document.write( "Question 968366: Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).\r
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document.write( "The test statistic in a left-tailed test is z = -2.05 \n" );
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Algebra.Com's Answer #591751 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). \n" ); document.write( "The test statistic in a left-tailed test is z = -2.05 \n" ); document.write( "------ \n" ); document.write( "p-value = p(z < -2.05) = normalcdf(-100,-2.05) = 0.0218 \n" ); document.write( "---- \n" ); document.write( "Since the p-value is less than 5%, reject Ho. \n" ); document.write( "----------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "------------- \n" ); document.write( " |