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) $$\overline{x}=11.75$$
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?
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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