Answers
A)
Null Hypothesis: Population mean radiation level (for the three locations) could be equal
Alternative Hypothesis: At least one of the population mean radiation level (for the three locations) could be unequal
B)
Number of Treatment (t) = 3 n = 24
T1 (Sum of Location A) = 636, T2(Sum of Location B) = 702, T3(Sum of Location C) = 643
G = grand total = 1981
CF = Correction Factor = G2/N = 19812 / 24 = 163515
Where, r = 8
SSTR = (1/8) * 
(6362 + 7022 + 6432) - 163515
SSTR = 328.6
SSE = TSS - SSTR
SSE = 656 - 328.6
SSE = 327.4
MSSTR = SSTR/t-1 = 328.6 / 3-1 = 164.3
MSSE = SSE / n-t = 327.4 /24-3 = 15.6
F = MSSTR / MSSE = 164.3 / 15.6 = 10.54
P-value: 0.0007 ............................From F table
P-value < 
, i.e. 0.0007 < 0.05, That is Reject Ho at 5% level of significance.
Therefore, At least one of the population mean radiation level (for the three locations) could be unequal
| ANOVA | |||||
| Source of Variation | SS | df | MS | F | P-value |
| Between Groups | 328.6 | t-1=3-1 =2 | 164.3 | 10.54 | 0.0007 |
| Within Groups | 327.4 | n-t=24-3 =21 | 15.6 | ||
| Total | 655.9583 | n-1=24-1 =23 |
Test statistic: F = 10.54
P-value: 0.0007 ............................From F table
ANSWER: A. Yes
Yes, Reject the Null Hypothesis.
ANSWER: B. No
No, population means Not equal
TSS =ΣΣΥ – CF = 164171 – 163515 = 656
SST - Στη - cr-
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