Question

The data in the table below presents the hourly quantity of production for three lines of production processes over the first 4 days in XYZ Company. Answer the questions based on the Excel Output given below. Day Process 1 Process 2 Process 3 1 33 33 28 2 30 35 36 3 28 30 30 4 29 38 34 ANOVA: Single Factor SUMMARY Groups Count Sum Average Variance Process 1 4 120 30 4.66667 Process 2 4 136 34 11.3333 Process 3 4 128 32 13.3333 ANOVA Source of Variation SS df MS F P value Between Groups 32 ? ? ? Within Groups 88 ? ? Total 120 11

a. State the null and alternative hypothesis for single factor ANOVA. (1 mark)

b. State the decision rule (α = 0.05).

c. Calculate the test statistic.

d. Make a decision. (2 marks

Answer 1

Answer:-

Given That:-

The data in the table below presents the hourly quantity of production for three lines of production processes over the first 4 days in XYZ Company.

Given,

here

k = 3

n = 12

= 3 - 1

= 2

= 12 - 3

= 9

= 32/2

= 16

= 88/9

= 9.7778

= 16/9.7778

= 1.6364

Source of variation	SS	df	MS	F	P
Between groups	32	2	16	1.6364	0.2210
Within groups	88	9	9.7778
Total	100	11

a. State the null and alternative hypothesis for single factor ANOVA. (1 mark)

(There is no Significant difference between samples)

(There is significant difference between samples)

b. State the decision rule (α = 0.05). (4 marks)

Critical value

Decision Rule:-

Reject if   Otherwise do not reject

c. Calculate the test statistic. (3 marks)

Test statistic (F) = 1.6364

d. Make a decision. (2 marks

Here F(1.6364) < (4.2565) so we fail to reject

So there is no significance difference between samples.

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