Question

In: Statistics and Probability

A researcher wanted to study the effect of type of diet on weight gain in mice....

A researcher wanted to study the effect of type of diet on weight gain in mice. The researcher was interested in 3 specific diets: standard mouse chow diet, “junk” food diet, and an organic diet.  A total of 54 mice were randomly assigned to receive one of the 3 types of diets (18 in each group). The outcome variable (response variable) was weight gain (in grams) over a 1 month period. The 3 data sets provided below are for the same study conducted by different researchers from completely different labs. For the sake of comparison, we will assume that they all followed the exact same procedures and that the only difference is the samples. For each study (A, B, and C), determine whether it would be appropriate to use a one-way ANOVA to analyze the data. If it is, conduct the analysis and any additional analysis that might be necessary. If it does not seem appropriate to use a one-way ANOVA, use an appropriate alternative, including any posthoc tests.  Remember, we are interested in determining whether the diet groups differ with respect to weight gain. Use a significance level of .05.

Study C

stand3<-c(10.27 , 9.57,  9.11, 9.38,  9.29, 12.45, 10.21, 10.02,  9.21,  9.00, 11.93,

         10.92,  9.59, 10.56,  9.53, 9.01,  9.24, 10.71)

junk3<-c(11.18, 10.74, 11.15, 12.50, 13.88, 11.73, 10.67, 11.47, 12.26, 12.33, 11.86, 14.10,

         12.33, 13.29, 10.73, 10.63, 10.97, 14.16)

organic3<-c(10.73, 10.19, 10.35, 11.12,  9.25, 10.02, 11.04,  9.95, 11.49, 9.65, 10.49,  9.83,

            9.58 ,11.85,  9.97, 10.44,  9.08 ,10.64)

a. Based on graphs and statistical tests, does it seem reasonable to assume normality?

b. Based on statistical tests, does it seem reasonable to assume equal variances?

c. Given the evidence regarding the ANOVA assumptions, which statistical result would be used to determine whether there is an effect for type of diet?

d. Assuming there is an overall effect, which groups differ, if any?

Solutions

Expert Solution

a.

From the above graph and p-value<0.05 so it does not seem reasonable to assume normality.

b.

Test for Equal Variances: weight gain versus type of diet

95% Bonferroni confidence intervals for standard deviations

type
of
diet N Lower StDev Upper
J 18 1.49792 2.11917 3.50651
O 18 0.52750 0.74628 1.23484
S 18 0.70570 0.99839 1.65199


Bartlett's Test (Normal Distribution)
Test statistic = 19.68, p-value = 0.000


Levene's Test (Any Continuous Distribution)
Test statistic = 2.30, p-value = 0.110

Since from a we conclude that the normality assumption does not hold then from Levene's Test since p-value>0.05,  it seems reasonable to assume equal variances.

c.

One-way ANOVA: weight gain versus type of diet

Source DF SS MS F P
type of diet 2 66.68 33.34 16.55 0.000
Error 51 102.76 2.01
Total 53 169.44

S = 1.419 R-Sq = 39.35% R-Sq(adj) = 36.97%


Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev -------+---------+---------+---------+--
J 18 12.499 2.119 (------*------)
O 18 10.315 0.746 (------*------)
S 18 10.000 0.998 (------*------)
-------+---------+---------+---------+--
10.0 11.0 12.0 13.0

Pooled StDev = 1.419

From ANOVA test we observe that p-value<0.05 so the effect of diets are significantly different.

d.

Grouping Information Using Tukey Method

type
of
diet N Mean Grouping
J 18 12.499 A
O 18 10.315 B
S 18 10.000 B

Means that do not share a letter are significantly different.


Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of type of diet

Individual confidence level = 98.05%


type of diet = J subtracted from:

type
of
diet Lower Center Upper ----+---------+---------+---------+-----
O -3.325 -2.184 -1.043 (------*-------)
S -3.640 -2.499 -1.358 (------*-------)
----+---------+---------+---------+-----
-3.0 -1.5 0.0 1.5


type of diet = O subtracted from:

type
of
diet Lower Center Upper ----+---------+---------+---------+-----
S -1.456 -0.315 0.826 (-------*-------)
----+---------+---------+---------+-----
-3.0 -1.5 0.0 1.5

From above Tukey's C.I., we observed that Junk3 is significantly different since C.I.s of (J-O) and (J-S) do not contain zero.


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