Question

A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observational study is conducted where the zoologist observes the gestation period of 50 randomly selected felines in each species. A one-way ANOVA test is run to test the hypothesis that the average gestation period is the same for each of the four species. A level of significance of 0.05 is used, and a P-value of 0.01 is calculated. Based on this result, the conclusion that can be drawn is that:
a. all of the average gestation periods are the same across all four species

b. at least some, but not all, of the gestation periods across all four species are the same

c. all of the average gestation periods across all four species are different

d. not all of the average gestation periods are the same across all four species

Answer 1

Answer:

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}= \mu_D$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C,D$

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have $p=4$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:  

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$  

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$  

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$  

And we have this property  

$SST=SS_{between}+SS_{within}$  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.