这是一份ed.ac.uk爱丁堡大学MATH08051作业代写的成功案例
$$
\begin{aligned}
&H_{0}: \mu=0.86 \
&H_{a}: \mu \neq 0.86
\end{aligned}
$$
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 $$ \begin{aligned} \bar{x} \pm t^{} \frac{s}{\sqrt{n}} &=5.0 \pm 2.365 \frac{3.63}{\sqrt{8}} \
&=5.0 \pm(2.365)(1.28) \
&=5.0 \pm 3.0 \
&=(2.0,8.0)
\end{aligned}
$$