Iron-deficiency anemia is the most common form of malnutrition in developing cou

Question

Iron-deficiency anemia is the most common form of malnutrition in developing countries, affecting about 50% of children and women and 25% of men. Iron pots for cooking foods had traditionally been used in many of these countries, but they have been largely replaced by aluminum pots, which are cheaper and lighter. Some research has suggested that food cooked in iron pots will contain more iron than food cooked in other types of pots. One study designed to investigate this issue compared the iron content of some Ethiopian foods cooked in aluminum, clay, and iron pots. One of the foods was yesiga wet, beef cut into small pieces and prepared with several Ethiopian spices. The iron content of four samples of yesiga wet cooked in each of the three types of pots is given below. The units are milligrams of iron per 100 grams of cooked food. data349.dat

(a) Make a table giving the sample size, mean, and standard deviation for each type of pot. Is it reasonable to pool the variances? Note that with the small sample sizes in this experiment, we expect a large amount of variability in the sample standard deviations.

Type of pot n x^^_ s s_(x^^_) Aluminum Clay Iron

(b) Run the analysis of variance. Report the F statistic with its degrees of freedom and P-value. What do you conclude? (Round your test statistic to two decimal places. Round your P-value to three decimal places.)

F =

P =

Conclusion: There is statistically significant difference between the three treatment means at the = .05 level.

Explanation / Answer

Here the dataset is,

There are three types of pot aluminium, clay and iron.

Here we have to test the hypothesis that,

H0 : mu1 = mu2 = mu3 Vs H1 : Atleast one of the mean is differ than 0.

Assume alpha = level of significance = 0.05

Here we have to test three means so we use one way anova.

We can use excel for one way anova.

steps :

ENTER data into EXCEL sheet --> Data --> Data Analysis --> Anova : Single Factor --> ok --> Input Range : select all the data --> Grouped by : columns --> select Labels in first row --> Alpha : 0.05 --> Output Range : select one empty cell --> ok

Output :

Here test statistic follows (F) distribution.

Test statistic (F) = 31.16

P-value = 9.01E-05 = 0.00009

Critical value = 4.256

Here test statistic > critical value and also P-value < alpha.

Reject H0 at 5% level of significance.

Conclusion : Atleast one of the mean is differ than 0.

aluminium clay iron 1.77 2.27 5.27 2.36 1.28 5.17 1.96 2.48 4.06 2.14 2.68 4.22

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