Design And Analysis Of Experiments 2

Design and Analysis of Experiments Student: Nguyen Thi Hang ID: R9605005

Mark:

Problem. 5-4. An article in Industrial Quality Control (1956, pp. 5-8) describes an experiment to investigate the effect of the type of glass and the type of phosphor on the brightness of a television tube. The response variable is the current necessary (in microamps) to to obtain a specified brightness level. The data are as follows:

Glass type

1

2

Phosphor type 1

2

3

280

300

290

290

310

285

285

295

290

230

260

220

235

240

225

240

235

230

a) Is there any indication that either factor influences brightness? Use α = 0.05; b) Do the two factors interact? Use α = 0.05; c) Analyse the residuals from this experiment.

Solution We use the Two-factor analysis of variance to indicate that either factor influences brightness. The equality of row treatment effects (factor A – Glass type) H0: τ1=τ2 = 0 H1: at least τi ≠ 0

The equality of column treatment effects (factor B – Phosphor type) H0: β1= β2 H1: at least βi ≠ 0

we are also interested in determining whether row and column treatments interact. Thus, we also wish to test H0: (τβ)ij = 0 for all i, j H1: at least one (τβ)ij ≠ 0 Using Minitab software with the following steps:

1. Input and prepare data for the test

2. Select Two-Way of ANOVA method

3. Set up the suitable parameters

we can use some options such as Graphs

Then, we can see the results from using the Minitab

(a) Is there any indication that either factor influences brightness? Use α = 0.05  F0 (glass type) = 273.79 > F0.05, 1, 12 = 4.75 or P = 0.000 < 0.05 thus we can conclude that the glass type affects on the brightness level of television tube.  F0 (Phosphor type) = 8.84 > F0,05, 2,12 = 3.89 of P = 0.004 < 0.05 thus the Phosphor type affects significantly on the brightness level of television tube (b) Do the two factors interact? Use α = 0.05  F0 (interaction) = 1.26 < F0.05, 2, 12 = 3.89 or P = 0.318 > 0.05 thus we can not reject H0 or there are no interaction effect between the glass type and the phosphor type. (c) Analyze the residual from this experiment From output of Minitab we get the following Figure:

Normal Probability Plot of the Residuals (response is Brightness level) 99

95 90

Percent

80 70 60 50 40 30 20 10 5

1

-15

-10

-5

0 Residual

5

10

15

From the normal probability plot of these residuals shows that doesn’t reveal anything particularly troublesome. Although the largest positive residual (15 at phosphor type 2 and glass type 2) does stand out some what from others.The standardized value of this residual is 15 / 52.8 = 2.06

Residuals Versus the Fitted Values (response is Brightness level) 15

Residual

10

5 0

-5

-10 220

230

240

250

260 270 Fitted Value

280

290

300

310

From plots the residuals versus the fitted values. This plots indicates that a mild tendency for the variance of the residuals to increase.

Residuals Versus Phosphor type (response is Brightness level) 15

Residual

10

5 0

-5

-10 1.0

1.5

2.0 Phosphor type

2.5

3.0

Residuals Versus Glass type (response is Brightness level) 15

Residual

10

5 0

-5

-10 1.0

1.2

1.4 1.6 Glass type

1.8

2.0

Two above figure plot the residuals versus Glass type and Phosphor type, respectively. Both plots indicates mild inequality of variance, with the treatment combination of Phosphor type 2 with glass type 2 possibly having larger variance than the others.