and a treatment is administered to each individual in the sample. After treatment sample mean is found to be 22.2 with SS=384. Calculate the t-statistic. Find the df. If we do 2-tailed test with a=0.05 what is the critial value of it?

SS =384

Variance = SS /(n-1) = 384/24 = 16

SD = 4

t = (22.2-20)/(4/5) = 2.2/0.8 = 22/8 =11/4 = 2.75

t value ( two tailed 0.05 ) = 2.064 for 24 degrees of freedom.

Since 2.75 > 2.064, we reject the null hypothesis that the mean is 20.

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