Question

Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coating on the wire electrode allows for cooling of the wire electrode core and provides an improved cutting performance. An article gave the following sample observations on total coating layer thickness (in µm) of eight wire electrodes used for WEDM. 21 16 29 36 42 25 24 25
Calculate a 99% CI for the standard deviation of the coating layer thickness distribution. (Round your answers to two decimal places.) , Is this interval valid whatever the nature of the distribution? Explain.

Yes, there are enough data points for this interval to be valid.

No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.

Answer 1

Answer:

Step-by-step explanation:

Hello!

You need to make a Confidence Interval for the population standard deviation of the coating layer tickness of wires.

The study variable is:

X: Coating layer tickness of a wire electrode.

Sample:

21; 16; 29; 36; 42; 25; 24; 25

n= 8

X[bar]= 27,25

∑xi= 218

∑xi²= 6424

S²= 69,07

S= 8.31

To make a Confidence interval to estimate a parameter, you need to choose the satistic that is more appropiate for it. The characteristics of a good Statistic is that is should have a known distribution, should contain the parameter and the estimator. The population standard deviation doesn't fit to any statistic, so to study it you have to do it trough the population variance. Any conclusions you make about the population variance, you can extrapolate to its square root.

The most appropiate statistic to study the population variance is the Chi-Square, if your variable is normally distributed, you can use it to make the confidence interval.

So with the given information I've ran a Shapiro- Wilks test to check the variable distribution. With a p-value= 0.7148, I can say that the variable has normal distribution.

The interval is:

$\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2} } ,\frac{(n-1)S^2}{X^2x_{n-1-,\alpha/2} }$

$X^2_{n-1; 1-\alpha /2} = X^2_{7; 0.995} =20.278$

$X^2_{n-1; \alpha /2} = X^2_{7; 0.005} = 0.9893$

$\frac{7*69,07}{20.278} } ,\frac{7*69,07}{0.989} }$

[23.84;488.87]

Hope it helps!