a). level of significance .05, df between = 4, df within = 61 F=2.52 p>.05______

Question

a). level of significance .05, dfbetween = 4, dfwithin = 61 F=2.52

p>.05__________, p<,05 ___________, p<.01___________

b) level of significance .01, dfbetween = 9, dfwithin= 33 F=3.01

p>.05___________, p<,05 ___________, p<.01___________

c) level of significance .05, dfbetween = 8, dfwithin = 45 F=2.16

p>.05___________, p<,05 __________, p<.01____________

d) level of significance .01, dfbetween = 3, dfwithin = 53 F=4.20

p>.05___________, p<,05 _______, p<.01____________

Explanation / Answer

Here what we have to caclculate it is not clear.But from these we can calculate the p-value of the test statistic

a)We have been given ,

numerator degree of freedom =4

Denominator degree of freedom =61

F-value =2.52       

Using calculator the p-value corresponding to this is is .0501

At the .05 level of significance it is significant            

b)

numerator degree of freedom =9

Denominator degree of freedom =33

F-value =3.01      

Using calculator the p-value corresponding to this is is .0098

At the .01 level of significance it is insignificant

c)

numerator degree of freedom =8

Denominator degree of freedom =45

F-value =2.16    

Using calculator the p-value corresponding to this is is .0492

At the .05 level of significance it is insignificant

d)

numerator degree of freedom =3

Denominator degree of freedom =53

F-value =4.2   

Using calculator the p-value corresponding to this is is .0097

At the .01 level of significance it is insignificant

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