Question

Preventing the production of defective items. It costs more toproduce defective items—because they must be scrappedor reworked—than it does to produce non-defective items.This simple fact suggests that manufacturers shouldensurethe quality of their products by perfecting theirproduction processes rather than through inspection of finishedproducts (Out of the Crisis,Deming, 1986). In orderto better understand a particular metal-stamping process, the manufacturer wishes to estimate the mean length of itemsproduced by the process during the past 24 hours.

a. How many parts should be sampled in order to estimatethe population means to within .1 millimetre (mm)with 90% confidence? Previous studies of this machinehave indicated that the standard deviation of lengthsproduced by the stamping operation is about 2 mm.

b. Time permits the use of a sample size no larger than100. If a 90% confidence interval for is constructedusing n= 100, will it be wider or narrower than wouldhave been obtained using the sample size determined in

part a? Explain.

c. If management requires that $\mathbf{\mu }$be estimated to within.1 mm and that a sample size of no more than 100 beused, what is (approximately) the maximum confidencelevel that could be attained for a confidence interval

Does that meet management's specifications?

Short answer

a. 1083 parts must be sampled to calculate the population averages with 90% confidence to within .1 millimeter (mm).

b. In component (b), the confidence interval is wider.

c. The greatest degree of confidence is 62 %, showing that the specification does not fulfill management requirements

Step-by-step solution

Given information

Previous research on this equipment found that the standard variation of lengths generated by the stamping process is around 2 mm.

(a) Finding the sample size

The standard deviation is 0.1 in this case.

The standard deviation value is two.

A 90 percent confidence interval's crucial value is$z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645$

$\begin{array}{rcl}SE& =& z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\\ n& =& {z^2}_{\alpha /2}\frac{{\sigma }^2}{S{E}^2}\\ n& =& 1.{645}^2\times \frac{2^2}{{0.1}^2}\\ n& =& 1082.41\\ n& \simeq & 1083\\ & & \end{array}$

To calculate the population means to within, 1083 parts must be sampled. 90 percent confidence in 1 millimeter (mm)

Step 3:(b) Checking confidence interval is wider or narrower

The formula for the confidence interval is

$Cl=\left(\overline{x}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}},\overline{x}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\right)$

As a result, as the sample size grows, the error portion reduces, as well as the width of the confidence interval grows, and as the sample size grows, the error part rises, as well as the width of the confidence interval grows.

As a result, the confidence interval built with n=100 will have a bigger confidence interval than a portion (a)

Step 4:(c) Finding the confidence level

The standard deviation is 0.1 in this case.

The standard deviation value is2.

The sample size is 100 people.

$$\begin{gathered} SE=z_{\alpha /2}\frac{\sigma }{\sqrt{n}} \\ {z}_{\alpha /2}=\frac{SE\times \sqrt{n}}{\sigma } \\ {z}_{\alpha /2}=\frac{0.1\times \sqrt{100}}{2} \\ {z}_{\alpha /2}=0.5 \end{gathered}$$

$$\begin{gathered} \frac{\alpha }{2}=0.308538 \left(\text{From} \text{standard} \text{normal} \text{distribution} \text{table}\right) \\ \alpha =0.617075 \\ \alpha \approx 0.62 \end{gathered}$$

The level of certainty is 62%.

The specification, in this case, does not fulfill management requirements.