Homework Answers
Answer
(A) Formula for the confidence interval between proportion difference is given as
![CI = (p_1-p_2) pm z*sqrt{[(p_1*(1-p_1))/n_1]+[(p_2*(1-p_2))/n_2]}](http://img.homeworklib.com/questions/b381d400-78d5-11ea-8c75-1fdf5d0c300b.png?x-oss-process=image/resize,w_560)
where p1 = 66% = 66/100= 0.66
p2= 41% = 41/100 = 0.41
sample size n1 = n2 = 1022/2 = 511
z score for 95% confidence interval is 1.96 (using z distribution table)
setting all the given values, we get
this gives us
on solving, we get
CI = (0.191,0.309)
converting it into percentages, we get
0.191*100 =19.1% and 0.309*100=30.9%
So, required confidence interval is 19.1% and 30.9%
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