Asked by pleasehelp
Q: Hours of TV
Less than average: Greater than average:
Mean 22.3 20.8
S.D 25.6 25.5
n 903 390
Use hypothesis testing procedures to determine if there is a statistically significant difference in the average number of days of church attendance across the TV groups. Use alpha of 0.01 , 2-tail.
My work:
H0: m1-m2=0
H1:m1-m2=0
t critical = 2.58 (I think since it is based on alpha)
t= 22.3-20.8/sqrt rt 25.6^2/903+25.5^2/390 = 0.96
I don't know if I'm doing this right.
Less than average: Greater than average:
Mean 22.3 20.8
S.D 25.6 25.5
n 903 390
Use hypothesis testing procedures to determine if there is a statistically significant difference in the average number of days of church attendance across the TV groups. Use alpha of 0.01 , 2-tail.
My work:
H0: m1-m2=0
H1:m1-m2=0
t critical = 2.58 (I think since it is based on alpha)
t= 22.3-20.8/sqrt rt 25.6^2/903+25.5^2/390 = 0.96
I don't know if I'm doing this right.
Answers
Answered by
John
Make sure you set H1: not equal to zero.
I got t = .969
Fail to reject the null hypothesis.. There is no statistical difference.
Note: if alpha is .01 then for a two-tailed test, you need to split the alpha so you have .005 in each tail.
I got t = .969
Fail to reject the null hypothesis.. There is no statistical difference.
Note: if alpha is .01 then for a two-tailed test, you need to split the alpha so you have .005 in each tail.
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