Question

Facility layout study. Facility layout and material flowpath design are major factors in the productivity analysisof automated manufacturing systems. Facility layout is concerned with the location arrangement of machines and buffers for work-in-process. Flow path design is concerned with the direction of manufacturing material flows (e.g., unidirectional or bidirectional; Lee, Lei, and Pinedo, Annals of Operations Research, 1997). A manufacturer of printed circuit boards is interested in evaluating two alternative existing layout and flowpath designs. The output of each design was monitored for 8 consecutive working days. The data (shown here) are saved in the file. Design 2 appears to be superior to Design 1. Do you agree? Explain

fully.

Working days Design1(Units) Design2(Units)
8/16 1,220 1,273 8/17 1,092 1,363 8/18 1,136 1,342 8/19 1,205 1,471 8/20 1,086 1,229 8/23 1,274 1,457 8/24 1,145 1,263 8/25 1,281 1,368

Short answer

\(Ha\) is true would be design-2 is greater than design1.

Step-by-step solution

Given information

Using evidence, it is necessary to address whether Design 1 is superior to Design 2.

\(\begin{aligned}{l}n1 &= 8\\n2 &= 8\end{aligned}\)

Explaining the variance

The variance measures variability. The amount of scattered is indicated by variance.

The variance is more significant with the mean and the more dispersed the data.

Checking whether design2 is greater than design1

\(\begin{aligned}{l}{\sigma ^2}1 &= Variance\,of\,design\,1\\{\sigma ^2}2 &= \,Variance\,of\,design\,2\end{aligned}\)

The test hypothesis is as follows.

\(\begin{aligned}{l}H0:\,{\sigma ^2}1 < {\sigma ^2}2\\Ha:\,{\sigma ^2}1 &= {\sigma ^2}2\end{aligned}\)

Denominator\({S^2}2\) and numerator\({S^2}1\) both

\(df = n1 - 1\)

\(v1 = n1 - 1\)

\(\begin{aligned}{l}df &= 8 - 1\\df &= 7\end{aligned}\)

\(df = n2 - 1\)

\(v2 = n2 - 1\)

\(\begin{aligned}{l}df &= 8 - 1\\df &= 7\end{aligned}\)

Design1\(\left( x \right)\)	\(xi - \bar x\)	\({\left( {xi - \bar x} \right)^2}\)	Design2\(\left( y \right)\)	\(yi - \bar y\)	\({\left( {yi - \bar y} \right)^2}\)
1,220	40.125	1,610.01	1,273	-81.5	6,642.25
1,092	-87.875	7,722.01	1,363	8.5	72.25
1,136	-43.875	1,925.01	1,342	-12.5	156.25
1,205	25.125	631.26	1,471	116.5	13,572.25
1,086	-93.875	8,812.51	1,299	-55.5	3,080.25
1,274	94.125	8,859.51	1,457	102.03	10,410.1209
1,145	-34.875	1,261.26	1,263	-91.5	8372.25
1,281	101.125	10,226.26	1,368	13.5	182.25
Total:9,439		Total:41,047.83	Total:10,836		Total:42,487.879

\(\begin{aligned}{l}\bar x &= \frac{{\sum xi}}{n}\\\bar x &= \frac{{9,439}}{8}\\\bar x &= 1,179.875\end{aligned}\)

\(\begin{aligned}{l}S{1^2} &= \frac{{\sum {{\left( {xi - \bar x} \right)}^2}}}{{n - 1}}\\S{1^2} = \frac{{41,047.83}}{7}\\S{1^2} &= 5,863.97\end{aligned}\)

\(\begin{aligned}{l}\bar y &= \frac{{\sum yi}}{n}\\\bar y &= \frac{{10,836}}{8}\\\bar y = 1,354.5\end{aligned}\)

\(\begin{aligned}{l}S{2^2} &= \frac{{\sum {{\left( {yi - \bar y} \right)}^2}}}{{n - 1}}\\S{2^2} &= \frac{{42,487.879}}{7}\\S{2^2} = 6,069.697\end{aligned}\)

The test statistics will therefore be.

\(\begin{aligned}{l}F &= {\textstyle{{Larger\,\,sample\,\,variance} \over {Smaller\,\,sample\,\,variance}}}\\F &= \frac{{{S^2}1}}{{{S^2}2}}\\F &= \frac{{41,047.83}}{{42,487.879}}\\F &= 0.96610\end{aligned}\)

Reject\(H0:\,{\sigma ^2}1 = {\sigma ^2}2\)for F=0.966 When the calculated design-2 exceeds the design1 .So, reject\(H0:\,{\sigma ^2}1 = {\sigma ^2}2\)

Therefore, this test indicates that both variances will differ. The probability of seeing a value of F at least as similar to \(Ha:{\sigma ^2}1 < {\sigma ^2}2\) as F=0.96610 .So, \(Ha\) is true would be design-2 is greater than design1.