document.write( "Question 1043809: Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the difference between two population proportions p 1p1 and p 2p2 at the given level of significance alphaα using the given sample statistics. Assume the sample statistics are from independent random samples.
\n" ); document.write( "Claim p^1= P^2,£=.001
\n" ); document.write( "Sample statistic x^1=98 n^1=143 and x^2=32 n^2=189
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #658999 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
I am assuming alpha =0.001
\n" ); document.write( "Ho:p1-p2=0
\n" ); document.write( "Ha:p1-p2 ne 0
\n" ); document.write( "alpha is 0.001
\n" ); document.write( "test stat is a z.
\n" ); document.write( "Critical value is |z|>3.29
\n" ); document.write( "p1=0.6853
\n" ); document.write( "p2=0.1693
\n" ); document.write( "pooled p=130/332=0.392
\n" ); document.write( "I can use a normal sampling distribution given the sample size and the proportions themselves. They were drawn randomly.
\n" ); document.write( "The SE is sqrt [0.392*0.608{(1/143)+(1/189)}]
\n" ); document.write( "That is 0.0541
\n" ); document.write( "z=(p1-p2)/SE=0.516/0.0541=9.54
\n" ); document.write( "This is highly significant at a level far less than 0.001. This is not surprising, given the reasonable size of the samples and the large differences in the point estimates of each proportion.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );