YOU WERE LOOKING FOR: Using The Chi Square Test In Ecology Answers
A population is at Hardy-Weinberg equilibrium for a gene if five conditions are met; random mating, no mutation, no gene flow, no natural selection, and large population size. Under these circumstances, the allele frequencies for a population are...
For the first example, we'll use a simple data set not generated by a simulation. In this case, there are 50 total individuals in the population; 10 are red, 10 are purple, and 30 are blue. These are the observed values for the chi-squared analysis....
See this blog post for an explanation of the simulation. In this run, there is no selection against any of the phenotypes and there is no mutation chance. The population size is set to Simulation results At the end of the simulation run, the red allele frequency is 0. The frequencies for the phenotypes are 0. Simulation frequencies We can use the phenotype frequencies and the total population number to find the number of individuals for each phenotype. These numbers are the observed values for the chi-squared calculation. Calculating observed values based on frequencies The next step is to find the expected values. If the population is at H-W equilibrium, the phenotype values calculated from the allele frequencies will be close to the observed phenotype values.
The expected frequency of the red individuals based on the H-W equation is the frequency of the red allele 0. Then multiply by to get the expected value. The work for the expected values is shown below. Calculation of expected values At this point, the chi-squared value can be determined by plugging the observed and expected values into the chi-squared equation. In this case, the chi-squared value is 1. Chi-squared calculation Our value of 1. This means that the final population distribution is consistent with the H-W equations.
One more example, again using the population genetics simulation to generate data. This time it is set to test a heterozygote advantage situation. The other variables are the same as the previous example. Simulation results, heterozygote advantage Again, we have to start by finding the observed numbers based on the phenotype frequencies, and the expected numbers based on the allele frequencies. Here is the chi-squared calculation for this example: Chi-squared calculation The chi-squared value is This is clearly more than the critical value, and so the hypothesis that the observed and expected are equivalent is rejected. This indicates that the population is not at H-W equilibrium. Note that the question that prompted this post asked for the chi-squared value only not the identification of df or critical value.
Specifically, for each cell, its row marginal is multiplied by its column marginal, and that product is divided by the sample size. Table 3 provides the results of this calculation for each cell. Table 3. Cell expected values and cell Chi-square values. For example, a 2 x 2 table has 1 df. A Chi-square table of significances is available in many elementary statistics texts and on many Internet sites. This is a result of the observed value being 23 while only Therefore, this cell has a much larger number of observed cases than would be expected by chance. Cell 1 reflects the number of unvaccinated employees who contracted pneumococcal pneumonia. This means that the number of unvaccinated people who contracted pneumococcal pneumonia was significantly greater than expected.
This means that a significantly lower number of vaccinated subjects contracted pneumococcal pneumonia than would be expected if the vaccine had no effect. Therefore the company can conclude that there was no difference between the two groups for incidence of non-pneumococcal pneumonia. It can be seen that for both groups, the majority of employees stayed healthy. The meaningful result was that there were significantly fewer cases of pneumococcal pneumonia among the vaccinated employees and significantly more cases among the unvaccinated employees.
As a result, the company should conclude that the vaccination program did reduce the incidence of pneumoccal pneumonia. Most researchers inspect the table to estimate which cells are overrepresented with a large number of cases versus those which have a small number of cases. Chi-square and closely related tests One might ask if, in this case, the Chi-square was the best or only test the researcher could have used. Nominal variables require the use of non-parametric tests, and there are three commonly used significance tests that can be used for this type of nominal data. The first and most commonly used is the Chi-square. Table 4. Example of a table that violates cell expected values. The third test is the maximum likelihood ratio Chi-square test which is most often used when the data set is too small to meet the sample size assumption of the Chi-square test. As exhibited by the table of expected values for the case study, the cell expected requirements of the Chi-square were met by the data in the example.
Specifically, there are 6 cells in the table. This table meets the requirement that at least 5 of the 6 cells must have cell expected of 5 or more, and so there is no need to use the maximum likelihood ratio chi-square. Suppose the sample size were much smaller. Suppose the sample size was smaller and the table had the data in Table 4. This table should be tested with a maximum likelihood ratio Chi-square test. When researchers use the Chi-square test in violation of one or more assumptions, the result may or may not be reliable. Second, the appropriate test may produce a significant result while the inappropriate test provides a result that is not statistically significant, which is a Type II error. Third, the appropriate test may provide a non-significant result while the inappropriate test may provide a significant result, which is a Type I error.
Finding a significant difference merely means that the differences between the vaccinated and unvaccinated groups have less than 1. That is, there are 1. That is a sufficiently remote probability of error that in this case, the company can be confident that the vaccination made a difference. While useful, this is not complete information. It is necessary to know the strength of the association as well as the significance. Statistical significance does not necessarily imply clinical importance. Clinical significance is usually a function of how much improvement is produced by the treatment. There is, however, a more standardized strength test for the Chi-Square. Statistical strength tests are correlation measures. It is easily calculated with the following formula: Where n is the number of rows or number of columns, whichever is less. For the example, the V is 0.
