document.write( "Question 599452: After a losing season, there is a great uproar to fire the head football coach. In a random sample of 250 college alumni, 105 favor keeping the coach. Test at the .01 level of significance whether the proportion of alumni who support the coach is less than 50 percent.\r
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document.write( "State the null hypothesis and the alternate hypothesis\r
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document.write( "State the decision rule for .01 significance level.\r
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document.write( "Compute the value of the test statistic \n" );
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Algebra.Com's Answer #378996 by stanbon(75887) ![]() You can put this solution on YOUR website! After a losing season, there is a great uproar to fire the head football coach. In a random sample of 250 college alumni, 105 favor keeping the coach. Test at the .01 level of significance whether the proportion of alumni who support the coach is less than 50 percent. \n" ); document.write( "State the null hypothesis and the alternate hypothesis \n" ); document.write( "State the decision rule for .01 significance level. \n" ); document.write( "Compute the value of the test statistic \n" ); document.write( "------------------------------ \n" ); document.write( "Ho: p >= 1/2 \n" ); document.write( "Ha: p < 0.5 (claim) \n" ); document.write( "---- \n" ); document.write( "p-hat = 105/250 = 0.42 \n" ); document.write( "---- \n" ); document.write( "Test statistic: z(0.42) = (0.42-0.5)/sqrt[0.5*0.5/250] = -2.5298 \n" ); document.write( "------ \n" ); document.write( "p-value = P(z < -2.5298) = 0.0057 \n" ); document.write( "---- \n" ); document.write( "Since the p-value is less than 1%, reject Ho. \n" ); document.write( "Conclusion: The test results support the claim. \n" ); document.write( "==================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "==================== \n" ); document.write( " |