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 \(p_{1}=0.25, p_{2}=0.50,\) and \(p_{3}=0.25 .\) 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 \(\alpha=0.05\). 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 the 200 plant sample size, the Chi-square statistic of 4.63 is less than the critical Chi-square value of around 5.99, so \(H_{0}\) is not rejected. The same conclusion applies for the 300 plant sample size since the Chi-square statistic remains the same despite the larger sample.

Step-by-step solution

Understanding the chi-square goodness-of-fit test

A chi-square goodness-of-fit test compares observed data to an expected data pattern or distribution. The null hypothesis, \(H_{0}\), proposes that there is no significant difference between the observed and expected datasets. The level of significance, \(\alpha\), determines the rejection region for \(H_{0}\). If the calculated chi-square statistic falls in this region, we reject \(H_{0}\), otherwise we fail to reject \(H_{0}\).

Carrying out the test for 200 plants

For the test with 200 plants, the chi-square statistic has already been given, \(X^{2}=4.63\). We compare this with the critical chi-square value corresponding to \(\alpha = 0.05\) and degrees of freedom \(df = n - 1 = 3 - 1 = 2\) (where n is the number of categories of phenotypes). From the chi-square distribution table, the critical value for \(\alpha = 0.05\) and \(df = 2\) is approximately 5.99. Given that \(X^{2}\)< 5.99, we fail to reject \(H_{0}\). Therefore, there's not enough evidence to say the theory is incorrect based on this data sample.

Carrying out the test for 300 plants

In the second scenario, the sample size has increased to 300 plants, but the \(X^{2}\) value remains the same. The larger the sample size, the better the estimate of the population it provides, which means potential for a more precise test. But as the chi-square value doesn't change in this case, the interpretation remains the same: we still fail to reject \(H_{0}\). It seems by increasing the sample size, the conclusions from part (a) aren't significantly impacted.