# 统计 Statistics MATH08051

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\begin{aligned} &H_{0}: \mu=0.86 \ &H_{a}: \mu \neq 0.86 \end{aligned}
at the $1 \%$ level of significance. What is the power of this test against the specific alternative $\mu=0.845$ ?
The test rejects $H_{0}$ when $|z| \geq 2.576$. The test statistic is
$$z=\frac{\bar{x}-0.86}{0.0068 / \sqrt{3}}$$

Some arithmetic shows that the test rejects when either of the following is true:
$$\begin{array}{ll} z \geq 2.576 & \text { (in other words, } \bar{x} \geq 0.870 \text { ) } \ z \leq-2.576 & \text { (in other words, } \bar{x} \leq 0.850 \text { ) } \end{array}$$
These are disjoint events, so the power is the sum of their probabilities, computed assuming that the alternative $\mu=0.845$ is true. We find that
\begin{aligned} P(\bar{x} \geq 0.87) &=P\left(\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \geq \frac{0.87-0.845}{0.0068 / \sqrt{3}}\right) \ &=P(Z \geq 6.37) \doteq 0 \end{aligned}

## MATH08051  COURSE NOTES ：

The sample mean is $\bar{x}=5$ and the standard deviation is $s=3.63$ with degrees of freedom $n-1=7$. The standard error is
$$\mathrm{SE}_{\bar{x}}=s / \sqrt{n}=3.63 / \sqrt{8}=1.28$$
From Table D we find $t^{}=2.365$. The $95 \%$ confidence interval is \begin{aligned} \bar{x} \pm t^{} \frac{s}{\sqrt{n}} &=5.0 \pm 2.365 \frac{3.63}{\sqrt{8}} \ &=5.0 \pm(2.365)(1.28) \ &=5.0 \pm 3.0 \ &=(2.0,8.0) \end{aligned}