Question

Studies of mimicry in predator-prey systems often use plastic/clay replicas of prey species to determine if predators preferentially avoid the mimic and model, just the model, just the mimic, or even additional species not thought to be the model or the mimic. Replicas are placed for days to weeks in habitats thought to contain predators, models, and mimic, and researchers then count the number of “wounds” or missing replicas as an indication of whether models and mimics are equally avoided/attacked. One example of this model-mimic pair is the coral snake and scarlet snake; the non-venomous scarlet snake is thought to mimic the venomous coral snake. Franca et al. (2017) created six styles of replicas (N = 100 for each style) that were increasingly poor mimics of the coral snake (see figure, A is best mimic through F being the worst mimic), and they recorded the number of attacks on each style of replica by avian and mammalian predators. Use their data to determine if the better mimics are attacked more.

Number of attacks: A = 8; B = 12, C = 14, D = 18, E = 20, F = 30

For each question, be sure to clearly show all your work, including a table that you created to organize your data, your null and alternative hypotheses, calculation of your test statistic, your table value, and a final written conclusion and the statistical basis for that conclusion.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: All the mimics are attacked equally.

Alternative hypothesis: The better mimics are attacked more..

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 6 - 1
D.F = 5
(E_i) = n * p_i


X² = 17.294

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 5 degrees of freedom is more extreme than 17.294.

We use the Chi-Square Distribution Calculator to find P(X² > 17.294) = 0.004

Interpret results. Since the P-value (0.004) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that  better mimics are attacked more.