For any correlation, a value of 0. It should be noted that a relatively weak correlation is all that can be expected when a phenomena is only partially dependent on the independent variable. In the case study, five vaccinated people did contract pneumococcal pneumonia, but vaccinated or not, the majority of employees remained healthy. Clearly, most employees will not get pneumonia. This fact alone makes it difficult to obtain a moderate or high correlation coefficient. The amount of change the treatment vaccine can produce is limited by the relatively low rate of disease in the population of employees. These are the factors the researcher should take into account when interpreting this statistical result. Summary and conclusions The Chi-square is a valuable analysis tool that provides considerable information about the nature of research data. It is a powerful statistic that enables researchers to test hypotheses about variables measured at the nominal level.
How to Do a Chi Square Report in APA In survey research, a Likert scale is an approach to response categories that measures the extent of a person's satisfaction or agreement with a specific set of statements or questions. Using a Likert Scale set of questions to measure customer service reactions on services or products is one way the measurement tool operates. This type of response category makes it easy to quantify survey responses thereby simplifying data analysis. A variety of options for analyzing Likert scale data exist including the chi square statistic. The chi square statistic compares survey respondents' actual responses to questions with expected answers to assess the statistical significance of a given hypothesis.
An example of chi square statistic might be examining whether two groups of people have varying opinions. The greater the level of deviation between actual and expected responses, the higher the Chi Square statistic will be. This deviation level indicates how much less the results fit the original hypothesis. There are two types of chi square statistic test: the chi-square goodness of fit test and the chi-square test for independence. Combine the response categories in your Likert scale. For example, if your Likert scale uses the response categories of strongly agree, agree, disagree, strongly disagree, neither agree nor disagree, combine the agree and strongly agree responses into one category and the disagree and strongly disagree into another.
This gives you three categories of responses: agree, disagree and neither. Run the chi square statistical test, using your spreadsheet program or statistical software. For example, to find the test in Excel, simply click the Formula tab at the top of your spreadsheet. Then choose More Functions and select Statistical; which displays the variety of available procedures followed by a section of the chitest or chi square procedure.
Clicking on a cell and dragging the mouse over the range of data you want analyzed tells Excel the data on which to conduct the chi square test. Next, examine the results of the chi square test generated by a spreadsheet or statistical program. When reviewing results, pay close attention to the size of the chi square statistic and the level of statistical significance. A higher chi square statistic indicates greater variation between observed and expected responses.
Most spreadsheet and statistical programs use a significance level of. Interpret the results of your analysis. Analysis at this point involves looking at whether the chi square indicates a statistically significant relationship that exists without revealing information about the strength of that statistical relationship. Because there may also be a number of survey takers that leave answers blank, the chi square test is a good option because you can add a row for "no answer".
While comparative analysis examines the data relationship, the chi square makes it possible to also measure non-reponse sampling errors and their corresponding relationship as well. Related Articles.
There are four possible outcomes, and we lose a degree of freedom because of finite sampling. However, it turns out that we lose two more degrees of freedom. This is because the expected values in the chi-square test were based, in part, on the observed values. Put another way: if we had different observed values, we would have calculated different expected values, because the allele frequencies were calculated from the data. We lose one degree of freedom for each independent parameter calculated from the data used to then calculate the expected values. We calculated two independent parameters: the frequency of the A allele and the frequency of the B allele. However, these are automatically 1. Our test statistic value of 0. Thus, we can say that 0. Note that we observed more parental offspring than expected. That is, we expected Regardless of the outcome of the chi-square test of independence, we would not have been allowed to reject the hypothesis of independent assortment if we had observed more recombinant than parental offspring.
One final note on this last test. Let's say we'd chosen to do the old-fashioned test. We would have expected 25 of each phenotype. We'd have three degrees of freedom, and would find that 0. We still wouldn't have rejected the hypothesis of independent assortment. But it won't always be that way. As above, an individual with the AaBb genotype is mated with an individual with the aabb genotype. Offspring are observed in the following numbers: AB, 97 ab, 78 Ab and 71 aB.
Should we reject the hypothesis that the alleles of the A and B genes are independently assorting? Solution Hardy-Weinberg Equilibrium. In a real population of interbreeding organisms, the different alleles of a gene may not be represented at equal frequencies. This doesn't mean there's something amiss with respect to Mendel's laws. The individual crosses that produced the offspring would be expected, in general, to follow Mendel's laws, but many other factors determine the frequencies of alleles. Some alleles may confer, on average, a selective advantage. Some alleles may leave or enter the population disproportionately emigration and immigration. One allele might mutate into the other more often than the reverse. And, finally, individuals with certain alleles might, just by chance, survive and leave more offspring, a phenomenon we call "genetic drift.
No comments:
Post a Comment