Six samples of each of four types of cereal grain grown in a certain region were

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 (micro gram of thiamin / gram of grain).

Wheat: 5.2, 4.5, 6.0, 6.1, 6.7, 5.8

Barley: 6.5, 8.0, 6.1, 7.5, 5.9, 5.6

Maize: 5.8, 4.7, 6.4, 4.9, 6.0, 5.2

Oats: 8.3, 6.1, 7.8, 7.0, 5.5, 7.2

Conduct an ANOVA test at a significance level of 0.05 to determine if the average thiamin content is differnet in the four cereals. If you determined that the thiamin content of at least one group is different than other groups, conduct Tukey's test to determine the pair of group with significantly different thiamin content.

(Include the null hypothesis, alternative hypothesis and conclusion along with the analysis)

Explanation / Answer

Null HYpothesis : H0 : There is no significant difference in average thiamin content in given 4 cereals.

ALternative Hypothesis : Ha : There is significant difference in average thiamin content in given 4 cereals.

I am doing ANOVA testing for single factor in Excel and if atleast one group would be diffeent, i will peform tuley post ad hoc HSD test here.

Here it is the ANOVA table

Here we can see from F - ratio that F > Fcritical so that we can reject the null hypothesis and can say that there is significant difference in thaamin content in at least one group.

Now we will perform Tukey's test for all four means and we will se which two means are different.

First mean difference table is

So

Here Q for every pair of mean is Q = (xmax - xmin)/ sqrt [ MSW/n]

Here Xmax = sample mean which is higher among two

Xmin = sample mean which is lower among two

MSW = 0.7568 and n = number of treatments per group = 6

(i) For Wheat and Barley

Q = (xmax - xmin)/ sqrt [ MSW/n] = 0.8833/ sqrt [ 0.7568/6] = 2.4871

(ii) FOr Wheat and Maize

Q =  (xmax - xmin)/ sqrt [ MSW/n] = 0.2167/ sqrt [ 0.7568/6] = 0.6106

(iii) For Wheat and Oats

Q =  (xmax - xmin)/ sqrt [ MSW/n] = 1.2667/ sqrt [ 0.7568/6] = 3.5666

(iv) FOr Barley and maize

Q =  (xmax - xmin)/ sqrt [ MSW/n] = 1.1/ sqrt [ 0.7568/6] = 3.0972

(v) FOr Barley and Oats

Q =  (xmax - xmin)/ sqrt [ MSW/n] = 0.3833/ sqrt [ 0.7568/6] = 1.0792

(vi) FOr Maize and Oats

Q =  (xmax - xmin)/ sqrt [ MSW/n] = 1.4833/ sqrt [ 0.7568/6] = 4.1765

so from Q - table we can check critical Q - value for number of treatments =4 and alpha = 0.05 and dF = 20

Qcritical = 3.96

so we can see that thiamin content is different only in the pair of oats and maize are statistically different from each other.

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Wheat 6 34.3 5.716667 0.589667 Barley 6 39.6 6.6 0.904 Maize 6 33 5.5 0.448 Oats 6 41.9 6.983333 1.085667 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 8.983333 3 2.994444 3.956544 0.022934 3.098391 Within Groups 15.13667 20 0.756833 Total 24.12 23

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