Metalworks, a supplier of fabricated industrial parts, wants to determine if the average output rate for a particular component is the same across the three work shifts. However, since any of four machines can be used, the machine effect must be controlled for within the sample. The accompanying table shows output rates (in units) for the previous day. (You will need to calculate the t-statistic, for the given α and df per the ANOVA Error row.)
| Machine (Factor B) | Shift (Factor A) | ||
| 1 | 2 | 3 | |
| A | 1392 | 1264 | 1334 |
| B | 1228 | 1237 | 1107 |
| C | 1173 | 1108 | 1186 |
| D | 1331 | 1342 | 1387 |
a-1. Construct an ANOVA table
a-2. If significant differences exist across the machines, use Fisher’s LSD method at the 5% significance level to determine which machines have different average output rates. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)
ANSWER::
| MSE= | 3362.806 | ||
| df(error)= | 6 | ||
| number of treatments = | 4 | ||
| pooled standard deviation=Sp =√MSE= | 57.990 |
| critical t with 0.05 level and 6 df= | 2.447 |
| Fisher's (LSD) =(t)*(sp*√(1/ni+1/nj) = | 115.86 |
| Lower bound | Upper bound | differ | |||
| (xi-xj ) | ME | (xi-xj)-ME | (xi-xj)+ME | ||
| μ1-μ2 | 139.333 | 115.857 | 23.48 | 255.19 | significant difference |
| μ1-μ3 | 174.33 | 115.86 | 58.48 | 290.19 | significant difference |
| μ1-μ4 | -23.33 | 115.86 | -139.19 | 92.52 | not significant difference |
| μ2-μ3 | 35.00 | 115.86 | -80.86 | 150.86 | not significant difference |
| μ2-μ4 | -162.67 | 115.86 | -278.52 | -46.81 | significant difference |
| μ3-μ4 | -197.67 | 115.86 | -313.52 | -81.81 | significant difference |
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