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From the given data:

Null hypothesis: $H_0:渭=5.0$

(The average length of rods generated by the vendor's manufacturer is 5.0 mm.)

Alternative hypothesis:

(The average length of rods generated by the vendor's manufacturer is 5.0 mm.)

The vendor supplies:

The significance level:

The test statistic z is given by:

$z=\frac{\overline{x}-渭}{{\displaystyle \frac{s}{\sqrt{n}}}}$

The resultant output are as follows:

One sample Z

$Testofmu=5vs<5$

The assumed standard deviation = 0.01

From the above result, we have

$z-statistic=-2.50andp-value=0.006$

The $p-value$ is smaller than the level of significance, in this case, and we must reject the null hypothesis as well as declare that there is adequate evidence to back up the assertion that the mean lengths of their odds created by the vendor's manufacturing technique are actually 5.0mm.

We calculate the probability that the vendee's test reject to fail the null hypothesis where it is genuine, and it can be as follows:

The sample means as follows:

$$\begin{gathered} =\frac{\overline{x}-渭}{{\displaystyle \frac{s}{\sqrt{n}}}}鈮�-z_{0.05} \\ =\frac{\overline{x}-5.0}{{\displaystyle \frac{0.001}{\sqrt{100}}}}鈮�-1.645 \\ =\overline{x}鈮�(-1.645脳0.001)+5.0 \\ =\overline{x}鈮�4.998 \end{gathered}$$

Based on the conclusions obtained in the preceding section, the chance that the vendee's test will reject the null hypothesis when it is true is classified as a type I mistake because it indicates that reject the null hypothesis when it is true $渭=5.0$

We know that the level of significance is referred to as the chance of Type I error, and it is calculated as $伪=5.0$

The probability of rejecting the null hypothesis it is actual:

Type-II error $\left(尾\right)$

$$\begin{gathered} P(Accepting\overline{)H_0}{H}_{1}istrue)=P(\overline{x}鈮�4.998\overline{)4}渭=4.9975 \\ =P\left(\frac{\overline{x}-渭}{未/\sqrt{n}}鈮�\frac{4.9984-4.9975}{0.001}\right) \\ =P(z鈮�0.855\left)\right(NORMDIST(z,cumulative)) \\ =1-0.8037 \\ Typellerror(尾)=0.1963 \end{gathered}$$

Based on the conclusions obtained in the preceding section, the chance that the vendee's test will reject the null hypothesis when it is true is classified as a type I mistake $渭=5.0$because it indicates that reject the null hypothesis when it is correct. We know that the degree of significance is referred to as the likelihood of Type I error, which is obtained as a result $伪=0.05$.

Here, we need to find out the power of the test to detect a departure of 0.0025 mm below the specified mean rod length of 5.0 mm is given by:

From part (a), we know the $尾$value as 0.1963.

Power of the test (1- $尾$):

$$\begin{gathered} 1-尾=1-0.1962 \\ 1-尾=0.8037 \end{gathered}$$

Therefore, the power of the test value is $0.8037$