Question

Alternate Homework 13 14 ANOVA Problems 2 A nuclear reactor experiences a radiation leak. The following data is measurements of radiation at three locations near the reactor. Perform a hypothesis test at 95% confidence to determine if the population mean radiation levels (for the three locations) could be equal. Then answer the questions below. Location A Location B Location Write your null and alternate hypothesis. Give the value of your test statistic. Show your p-value. Do you reject the null hypothesis? Answer "yes" or "no". Could the population means be equal? Answer "yes" or "no".

Answer 1

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

TSS =ΣΣΥ – CF = 164171 – 163515 = 656

SST - Στη - cr-

Where, r = 8

SSTR = (1/8) * $\sum$ (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 < $\alpha$ , 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