document.write( "Question 1085207: A RANDOM SAMPLE OF 400 MEMBERS OF LABOR FORCE IN A FIVE STATE REGION SHOWED THAT 32 WERE UNEMPLOYED. CONSTRUCT THE 95% CONFIDENCE INTERVAL FOR THE PROPORTION UNEMPLOYED IN THE REGION. \n" ); document.write( "
Algebra.Com's Answer #699275 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! The interval width is z(0.975)*sqrt (0.08*0.92)/400; sqrt term=0.01356. \n" ); document.write( "The 0.975 is the z-value with 0.025 on each end or 95%. The sqrt is of p*(1-p)/n, and p is 32/400=0.08, the point estimate. \n" ); document.write( "0.0266 is the interval width. \n" ); document.write( "That is above and below the point estimate \n" ); document.write( "(0.0534, 0.1066) is 95% CI\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |