a researcher uses anova to evaluate the mean difference between two treatments a

Question

a researcher uses anova to evaluate the mean difference between two treatments and obtains F(1,22)= 4.00. if the researcher had uses an independent measures t test to evaluate the data, what t value would be obtained A researcher uses an ANOVA to evaluate we mean difference between two treatments and obtains F(1, 22) = 4.00. If the researcher had used an independent-measures t test to evaluate the data, what t value would be obtained? 0 t(22) = 16.00 t(23) = 16.00 O t(22)- 2.00 O t(23) 2.00 2013

Explanation / Answer

let mean idfference=mu

to test H0: mu=0 vs H1:not H0

F statistic is n(y_bar -k)^2/s^2

and t statisctic is (y_bar-k)*sqrt(n)/s

so we can see that F=t^2

so t =2

in F statistic numerator (sqrt(n)(y_bar -k)/sigma)^2 follows chi square 1

denominator (n-1)s^2/sigma^2 follows chi square 22

so t

`t_22

so answer is

t_22=2

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