## Related questions with answers

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

$H_0: μ = 1.5$

versus

$H_1: μ ≠ 1.5$

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?

Solution

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

#### a)

We need to test

$H_0: \mu=1.5$

versus

$H_1:\mu\ne 1.5$

using $\alpha=0.01$.

Test statistic:

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

Now is

$\alpha=0.01 \Rightarrow z_{\frac{\alpha}{2}}=z_{\frac{0.01}{2}}=2.58.$

Rejection Criterion for Fixed-Level Tests: If $-z_\alpha\leq z_0\leq z_\alpha$ then we should fail to reject $H_0$. Therefore

$-1.25= z_0>z_\alpha=-2.58,$

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

#### b)

The P-value for this test is

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

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

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