document.write( "Question 1153955: GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.55 mg of mercury. A sample of 40 bulbs shows a mean of 3.62 mg of mercury. The standard deviation is 0.23mg and I found the test statistic which is 1.92 but I need help Finding the p-value. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) I don't understand how to do it in excel. \n" ); document.write( "
Algebra.Com's Answer #776400 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "You have the correct z test statistic. Nice work so far.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now you'll use the NORM.DIST function in excel. More info can be found here\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The general template of the function is \n" ); document.write( "=NORM.DIST(x, mean, standard_dev, cumulative) \n" ); document.write( "In your case, you will type in \n" ); document.write( "=NORM.DIST(1.92, 0, 1, 1) \n" ); document.write( "The x is the z score you got \n" ); document.write( "the mean and standard deviation are 0 and 1 respectively \n" ); document.write( "Cumulative is set to 1 to indicate we want the area under the normal curve. You can use \"TRUE\" in place of \"1\" for the cumulative part. So your command could look like this: =NORM.DIST(1.92, 0, 1, TRUE)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The result of that excel calculation is approximately 0.97257105 \n" ); document.write( "This is the area under the curve to the left of z = 1.92 \n" ); document.write( "Subtract this from 1 (see note below) to get \n" ); document.write( "1-0.97257105 = 0.02742895 \n" ); document.write( "which is also approximate \n" ); document.write( "This is the approximate area under the standard normal curve that is to the right of z = 1.92\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The p value is approximately 0.0274\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Note: the reason why I knew to subtract from 1 is because of the hypothesis. The null hypothesis is \n" ); document.write( " \n" ); document.write( " |