Question

1.        A researcher wants to determine the effects that smoking has on resting heart rate. She randomly selects 7 individuals from 3 categories: nonsmokers, light smokers (fewer than 10 cigarettes per day), and heavy smokers (10 or more cigarettes per day) and obtains the following heart rate data (beats per minute):

nonsmokers	light smokers	heavy smokers
70	67	79
58	75	80
51	65	77
56	78	77
53	62	86
53	70	68
65	73	83

Use your calculator to complete the one-way ANOVA table. Round to 4 decimal places when necessary. You can copy the table and paste it into your document.

source of variation	sum of squares (SS)	df	mean square (MS)	F
factor (between)
error (within)
total

2.        An environmentalist wanted to determine if the mean acidity of rain differed among Alaska, Florida, and Texas. He randomly selected 6 rain dates at each of the 3 locations and obtained the following data (pH of rain):

Alaska	Florida	Texas
5.41	4.87	5.46
5.39	5.18	5.89
4.90	4.52	5.57
5.14	5.12	5.15
4.80	4.89	5.45
5.24	5.06	5.30

Conduct a full hypothesis test. Be sure to:

• check conditions

1. state hypotheses
2. find the test statistic and df (both)
3. state significance level
4. find p-value and draw/shade the graph
5. make a decision
6. write a conclusion

Answer 1

Question 1:

Hypothesis:

Ho: 1 = 2 = 3

Ha: At least one of the mean is different

CALCULATION:

Number of Treatment, t = 3 n = 21

Degrees of Freedom (df):

df_Treatment = t-1 = 3-1 = 2

df_Total = n-1 = 21-1 = 20

df_Error = df_Total - df_Treatment = 20-2 = 18

T1 (sum of nonsmokers) = 406, T2(Sum of lightsmokers) = 490, T3(Sum of Heavysmokers) = 550

G = Grand Total = 1446

CF = Correction Factor = G2/N = 14462 / 21 = 99567.43

= (70)2 + (67)2 + ......................+ (83)2 - 99567.43

= 2180.5714

TSS = 489820.667

SSTR = (1 / 7) * [(406)2 + (490)2 + (550)2] - 99567.43

SSTR = 1494.8571

SSE = TSS - SSTR

= 2180.5714 - 1494.8571

SSE = 685.7143

MSSTR = SSTR/t-1 = 1494.8571 / 3-1 = 747.4286

MSSE = SSE / n-t = 685.7143 / 21-3 = 38.0952

F = MSSTR / MSSE = 747.4286 / 38.0952 = 19.6200

Ftabulate = F,(t-1,n-t) = F0.05,(2,18) = 3.5546                            ..............Using F table

ANOVA TABLE:

Conclusion:

Test statistic > Critical value, i.e. 19.6200 > 3.5546, That is Reject Ho at 5% level of significance.

Therefore, At least one of the mean is different.