document.write( "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? \r
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document.write( "a. State the hypotheses and identify the claim.
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document.write( "b. Find the critical value(s).
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document.write( "c. Compute the test value.
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document.write( "d. Make the decision.
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document.write( "e. Summarize the results. \n" );
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Algebra.Com's Answer #785730 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! The difference between two proportions test is \n" ); document.write( "z=(p1-p2)/sqrt(p1(1-pi)/n1+p2(1-p2)/n2))\r \n" ); document.write( "\n" ); document.write( "The half-interval of the CI is the z*SE, where z(0.95) =1.645 for a 90% CI \n" ); document.write( "That is added to and subtracted from the difference of the means, which is -0.03 if we make it p(A)-p(B)\r \n" ); document.write( "\n" ); document.write( "Ho:p(A)-p(B)=0 \n" ); document.write( "Ha: p(A)-p(B) does not equal 0 \n" ); document.write( "alpha=0.10 p{reject Ho|Ho true} \n" ); document.write( "SE=sqrt (.05*.95/80+.08*.92/50) \n" ); document.write( "=0.04545 \n" ); document.write( "the z value is -0.66, p-value=0.49 so Ho is not rejected\r \n" ); document.write( "\n" ); document.write( "The half-interval is z*SE=1.645*0.0454=0.075 \n" ); document.write( "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,\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |