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Tests of Hypotheses for a Single Sample 45:06
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A textile fiber manufacturer is investigating a new
drapery yarn, which the company claims has a mean thread
elongation of 12 kilograms with a standard deviation of 0.5
kilograms. The company wishes to test the hypothesis
$$H_{0} : \mu=12$$ against $$H_{1} : \mu<12,$$ using a random sample of
four specimens.

(a) What is the type I error probability if the critical region is
defined as $$\overline{x}<11.5$$ kilograms?
(b) Find $$\beta$$ for the case where the true mean elongation is
11.25 kilograms.
(c) calculate the $$P$$ -value if the observed statistic is
(a) $$\overline{x}=11.25 \quad$$ (b) $$\overline{x}=11.0 \quad$$ (c) $$\overline{x}=11.75$$

$$\mu=12 \mathrm{kg}$$

$$\sigma=0.5$$

$$H_{0} : \mu=12$$ against $$H_{1} : \mu<12$$

$$n=4$$

(a) $$\overline{x}<11.5$$

$$\alpha=P(\overline{x}<11.5)=P\left(\frac{\overline{x}-\mu}{\sigma  / \sqrt{n}} \leq \frac{11.5-12}{\frac {0.5}{\sqrt 4}}\right)$$

$$=p(Z \leq -2)$$
$$=0.02275$$

(b) $$\beta$$ when $$\mu=11.25$$

$$\beta=P\left(\text { accept } H_{0}\right)=P(\overline{x}>11.5)$$

$$=P\left(Z>\frac{11.5 - 11.25}{\frac {0.5}{\sqrt 4}}\right)$$

$$\beta=P(Z>1)=1-P(Z<1)$$

$$=0.15866$$

(c) $$P=\phi\left(Z_{0}\right)$$

(a) $$P=\phi\left(\frac{11.25-12}{0.5 / \sqrt{4}}\right)=\phi (-3)$$

$$=0.00135$$

(b) $$P=\phi\left(\frac{11-12}{\frac {0.5}{\sqrt{4}}}\right)=\phi (-4)=0$$

(c) $$P=\phi \left(\frac{11.75-12}{0.5 / \sqrt{4}}\right)=\phi(-1)$$

$$=0.15866$$

A manufacturer is interested in the output voltage of a
power supply used in a PC. Output voltage is assumed to be
normally distributed, with standard deviation 0.25 Volts, and
the manufacturer wishes to test $$H_{0} : \mu=5$$ Volts against
$$H_{1} : \mu \neq 5$$ Volts, using $$n=8$$ units.

(a) The acceptance region is $$4.85 \leq \overline{x} \leq 5.15$$ . Find the value
of $$\alpha .$$
(b) Find the power of the test for detecting a true mean output
voltage of 5.1 Volts.
(c) suppose that the manu facturer wants the type I error
probability for the test to be $$\alpha=0.05 .$$ Where should
the acceptance region be located?

$$\sigma=0.25 \mathrm{v} \quad \mathrm{n}=8$$

$$H_{0}: \mu=5 v$$
$$H_{1} : \mu \neq 5$$

(a) $$4.85 \leq \overline{x} \leq 5.15$$

$$\alpha ?!$$

$$\alpha=P(\overline{x}<4.85)+P(\overline{x}>5.15)$$

$$P\left(Z <\frac{4.85-5}{0.25 /\sqrt 8}\right)+\left(Z>\frac{{5.15}-5}{0.25/\sqrt{8}}\right)$$

$$=P(Z<-1.7)+P(Z>1.7)$$

$$\alpha = 0.08914$$

(b) $$\mu=5.1,$$ The Power

Power $$=1- \beta$$

$$\beta=P(4.85 \leq \overline x \leq 5.15)$$

$$=P\left(\frac{4.85-5.1}{0.25/\sqrt 8} \leq Z \leq \frac{5.15 -5.1}{\frac{0.25}{\sqrt{8}}}\right)$$

$$=P(-2.83 \leq Z \leq 0.566)$$

$$=P(Z \leq 0.566)-P(Z \leq -2.83)$$

$$=0.71566-0.00233$$

$$=0.71333$$

Power $$=1-\beta=0.2867$$

(c) $$\alpha=0.05$$

$$\frac{\alpha}{2}=0.025$$

$$0.025=P\left(Z \leq \frac{c-5}{0.25 / \sqrt{8}}\right)$$

$$Z=\pm 1.96$$

$$\frac{c-5}{0.25 / \sqrt{8}}=-1.96 \rightarrow c=4.83$$

$$=1.96 \rightarrow c=5.17$$

$$(4.83\leq \overline{x} \leq 5.17)$$

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