Question

The following data was reported on total Fe for four types of iron formation (1 = carbonate, 2 = silicate, 3 = magnetite, 4 = hematite).

1: 21.0 28.1 27.8 27.0 27.7 25.2 25.3 27.1 20.5 31.1

2: 26.3 24.0 26.2 20.2 23.8 34.0 17.1 26.8 23.7 24.8

3: 30.4 34.0 27.5 29.4 27.5 26.2 29.9 29.5 30.0 35.9

4: 37.2 44.2 34.1 30.3 31.7 33.1 34.1 32.9 36.3 25.7

Carry out an analysis of variance F test at significance level 0.01. State the appropriate hypotheses. H0: μ1 = μ2 = μ3 = μ4 Ha: all four μi's are unequal H0: μ1 = μ2 = μ3 = μ4 Ha: at least two μi's are unequal H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 Ha: all four μi's are equal H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 Ha: at least two μi's are equal

Summarize the results in an ANOVA table. (Round your answers two decimal places.) Source df Sum of Squares Mean Squares f Treatments Error Total

Give the test statistic. (Round your answer to two decimal places.) f =

What can be said about the P-value for the test? P-value > 0.100 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 P-value < 0.001

State the conclusion in the problem context. Fail to reject H0. There is sufficient evidence to conclude that the total Fe differs for at least two of the four formations.

Fail to reject H0. There is insufficient evidence to conclude that the total Fe differs for the four formations.

Reject H0. There is insufficient evidence to conclude that the total Fe differs for the four formations.

Reject H0. There is sufficient evidence to conclude that the total Fe differs for at least two of the four formations.

Answer 1

The statistical software output for this problem is:

Hence,

Hypotheses: Option B is correct.

ANOVA table:

Source	df	Sum of Squares	Mean Squares	f
Treatments	3	520.62	173.54	11.15
Error	36	560.52	15.57
Total	39	1081.14

Test statistic = 11.15

P-value < 0.001

Conclusion: Option D is correct.