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Tests of Hypotheses for a Single Sample 45:06
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A bearing used in an automotive application is suppose
to have a nominal inside diameter of 1.5 inches. A random sam-
ple 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 $$\sigma=0.01$$ inch.

(a) Test the hypotheses $$H_{0} : \mu=1.5$$ versus $$H_{1} : \mu \neq 1.5$$ using
$$\alpha=0.01$$
(b) Compute the power of the test if the true mean diameter is
1.495 inches.
(c) What sample size would be required to detect a true mean
diameter as low as 1.495 inches if we wanted the power of
the test to be at least 0.9$$?$$

$$\sigma=0.01 \mathrm{inch}$$
$$n=25$$
$$\overline{x}=1.4975$$
$$\alpha=0.01$$

(a) (1) Parameter of interest: $$\mu$$

(2) $$H_{0} : \mu=1.5 \text { inch }$$

(3) $$H_{1} : \mu \neq 1.5$$

(4) Test statistics: $$Z_{0}=\frac{\overline{x}-\mu_{0}}{\sigma / \sqrt{n}}$$

$$Z_{0}=\frac{1.4975 - 1.5}{0.01 / \sqrt{25}}=-1.25$$

(5) P-value $$\rightarrow P=2\left[1-\phi\left(Z_{0}\right)\right]=0.2113>\alpha \rightarrow \text {accept} \ H_{0}$$

(b) $$\mu=1.495$$

power $$=1-\beta$$

$$\delta=1.495-1.5=-0.005$$

$$\beta=\phi\left(Z_{\frac {\alpha}{2}} \cdot \frac{\delta \sqrt{n}}{\sigma}\right) \cdot \varphi\left(-Z_{\frac {\alpha}{2}}-\frac{\delta \sqrt{n}}{\sigma}\right)$$

$$Z_{0.005} \rightarrow 2.58$$

$$\beta=\phi(5.08)-\phi(-0.08)$$

$$=1-0.46812$$

$$=0.53188$$

power $$=1-\beta$$

$$=0.46812$$

(c) $$\mu=1.495$$

power $$= 0.9$$

$$n=\frac{\left(Z_{\frac {\alpha}{2}}+Z_{\beta}\right)^{2} \sigma^{2}}{\delta^{2}}$$

$$\delta=1.495-1.5=-0.005$$

$$Z_{\alpha / 2}=Z_{0.005}=2.58$$

power $$=1-\beta=0.9$$

$$\Rightarrow \beta=0.1$$

$$Z_{\beta}=Z_{0.1}=1.29$$

$$n=\frac{(2.58+1.29)^{2}(0.01)^{2}}{(-0.005)^{2}}$$

$$=59.908$$

$$n \cong 60$$

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