Question

A certain genetic characteristic of a particular plant can appear in one of three forms (phenotypes). A researcher has developed a theory, according to which the hypothesized proportions are \(\pi_{1}=.25, \pi_{2}=.50\), and \(\pi_{3}=\) .25. A random sample of 200 plants yields \(X^{2}=4.63\). a. Carry out a test of the null hypothesis that the theory is correct, using level of significance \(\alpha=.05\). b. Suppose that a random sample of 300 plants had resulted in the same value of \(X^{2} .\) How would your analysis and conclusion differ from those in Part (a)?

Short answer

For both (a) and (b), we fail to reject the null hypothesis that the researcher's theory is correct.

Step-by-step solution

Compute Degrees of Freedom

The number of degrees of freedom is given by the number of categories minus 1. Since there are three possible phenotypes, the degrees of freedom (dof) is \(3 - 1 = 2\).

Conduct the Hypothesis Test for (a)

The null hypothesis is that the observed proportions for each phenotype are as stated by the researcher. We compare the observed test statistic \(X^{2}=4.63\) with the Chi-square distribution table value at \(\alpha = 0.05\) which is approximately 5.99. Since 4.63 < 5.99, we fail to reject the null hypothesis.

Conduct the Hypothesis Test for (b)

Even if the sample size increased to 300, the procedure stays the same. If the test statistic were still \(X^{2}=4.63\) it would be compared to the Chi-square distribution table value at \(\alpha = 0.05\) and the same decision would be reached - we fail to reject the null hypothesis.

Key concepts

Chi-Square Distribution

The Chi-Square distribution is a fundamental concept in statistics that is used to test hypotheses concerning categorical data. When we want to determine if the observed frequencies of a certain event match the expected frequencies, we use a Chi-Square test. The distribution itself is skewed to the right and its shape depends on the degrees of freedom (dof).

For example, when a researcher wants to test a theory regarding the expected proportions of a plant's phenotypes, the Chi-Square distribution provides a way to assess whether the observed data deviates significantly from what was expected. By comparing the computed Chi-Square test statistic to critical values from the Chi-Square distribution, we can make inferences about the population.

Null Hypothesis

In hypothesis testing, the null hypothesis (\(H_{0}\)) is a statement of no effect or no difference. It serves as a benchmark that assumes the observed data is purely a result of random variation.

In our plant phenotypes example, the null hypothesis states that the actual proportions of the three plant forms are the same as the hypothesized ones (\( \frac{1}{4}, \frac{1}{2} \text{, and } \frac{1}{4} \text{ respectively}\)). The alternative hypothesis (\(H_{1}\)) would claim that there is a significant difference between the observed and expected frequencies. The goal of the Chi-Square test is to determine whether the observed data provide enough evidence to reject the null hypothesis.

Degrees of Freedom

Degrees of freedom (dof) play a crucial role in the context of statistical testing. They are a function of the amount of independent information available to estimate a parameter. In the case of the Chi-Square test, the degrees of freedom are calculated as the number of categories minus 1.

In our exercise, with three phenotypes, the degrees of freedom would be 3 - 1 = 2. Degrees of freedom are important because they help to determine the appropriate critical value from the Chi-Square distribution, which we compare our test statistic against. A larger number of degrees of freedom generally indicates more variability in the data.

Statistical Significance

Statistical significance is a measure of whether the results obtained from a test are likely to be due to something other than random chance. The alpha level (\( \text{denoted as } \boldsymbol{\beta} \text{ or } \boldsymbol{\beta} \text{-level}\)) represents the threshold probability for rejecting the null hypothesis.

In our scenario, the significance level is set at 0.05, suggesting a 5% risk of concluding that a difference exists when there is no actual difference. If the calculated Chi-Square statistic is less than the critical Chi-Square value at this level, the result is not statistically significant, and we fail to reject the null hypothesis.