Question

Six samples of each of four types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data (µg/g). Wheat 5.1 4.6 6.0 6.0 6.7 5.7 Barley 6.5 8.0 6.1 7.5 5.8 5.5 Maize 5.7 4.6 6.4 4.9 6.0 5.3 Oats 8.2 6.0 7.9 7.1 5.4 7.1 Does this data suggest that at least two of the grains differ with respect to true average thiamin content? Use a level α = 0.05 test. State the appropriate hypotheses. H0: μ1 = μ2 = μ3 = μ4 Ha: at least two μi's are unequal H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 Ha: at least two μi's are equal H0: μ1 = μ2 = μ3 = μ4 Ha: all four μi's are unequal H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 Ha: all four μi's are equal Compute the test statistic value. (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 not significant evidence that at least two of the grains differ in average thiamin content. Reject H0. There is significant evidence that at least two of the grains differ in average thiamin content. Fail to reject H0. There is significant evidence that at least two of the grains differ in average thiamin content. Reject H0. There is not significant evidence that at least two of the grains differ in average thiamin content.

Answer 1

We can directly use here one way anova by Excel.

Step1) Enter data in Excel.

Step 2) Data >>Data analysis >> One way anova >>Select data >>click on label in first row >>ok

Null and alternative hypothesis

H0: μ1 = μ2 = μ3 = μ4 vs Ha: at least two μi's are unequal

test statistic F value is =3.74

P value is 0.028

0.010 < P-value < 0.050

Reject H0. There is significant evidence that at least two of the grains differ in average thiamin content.