document.write( "Question 1112728: Help me to Solve this A Confidence Interval for a Population Proportion Problem ! Need help for question B\r
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document.write( "The Fox TV network is considering replacing one of its prime-time crime investigation shows with a new family-oriented comedy show. Before a final decision is made, network executives commission a sample of 350 viewers. After viewing the shows, 250 indicated they would watch the new show and suggested it replace the crime investigation show.\r
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document.write( "A) Estimate the value of the population proportion. (Round your answers to 3 decimal places.) (i already found the answer i think its 0.714)\r
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document.write( "B)Develop a 90% confidence interval for the population proportion. (Use z Distribution Table.) (Round your answers to 3 decimal places.)
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document.write( "Confidence interval ........ and ........, \n" );
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Algebra.Com's Answer #727797 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! B) standard error(SE) = square root(0.714 * (1-0.714)/350) = 0.024 \n" ); document.write( ": \n" ); document.write( "alpha(a) = 1 - (90/100) = 0.10 \n" ); document.write( ": \n" ); document.write( "critical probability(p*) = 1 - (a/2) = 0.95 \n" ); document.write( ": \n" ); document.write( "critical value(CV) is the z-score corresponding to p*, CV = 1.645 \n" ); document.write( ": \n" ); document.write( "margin of error(ME) = CV * SE = 1.645 * 0.024 = 0.0395 approximately 0.040 \n" ); document.write( ": \n" ); document.write( "90% confidence interval is 0.714 + or - 0.040 or (0.674, 0.754) \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |