document.write( "Question 1083054: A sample of 44 observations is selected from one population with a population standard deviation of 4.9. The sample mean is 102.0. A sample of 45 observations is selected from a second population with a population standard deviation of 3.8. The sample mean is 100.6. Conduct the following test of hypothesis using the 0.02 significance level. \r
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document.write( "I need help with # b the intervals and e the p-value\r
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document.write( "a. This is a -tailed test.
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document.write( "b.
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document.write( "State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)
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document.write( " The decision rule is to reject H0 if z is outside the interval (),().
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document.write( "c. Compute the value of the test statistic. (Round your answer to 2 decimal places.)
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document.write( " Value of the test statistic 1.51
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document.write( "d. What is your decision regarding H0?
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document.write( " Do Not reject
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document.write( "e. What is the p-value? (Round your answer to 4 decimal places.)
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document.write( " p-value \r
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Algebra.Com's Answer #697144 by Boreal(15235) You can put this solution on YOUR website! It is a 2-tailed test. \n" ); document.write( "Reject Ho if |z|>2.33. The absolute value of z is a nice shortcut to say if z is <-2.33 or >+2.33 one will reject Ho. \n" ); document.write( "z=1.51 \n" ); document.write( "p-value is 0.1326. \n" ); document.write( "With a calculator, that comes up. Without, one looks up the z-value for 1.51 in a table and that is one sided probability. For two sided probability, multiply the z-value by 2. \n" ); document.write( " |