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Question

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H$_0$: μ = 100 versus H$_1$: μ ≠ 100 with a sample of n = 9 specimens. a. If the acceptance region is defined as 98.5 ≤ $\overline{x}$ ≤ 101.5, find the type I error probability α. b. Find β for the case in which the true mean heat evolved is 103. c. Find β for the case where the true mean heat evolved is 105. This value of β is smaller than the one found in part (b). Why?

Solution

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The probability of making a $\textbf{type I error}$ is

\begin{align} \alpha=P(\text{type I error})=P(\text{ reject H_0 when H_0 is true}). \end{align}

Sometimes the type I error probability is called the significance level, the $\alpha$-error, or the size of the test.

The mean is thought to be $\mu=100$, and the standard deviation is $\sigma=2$. You wish to test $H_0:\mu = 100$ versus $H_1:\mu\ne 100$ with a sample of $n = 9$ specimens.

a)

The acceptance region is defined as

$98.5\leq\bar{x}\leq 101.5,$

than Equation (1), the type I error probability is

\begin{align*} \alpha&=P(98.5>\bar{x}\, \text{when}\, \mu=100)+ P( \bar{x}> 101.5\, \text{when}\, \mu=100)\\ &=P\left(\frac{98.5-100}{2/\sqrt{9}}>\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\, \text{when}\, \mu=100\right)+\left(\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}>\frac{101.5-100}{2/\sqrt{9}}\right)\\ &\overset{(\ast)}{=}P(z> 2.25)+P(z<-2.25)\\ &=1-P(z\leq2.25)+1-P(z\geq-2.25). \end{align*}

$(\ast)$: The distribution of random variable

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$

is approximately normal with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}.$

Therefore, from Appendix Table III, the type I error probability is

\begin{align*} \alpha\overset{(\ast\ast)}{=}2-2P(z\leq 2.25)=\boxed{0.024448.} \end{align*}

$(\ast\ast)$: Based on the symmetry of the normal distribution

$P(z\leq 2.25)=P(z\geq -2.25).$

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