Question

1. Researchers conducted a study looking at the effects of cooking container type (aluminum, clay, iron) and food type (meat, legumes, vegetables) on the amount of iron present in the food after cooking. They cooked the same weight of each of the 3 food types for a similar amount of time in each of the cooking container types and replicated each. See columns “type”, “food” and “iron” on the ANOVA_2Way_data file. Does iron content vary by food type and container type after the food has been cooked?

1. What are the null and alternate hypotheses?

2. Check the assumption of normality of the response with a histogram

3. Run the 2-Way ANOVA (including the interaction) and interpret the Sums of Squares, Mean Squares, F statistics, and P values

4. Make an appropriate interaction plot

5. Interpret the results and determine if you accept or reject the null hypotheses

   type	food	iron
   Aluminum	meat	1.77
   Aluminum	meat	2.36
   Aluminum	meat	1.96
   Aluminum	meat	2.14
   Clay	meat	2.27
   Clay	meat	1.28
   Clay	meat	2.48
   Clay	meat	2.68
   Iron	meat	5.27
   Iron	meat	5.17
   Iron	meat	4.06
   Iron	meat	4.22
   Aluminum	legumes	2.4
   Aluminum	legumes	2.17
   Aluminum	legumes	2.41
   Aluminum	legumes	2.34
   Clay	legumes	2.41
   Clay	legumes	2.43
   Clay	legumes	2.57
   Clay	legumes	2.48
   Iron	legumes	3.69
   Iron	legumes	3.43
   Iron	legumes	3.84
   Iron	legumes	3.72
   Aluminum	vegetables	1.03
   Aluminum	vegetables	1.53
   Aluminum	vegetables	1.07
   Aluminum	vegetables	1.3
   Clay	vegetables	1.55
   Clay	vegetables	0.79
   Clay	vegetables	1.68
   Clay	vegetables	1.82
   Iron	vegetables	2.45
   Iron	vegetables	2.99
   Iron	vegetables	2.8
   Iron	vegetables	2.92

Answer 1

a)

The null hypotheses:

Ho_1: The population means of Iron conent for food types are equal.

Ho_2: The population means of Iron content for container types are equal.

Ho_3: There is no interaction between the two factors - food & container types.

The alternative hypotheses:

Ha_1: The population means of Iron conent for food types are NOT equal.

Ha_2: The population means of Iron content for container types are NOT equal.

Ha_3: There is interaction between the two factors - food & container types.

b) Assumption of normality of the reponse with histogram

Explanation from the following three graphs:

i) Original data is positive skewed.

ii) Residuals of ANOVA model are negqtively skewed

iii) Normal Q-Q plot indicates slight deviation from normality

c)

2-way ANOVA output from R-programming:

i) Interpretation of Sum of Squares (SS):

A high sum of squares indicates that most of the values are farther away from the mean, and hence, there is large variability in the data. A low sum of squares refers to low variability in the set of observations.

Here, food SS is less than type SS , implies type data variation is higher and that of food.

ii) Interpretation of Mean Sum of Squares (MSS):

It is an average variation of independent variables.

Here, on an average type variability is the highest.

iii) Interpretation of F-statistic:

An F statistic is a value in an ANOVA table to find out if the means between two populations are significantly different.

Higher the F statistic value implies means are significantly different.

Here, for food( 34.456) and type(92.263) F statistic is high which implies means are significantly different.

iv) Interpretation of p-values:

For taking decision to Accept or Reject Null hypothesis p-value is being used.

Rule : If p-value < Alpha or Level of significance == > Then Reject Ho.

Decision :
food : p-value = 0.0000 < 0.05 ( default alpha ) ==> Reject Ho_1:

Conclusion: The population means of Iron conent for food types are NOT equal.

type : p-value = 0.0000 < 0.05 ( default alpha ) ==> Reject Ho_2:

Conclusion: The population means of Iron content for container types are NOT equal.

Interaction : p-value = 0.00425 < 0.05 ( default alpha ) ==> Reject Ho_3

Conclusion: There is interaction between the two factors - food & container types.

d) Interaction plot:

e) Interpret Results:

food : p-value = 0.0000 < 0.05 ( default alpha ) ==> Reject Ho_1:

Conclusion: The population means of Iron conent for food types are NOT equal.

type : p-value = 0.0000 < 0.05 ( default alpha ) ==> Reject Ho_2:

Conclusion: The population means of Iron content for container types are NOT equal.

Interaction : p-value = 0.00425 < 0.05 ( default alpha ) ==> Reject Ho_3

Conclusion: There is interaction between the two factors - food & container types.

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