document.write( "Question 362868: A manufacturer of desks has made 200 per week for the last year with a standard deviation (σ) of 16. Recently, a manufacturing change was made and now production is at 205 desks per week for the 50 randomly sampled days. Determine, at the 0.01 level of significance (α) using a 1 tail test if the change made any difference. \n" ); document.write( "
Algebra.Com's Answer #258664 by stanbon(75887) ![]() You can put this solution on YOUR website! A manufacturer of desks has made 200 per week for the last year with a standard deviation (σ) of 16. \n" ); document.write( "Recently, a manufacturing change was made and now production is at 205 desks per week for the 50 randomly sampled days. \n" ); document.write( "Determine, at the 0.01 level of significance (α) using a 1 tail test if the change made any difference. \n" ); document.write( "---- \n" ); document.write( "Ho: u <= 200 \n" ); document.write( "Ha: u > 200 \n" ); document.write( "---- \n" ); document.write( "t(205) = (205-200)/[16/sqrt(50)] = 2.2097 \n" ); document.write( "--- \n" ); document.write( "p-value = P(t > 2.2097 when df = 49) = 0.0159 \n" ); document.write( "-------------------- \n" ); document.write( "Conclusion: Since the p-value is greater than \n" ); document.write( "1%, fail to reject Ho. \n" ); document.write( "-- \n" ); document.write( "The test does not show the change made a difference \n" ); document.write( "at the 1% level of significance. \n" ); document.write( "============== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "======================================== \n" ); document.write( " \n" ); document.write( " |