document.write( "Question 142549: Exercise using Excel
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\n" ); document.write( "The School Committee members of a midsized New England city agreed that a strict
\n" ); document.write( "discipline code has caused an increase in the number of student suspensions. The
\n" ); document.write( "number of suspensions for September 1992 - February 1993 for a sample of the schools is provided below. {Data}
\n" ); document.write( "The average number of suspensions for the previous year mean{X bar} = 130.5 and the Standard deviation {STDEV}
\n" ); document.write( "was 158.2
\n" ); document.write( "(a) Set up the null and alternative hypothesis to test if the averave number of suspensions
\n" ); document.write( "has changed.
\n" ); document.write( "(b) Test your hypothesis with alpha = 0.05.
\n" ); document.write( "(c) Find the p value.
\n" ); document.write( "(d) Display the data in an Excel Chart to see if it is reasonable to assume that the underlaying population
\n" ); document.write( "distribution is normal.
\n" ); document.write( "(e) Based on the p value, what can you conclude about the average number of suspensions?
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\n" ); document.write( "Data:
\n" ); document.write( " Suspensions
\n" ); document.write( "Central 245 a) - c) mu = 195
\n" ); document.write( "MCDI 1 Step 1 H(0) mean equal 130.5
\n" ); document.write( "Chestnut 65 H(A) mean not equal 130.5
\n" ); document.write( "Duggan 133 Sigma is given as = 158.2
\n" ); document.write( "Kennedy 97 Step 2 Critical Values (Z) = +/-1.96 because alpha is 0.05
\n" ); document.write( "Forest Park 149 (n)^0.5 Z P(Z<-1.352) p = 2*P(Z<-1.352) = 2*0.0885
\n" ); document.write( "Putnam 1024 Step 3 Test Statistic Z 3.317 Calculate Z here in cell H25.
\n" ); document.write( "Kiley 56 p value = 0.177
\n" ); document.write( "Central Academy 254 Why is Z minus?
\n" ); document.write( "Commerce 114 Step 4/5 d) See Below.
\n" ); document.write( "Bridge 7 e) What do you think?
\n" ); document.write( "STDEV= 287.15
\n" ); document.write( " mu = 195.00
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\n" ); document.write( " ReOrder
\n" ); document.write( "MCDI 1
\n" ); document.write( "Bridge 7
\n" ); document.write( "Kiley 56
\n" ); document.write( "Chestnut 65
\n" ); document.write( "Kennedy 97
\n" ); document.write( "Commerce 114
\n" ); document.write( "Duggan 133
\n" ); document.write( "Forest Park 149
\n" ); document.write( "Central 245
\n" ); document.write( "Central Academy 254
\n" ); document.write( "Putnam 1024
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Algebra.Com's Answer #103767 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The School Committee members of a midsized New England city agreed that a strict
\n" ); document.write( "discipline code has caused an increase in the number of student suspensions. The number of suspensions for September 1992 - February 1993 for a sample of the schools is provided below. {Data}
\n" ); document.write( "The average number of suspensions for the previous year mean{X bar} = 130.5 and the Standard deviation {STDEV} was 158.2
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\n" ); document.write( "(a) Set up the null and alternative hypothesis to test if the average number of suspensions has changed.
\n" ); document.write( "Ho: u = 130.5
\n" ); document.write( "H1: u is not equal to 130.5\r
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\n" ); document.write( "(b) Test your hypothesis with alpha = 0.05.
\n" ); document.write( "Critical values for 2-tail test with alpha = 5%: z = +-1.96
\n" ); document.write( "Test statistic: z(195) = (195-130.5)/[158.2/sqrt(10)] = 1.2893...
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\n" ); document.write( "(c) Find the p value.
\n" ); document.write( "p-value = 2*P(1.2893 < z < 10) = 0.1973
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\n" ); document.write( "(d) Display the data in an Excel Chart to see if it is reasonable to assume that the underlaying population distribution is normal.
\n" ); document.write( "I'll leave that to you.
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\n" ); document.write( "(e) Based on the p value, what can you conclude about the average number of suspensions?
\n" ); document.write( "The p value is greater than 5% so fail to reject Ho.
\n" ); document.write( "The average number of suspensions has not changed.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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