Question

1. To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow. Temperature 50°C 60°C 70°C 31 30 24 21 31 29 33 34 29 36 23 31 29 27 32 a. Construct an analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary).

Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 2 0.0 Error 12 Total 14 b. Use a .05 level of significance to test whether the temperature level has an effect on the mean yield of the process. Calculate the value of the test statistic (to 2 decimals).

The p-value is What is your conclusion?

2. In a study conducted to investigate browsing activity by shoppers, each shopper was initially classified as a nonbrowser, light browser, or heavy browser. For each shopper, the study obtained a measure to determine how comfortable the shopper was in a store. Higher scores indicated greater comfort. Suppose the following data were collected.

		Light	Heavy
	Nonbrowser	Browser	Browser
	6	7	7
	7	8	9
	8	7	7
	5	6	9
	5	9	6
	6	6	8
	7	8	7
	6	7	9

a. Use x=.05 to test for a difference among mean comfort scores for the three types of browsers.

Compute the values identified below (to 2 decimals, if necessary).

Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
Mean Squares, Error

Calculate the value of the test statistic (to 2 decimals, if necessary).

_____

Use Fisher's LSD procedure to compare the comfort levels of nonbrowsers and light browsers. Use .

Compute the LSD critical value (to 2 decimals). _____

3.

The following data are from an experiment designed to investigate the perception of corporate ethical values among individuals who are in marketing. Three groups are considered: management, research and advertising (higher scores indicate higher ethical values).

Marketing Managers	Marketing Research	Advertising
10	9	11
9	9	12
8	8	11
9	8	10
10	9	11
8	8	11

a. Compute the values identified below (to 2 decimal, if necessary).

Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
Mean Squares, Error

b. Use x=.05 to test for a significant difference in perception among the three groups.

Calculate the value of the test statistic (to 2 decimals).

Using , determine where differences between the mean perception scores occur.

Calculate Fisher's LSD value (to 2 decimals).

Difference	Absolute Value	Conclusion
~x1 - ~x2
~x1-~x3
~x2-~x3

Answer 1

2)

excel data analysis tool for one factor anova,steps are: write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
non browser	8	50	6.25	1.071429
light	8	58	7.25	1.071429
heavy	8	62	7.75	1.36
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	9.333	2	4.667	4.000	0.0337	3.467
Within Groups	24.500	21	1.167
Total	33.833	23

a)

Sum of Squares, Treatment	9.33
Sum of Squares, Error	24.50
Mean Squares, Treatment	4.67
Mean Squares, Error	1.17

value of the test statistic =4.00

_____

Level of significance	0.0500
no. of treatments,k	3
DF error =N-k=	21
MSE	1.167
t-critical value,t(α/2,df)	2.0796

Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj)) = 1.12

if absolute difference of means > critical value,means are significnantly different ,otherwise not                      
                      

for  the comfort levels of nonbrowsers and light browsers , absolute mean difference is 1

absolute mean differece = 1 < LSD , so, there is no significant difference

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3)

excel data analysis tool for one factor anova,steps are: write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Marketing Managers	6	54	9	0.8
Marketing Research	6	51	8.5	0.3
Advertising	6	66	11.00	0.40
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	21.000	2	10.500	21.000	0.0000	3.682
Within Groups	7.500	15	0.500
Total	28.500	17

a)

Sum of Squares, Treatment	21.00
Sum of Squares, Error	7.50
Mean Squares, Treatment	10.50
Mean Squares, Error	0.50

test stat=21.00

b) Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj)) = 0.87

population mean difference		critical value			result
µ1-µ2	0.50	0.87			means are not different
µ1-µ3	2.00	0.87			means are different
µ2-µ3	2.50	0.87			means are different