Question

An Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity.

Source of Variation	SS			df				MS				F				P-value				F crit
Plant location		3.0075				3				1.0025				2.561				0.1199				3.862
Noise level		8.4075				3				2.8025				7.160				0.0093				3.863
Error		3.5225				9				0.3914
Total		14.9375

1. State the 2 null hypothesis for the F statistics listed above
2. At the 1% significance level, do you reject or fail to reject your null hypothesis?
3. At the 15% significance level, do you reject or fail to reject your null hypothesis?

Answer 1

a) for plant location

Null Hypothesis : H01 : There is no significant diffrence among the plant locations

Alternative hypothesis :Ha1 : There is significant diffrence among the plant locations

for noise level

Null Hypothesis : H02 : There is no significant diffrence among the noise levels

Alternative hypothesis :Ha2 : There is significant diffrence among the noise levels

Decision criteria : Reject null hypothesis if F > F critical

b) at 1% level of significance that is =0.01 F crtical value for (3,9) degrees of freedom = 6.9919

for plant location

F= 2.561

since F < F critical value =6.9919

we donot reject Null Hypothesis H01 and conclude that There is no significant diffrence among the plant locations

for noise level

F = 7.16

since F >  F critical value =6.9919

we reject Null Hypothesis H02  and conclude that There is significant diffrence among the noise levels

c)

at 15% level of significance that is =0.15 F crtical value for (3,9) degrees of freedom = 2.2644

for plant location

F= 2.561

since F > F critical value =2.2644

we reject Null Hypothesis H01 and conclude that There is significant diffrence among the plant locations

for noise level

F = 7.16

since F > F critical value =2.2644

we reject Null Hypothesis H02  and conclude that There is significant diffrence among the noise levels