In: Statistics and Probability
8) (1 point) Melanism in Moths. Melanism is the process by which animals produce melanin to darken body tissues and produce color variation. Researchers at Oxford University suspected increased levels of pollution in London may be influencing the evolution of Lepidoptera, a species of moth native to the area that took on one of two distinct camouflage schemes: "light" and "dark." They surveyed a sample of dark and light moths on birch trees and labelled each "conspicuous" if it was visible from 30 yards away or "inconspicuous" if it was not. The table below shows the number of moths of each color that were conspicuous and inconspicuous. Use this data to investigate the researchers' theory that increased levels of pollution might provide an advantage to "dark" moths over "light" ones.
Conspicuous | Inconspicuous | Total | |
Dark | 7 | 332 | 339 |
Light | 129 | 16 | 145 |
Total | 136 | 348 | 484 |
Conduct a test for the stated null and alternative hypotheses. Use αα = and round numeric answers to four decimal places.
H0H0: The variables use of color and detectability are independent.
HAHA: The variables use of color and detectability are not independent.
1. Calculate the test statistic for this hypothesis test. ? z t X^2 F =
2. Calculate the degrees of freedom for this hypothesis test.
3. Report the p-value for this hypothesis test out to four
decimal places.
p-value =
4. Based on the p-value, we have:
A. strong evidence
B. extremely strong evidence
C. little evidence
D. some evidence
E. very strong evidence
F. Reject H0H0
that the null model is not a good fit for our observed data.
We have to test for null hypothesis
against the alternative hypothesis
Under null hypothesis, expected frequencies are as follows.
Expected frequencyi (eij) | Detectability | Row total | ||
Conspicuous | Inconspicuous | |||
Use of color | Dark | 339*136/484 = 95.26 | 339*348/484 = 243.74 | 339 |
Light | 145*136/484 = 40.74 | 145*348/484 = 104.26 | 145 | |
Column total | 136.00 | 348.00 | 484 |
Our Chi-square test statistic is given by
Here,
Number of rows
Number of columns
Degrees of freedom
[Using R-code '1-pchisq(379.7007,1)']
We reject our null hypothesis if , level of significance.
We generally test for level of significance 0.10, 0.05, 0.01 or something like these.
So, we reject our null hypothesis.
1.
Test statistic is given by .
2.
Degrees of freedom is 1.
3.
P-value = 0
4.
Based on the p-value, we have (B) extremely strong evidence that the null model is not a good fit for our observed data.