Question

Manufacturers that practice sole sourcing. If a manufacturer (the vendee) buys all items of a particular type from a particular vendor, the manufacturer is practicing sole sourcing (Schonberger and Knod, Operations Management, 2001). As part of a sole-sourcing arrangement, a vendor agrees to periodically supply its vendee with sample data from its production process. The vendee uses the data to investigate whether the mean length of rods produced by the vendor's production process is truly 5.0 millimetres (mm) or more, as claimed by the vendor and desired by the vendee.

a. If the production process has a standard deviation of .01 mm, the vendor supplies n = 100 items to the vendee, and the vendee uses a = .05 in testing H₀: m = 5.0 mm against H_a: m < 5.0 mm, what is the probability that the vendee's test will fail to reject the null hypothesis when in fact m = 4.9975 mm? What is the name given to this Type of error?

b. Refer to part a. What is the probability that the vendee's test will reject the null hypothesis when m = 5.0? What is the name given to this Type of error?

c. What is the power of the test to detect a departure of .0025 mm below the specified mean rod length of 5.0 mm?

Short answer

⦁$TypeIIerror\left(\beta \right)=0.1963$

⦁ $TypeIIerror\left(\beta \right)=0.1963$

⦁ We know that the degree of significance is referred to as the likelihood of Type I error, and it is obtained as a result $\alpha =0.05$.

Step-by-step solution

(a) Type of error

From the given data:

Null hypothesis: $H_0:\mu =5.0$

(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:

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

The resultant output are as follows:

One sample Z

$Testofmu=5vs<5$

The assumed standard deviation = 0.01

N	Mean	SE Mean	95% Upper bound	Z	P
100	4.99750	0.00100	4.99914	-2.50	0.006

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:

$$\begin{gathered} =\frac{\overline{x}-\mu }{{\displaystyle \frac{s}{\sqrt{n}}}}\ge -z_{0.05} \\ =\frac{\overline{x}-5.0}{{\displaystyle \frac{0.001}{\sqrt{100}}}}\ge -1.645 \\ =\overline{x}\ge (-1.645\times 0.001)+5.0 \\ =\overline{x}\ge 4.998 \end{gathered}$$

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 $\mu =5.0$

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

The probability of rejecting the null hypothesis it is actual:

Type-II error $\left(\beta \right)$

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

:

(b) Reject the null hypothesis

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 $\mu =5.0$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 $\alpha =0.05$.

(c) The specified mean rod length of 5.0 mm

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 $\beta$value as 0.1963.

Power of the test (1- $\beta$):

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

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