document.write( "Question 1185090: A sales clerk in a departmental store claims that 45% of the shoppers entering the store leave without making a purchase. A random sample of 60 shoppers showed that 40 of them left without buying anything. Can we accept the clerk's claim at α=5% level of significance?\r
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document.write( "The test value is: \n" );
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Algebra.Com's Answer #815852 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! p hat is 0.667 because 40/60 shoppers left without buying \n" ); document.write( "Ho: p is =0.45 \n" ); document.write( "Ha: p is NE 0.45 \n" ); document.write( "alpha is 0.05 p{reject Ho|Ho true} \n" ); document.write( "test stat is a z, 2-way 1-proportion test \n" ); document.write( "reject Ho for |z|> 1.96 \n" ); document.write( "z=(p hat-p)/sqrt (p(1-p)/n) \n" ); document.write( "=-0.667/sqrt (0.45*0.55/60) \n" ); document.write( "=-0.217/0.0642 \n" ); document.write( "=-3.37 \n" ); document.write( "Reject Ho; the true proportion is not 0.45. \n" ); document.write( " |