document.write( "Question 1185131: A dietician read in a survey that 68.9% of adults in the U.S. do not eat breakfast at least 2 days a week. She believes that a larger proportion skip breakfast 2 days a week. To verify her claim, she selects a random sample of 71 adults and asks them how many days a week they skip breakfast. 49 of them report that they skip breakfast at least 2 days a week. Test her claim at
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document.write( "α= 0.05.\r
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document.write( "The test statistic is:______\r
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document.write( "The p-value is: _____ \n" );
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Algebra.Com's Answer #815923 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Assuming randomness and independence, \n" ); document.write( "p hat=49/71=0.690 \n" ); document.write( "Ho: p >0.690 \n" ); document.write( "Ha: p < = 0.690 \n" ); document.write( "alpha=0.05 p{reject Ho|Ho true} \n" ); document.write( "test stat is a z \n" ); document.write( "reject for z> 1.645 \n" ); document.write( "z=(0.690-0.689)/sqrt (0.689*0.311/71) \n" ); document.write( "=0.001/0.0549=+0.018 \n" ); document.write( "fail to reject Ho, insufficient evidence to support the claim. \n" ); document.write( "p-value is 0.493\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |