SOLUTION: Hello, please i need assistant in solving this question A survey found that 78% of the men questioned preferred computer-assisted instruction to lecture and 68% of the women prefe

Question 1201155: Hello, please i need assistant in solving this question
A survey found that 78% of the men questioned preferred computer-assisted instruction to lecture and 68% of the women preferred computer-assisted instruction to lecture. There were 100 randomly selected individuals in each sample. Find the 95% confidence interval for the difference of the two proportions.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
men p = .78
women p = .68
sample size for men and women the same at 100.
pooled proportion is (78 + 68) / (100 + 100) = .73
pooled standard error = sqrt(.73*.27*(1/100 + 1/100)) = .0627853486.
z-score = (.78 - .68) / .0627853486 = 1.59 rounded to 2 decimal places.
critical z-score is plus or minus 1.96.
test z-score in between those critical z-scores, so the resuls are not significant and you do not have any information to say that the men proportion and the women proportion are not the same.
this means that the observed difference is most likely due to variations in the mean of samples taken rather than any real difference.

i used a calculator and go the same results, so the results are accurate, given the assumption of pooled variance.
here's the results from the calculator at https://www.socscistatistics.com/tests/ztest/default2.aspx

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