The numbers of false fire alarms were counted each month at a number of sites. The results are given in the following table. Number of Alarms 37 42 26 36 36 31 35 35 44 25 26 42 Test the hypothesis that false alarms are equally likely to occur in any month. Use 10% level of significanc Procedure: Select an answer Assumptions: (select everything that applies) Month January February March April May June July August September October November December Independent samples At most 20% of the expected frequencies are less than 5 Equal population standard deviations Normal populations All expected frequencies are 1 or greater Simple random sample

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The numbers of false fire alarms were counted each month at a number of sites. The results are given in
the following table.
Number of Alarms
37
42
26
36
36
31
35
35
44
25
26
42
Test the hypothesis that false alarms are equally likely to occur in any month. Use 10% level of significance.
Procedure: Select an answer
Assumptions: (select everything that applies)
Month
January
February
March
April
May
June
July
August
September
October
November
December
Independent samples
At most 20% of the expected frequencies are less than 5
Equal population standard deviations
Normal populations
All expected frequencies are 1 or greater
Simple random sample
Transcribed Image Text:The numbers of false fire alarms were counted each month at a number of sites. The results are given in the following table. Number of Alarms 37 42 26 36 36 31 35 35 44 25 26 42 Test the hypothesis that false alarms are equally likely to occur in any month. Use 10% level of significance. Procedure: Select an answer Assumptions: (select everything that applies) Month January February March April May June July August September October November December Independent samples At most 20% of the expected frequencies are less than 5 Equal population standard deviations Normal populations All expected frequencies are 1 or greater Simple random sample
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What would the left and right critical values be? Also, it says your p-value answer is incorrect. Not sure if it is your calculations or just the program my professor is using.

June
July
August
35
Septemeber 44
October
25
26
42
November
December
Total:
415
31
35
34.5833
34.5833
34.5833
34.5833
34.5833
34.5833
34.5833
-3.5833
0.4167
0.4167
9.4167
-9.5833
-8.5833
7.4167
p 4. Testing Procedure: (Round the answers to 3 decimal places)
CVA
Provide the critical value(s) for the Rejection Region:
left CV is -1.3634x
nt: For a LT test, the rejection region is to the left of 21-a
o the left critical value is 21-a and the right critical value
oesn't exist (DNE). For a TT test, the rejection region is to
he left of 1- and to the right of , so the left critical
ue is ₁-# and the right critical value is . For a RT test,
he rejection region is to the right of a, so the left critical
value doesn't exist (DNE) and the right critical value is zo.
and right CV is 1.3634 X
=
12.84
0.1736
0.1736
88.6742
91.8396
73.673
55.0074
(O - E)²
Σ E
df=
0.3713
0.005
0.005
2.5641
2.6556
2.1303
1.5906
13.3277
11
OF
09
PVA
Compute the P-value of the test
statistic:
P-value is 0.2724 Xx
Hint: For a LT test, the p-value is the
area under the probability density curve
to the left of the test statistic, i.e.
P(X<zo). For a RT test, the p-value
is the area under the probability density
curve to the right of the test statistic,
Transcribed Image Text:June July August 35 Septemeber 44 October 25 26 42 November December Total: 415 31 35 34.5833 34.5833 34.5833 34.5833 34.5833 34.5833 34.5833 -3.5833 0.4167 0.4167 9.4167 -9.5833 -8.5833 7.4167 p 4. Testing Procedure: (Round the answers to 3 decimal places) CVA Provide the critical value(s) for the Rejection Region: left CV is -1.3634x nt: For a LT test, the rejection region is to the left of 21-a o the left critical value is 21-a and the right critical value oesn't exist (DNE). For a TT test, the rejection region is to he left of 1- and to the right of , so the left critical ue is ₁-# and the right critical value is . For a RT test, he rejection region is to the right of a, so the left critical value doesn't exist (DNE) and the right critical value is zo. and right CV is 1.3634 X = 12.84 0.1736 0.1736 88.6742 91.8396 73.673 55.0074 (O - E)² Σ E df= 0.3713 0.005 0.005 2.5641 2.6556 2.1303 1.5906 13.3277 11 OF 09 PVA Compute the P-value of the test statistic: P-value is 0.2724 Xx Hint: For a LT test, the p-value is the area under the probability density curve to the left of the test statistic, i.e. P(X<zo). For a RT test, the p-value is the area under the probability density curve to the right of the test statistic,
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