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
\n" ); document.write( "α= 0.05.\r
\n" ); document.write( "\n" ); document.write( "The test statistic is:______\r
\n" ); document.write( "\n" ); document.write( "The p-value is: _____
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Algebra.Com's Answer #815923 by Boreal(15235)\"\" \"About 
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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
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