Question 1161977: In a random sample of 80 workers from a factory in City A, it was found that 5% were unable to read, while in a random sample of 50 workers in city B, 8% were unable to read. Can it be concluded that there is a difference in the proportions of nonreaders in the two cities? Use alpha = .10. Find the 90% confidence interval for the difference of the two proportions. Does the confidence interval support your original conclusion?
a. State the hypotheses and identify the claim.
b. Find the critical value(s).
c. Compute the test value.
d. Make the decision.
e. Summarize the results.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The difference between two proportions test is
z=(p1-p2)/sqrt(p1(1-pi)/n1+p2(1-p2)/n2))
The half-interval of the CI is the z*SE, where z(0.95) =1.645 for a 90% CI
That is added to and subtracted from the difference of the means, which is -0.03 if we make it p(A)-p(B)
Ho:p(A)-p(B)=0
Ha: p(A)-p(B) does not equal 0
alpha=0.10 p{reject Ho|Ho true}
SE=sqrt (.05*.95/80+.08*.92/50)
=0.04545
the z value is -0.66, p-value=0.49 so Ho is not rejected
The half-interval is z*SE=1.645*0.0454=0.075
the interval is (-0.105, +0,.045). Because 0 is in the confidence interval, we are 90% confident that the true parameter of the difference is in the interval, and because 0 is in the interval, that is another way of saying the two proportions could be part of the same population and therefore not different,
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