the majority of the kids have scores that form a normal distribution with a mean of 50 and SD of 15. The sample has average of 53.8. Conduct 2-tail test with a=0.05.
A. Null hypothesis?B. Standard error?
C. Calculate z
D. Cohen's d?
xbar = 53.8
null hypothesis H0 : mu =50
Standard error =15/sqrt(100)=15/10=1.5
z= (xbar-mu)/standard error = (53.8-50)/1.5 = 3.8/1.5 = 7.6/3 =2.533
|z|>1.96, we reject the null hypothesis.
The mean of test scores is not 50
Cohen's d is for two populations.
d =(mean difference)/SD
d=(53.8-50)/15 =3.8/15 =0.2533
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