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)\"\" \"About 
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\n" ); document.write( "You have the correct z test statistic. Nice work so far.\r
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\n" ); document.write( "\n" ); document.write( "Now you'll use the NORM.DIST function in excel. More info can be found here\r
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\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
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\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
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\n" ); document.write( "\n" ); document.write( "The p value is approximately 0.0274\r
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\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 \"mu+%3C=+3.55\" (which is where the claim is being made) and the alternative hypothesis is \"mu+%3E+3.55\" which runs counter or opposite to the claim made in the null. The alternative hypothesis tells us we have a right tailed test.
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