Question

A questionnaire is developed to assess women's and men's attitudes toward using animals in research. One question asks whether animal research is wrong and is answered on a 7-point scale. Assume that in the population, the mean for women is 3, the mean for men is 2, and the standard deviation for both groups is 1.5. Assume the scores are normally distributed. If 12 women and 12 men are selected randomly, what is the probability that the mean of the women will be more than 1.5 points higher than the mean of the men?

Answer 1

here let A =Xbar -Ybar ; wher Xbar and Y bar are mean for 12 women and men

expected mean of A =E(Xbar)-E(Ybar)=3-2 =1

and std deviation of A =sqrt(1.5²/12+1.5²/12) =0.612

for normal distribution z score =(X-μ)/σ
here mean=       μ=	1
std deviation   =σ=	2.121

  probability that the mean of the women will be more than 1.5 points higher than the mean of the men:

probability =

P(X>1.5)

=

P(Z>0.82)=

1-P(Z<0.24)=

1-0.7939=

0.2061