Question

Transistor Manufacturing Study: A transistor manufacturer conducted an experiment to investigate the effects of three factors on PRODUCTivity (measured in thousands of dollars of items produced) per 40-hour week. The factors were as follows:

Length of work week in DAYS (two levels): Five consecutive 8-hour days or four consecutive 10-hour days

SHIFT (two levels): day or evening shift

Number of coffee BREAKS (three levels): 0, 1, or 2

The experiment was conducted over a 24-week period with the 2 × 2 × 3 = 12 treatments assigned randomly to the 24 weeks.

Day Shift Coffee Breaks 0 1 20 Night Shift Coffee Breaks 2 Length 4 days94 105 9690 102 103 0 97 106 91899798 Work 5 days 96 100 82 81 90 94 Week 92 103 88 84 92 96

How many cases in each treatment group?   

Load data set “TRANSISTOR1” into the active data file. Look at data view. Factors can be quantitative (numeric) or qualitative (string). (Data table given below)

Use SPSS “Analyze – General Linear Model – Univariate” to see if the PRODUCT (Dependent) means are different in the treatment groups, because of the interaction of SHIFT*DAYS*BREAKS, other interactions, or the main effects of the three factors. In the “Options” box, Display Means for: (OVERALL), Display “Descriptive Statistics” and “Homogeneity tests”. Run the ANOVA. If any interactions are significant, interpret main effects with caution.

From the output:

a) Are the error variances significantly different from normal? What could cause this?

b) Are any interactions significant?

c) Draw a picture of the significant interaction means.

e) Are there significant main effects separate from the interactions?

f) Which treatment condition had the highest productivity rating?

g) The lowest?

h) Which level of DAYS was most productive?

SHIFT	DAYS	BREAKS	PRODUCT
1	4	0	94
1	4	0	97
1	4	1	105
1	4	1	106
1	4	2	96
1	4	2	91
1	5	0	96
1	5	0	92
1	5	1	100
1	5	1	103
1	5	2	82
1	5	2	88
2	4	0	90
2	4	0	89
2	4	1	102
2	4	1	97
2	4	2	103
2	4	2	98
2	5	0	81
2	5	0	84
2	5	1	90
2	5	1	92
2	5	2	94
2	5	2	96

Answer 1

IN ANOVA since for all main effects and interaction p - value greaterthan 0.05 i.e as 5% lose we do not reject null hypothesis and none of these differ significantly also from normality test of residuals p - value is 0.000 for Shapiro will test hence data is not normal as residual do not follow normal distribution (I + took standardized residuals) (b) none of interaction are significance (c) none of the interaction differ significance so not possible (e) no, none of the main effect are significance (f) day shift, u day working and one break has highest productivity (g) night shifty 5 day and 0 break has lowest productivity.