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Question

In the manufacture of a certain scientific instrument great importance is attached to the life of a critical component. This component is obtained in bulk from two sources, A and B, and in a course of inspection, the lives of 1000 of the components from each source are determined. The following frequency tables are obtained:

Q1.

Life (hours) 1000 – 1020 1020 – 1040 1040 – 1060 1060 – 1080 1080 – 1100 1100 – 1120

Source A

No. of components

40

96

364

372

85

43

Life (hours) 1030 – 1040 1040 – 1050 1050 – 1060 1060 – 1070 1070 – 1080 1080 – 1090

Source A

No. of components

339

136

25

20

130

350

Q2.

Examine the effectiveness of the measures of dispersion with which you are familiar for comparing the dispersion of the two distributions

Answer 1

Source A

Freq SubGroup Mean Subtotal SubGroup Dev SubGroup Dev^2

40 1010 40400 -1996 3984016

96 1030 98880 -2870.4 8239196.16

364 1050 382200 -3603.6 12985932.96

372 1070 398040 3757.2 14116551.84

85 1090 92650 2558.5 6545922.25

43 1110 47730 2154.3 4641008.49

1000 1059900 -9.09495E-11 50512627.7

Group Mean 1059.9

Group Variance 50563.19089

Source B

Freq Subgroup Mean Subtotal Subgroup Dev Subgroup Dev^2

339 1035 350865 -25.16 633.0256

136 1045 142120 -15.16 229.8256

25 1055 26375 -5.16 26.6256

20 1065 21300 4.84 23.4256

130 1075 139750 14.84 220.2256

350 1085 379750 24.84 617.0256

1000 1060160 -0.96 1750.1536

Group Mean 1060.16

Group Variance 1.751905506

Subgroup A is normally distributed. INSIDE of that lives subgroup B, which is an INVERTED normal distribution.

As you can see, the variances will fail any F-Test of hypothesis about the variance. You can also do an ANOVA if you wish.