{1.4975-1.5}{\frac{0.01}{\sqrt{25}}}=-1.25.

Now is

α=0.01zα2=z0.012=2.58.\alpha=0.01 \Rightarrow z_{\frac{\alpha}{2}}=z_{\frac{0.01}{2}}=2.58.

Rejection Criterion for Fixed-Level Tests: If zαz0zα-z_\alpha\leq z_0\leq z_\alpha then we should fail to reject H0H_0. Therefore

1.25=z0>zα=2.58,-1.25= z_0>z_\alpha=-2.58,

and $\text{\textcolor{#4257b2}{ hypothesis H0:μ=3500H_0: \mu= 3500 is true.}}$

b)

The P-value for this test is

P-value=2[1Φ(z0)]=2[1Φ(1.25)]=2[10.894350]=0.2113\begin{align*} \text{P-value}&=2[1-\Phi(|z_0|)]\\ &=2[1-\Phi(1.25)]\\ &=2[1-0.894350]\\ &=\boxed{0.2113} \end{align*}

where Φ(z)\Phi(z) denotes the probability to the left of zz in the standard normal distribution.

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