Question 142549: 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 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
Answer by stanbon(75887)   (Show Source):
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
---------------------------------------
(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
-----------------------------------------
(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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