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