Question

An experiment was conducted in which two types of acorn squash were crossed. According to a genetic model, \(1 / 2\) of the resulting plants should have dark stems and dark fruit, \(1 / 4\) should have light stems and light fruit, and \(1 / 4\) should have light stems and plain fruit. The actual data were \(220,129,\) and 105 for these three categories. \({ }^{49}\) Do these data refute this model? Consider a chi-square test. (a) What is the value of the chi-square test statistic for investigating whether these data are consistent with the \(1 / 2,1 / 4,1 / 4\) probabilities model? (b) The \(P\) -value for the chi-square test is \(0.23 .\) If \(\alpha=0.10\), what is your conclusion regarding \(H_{0} ?\)

Short answer

The chi-square test statistic is calculated to be approximately 2.39. Since the p-value (0.23) is greater than the significance level (0.10), the null hypothesis cannot be rejected, and the data do not provide significant evidence against the genetic model.

Step-by-step solution

State the null hypothesis

The null hypothesis (\(H_0\)) posits that the observed frequencies of squash with various characteristics match the expected frequencies based on the genetic model. In this case, that model predicts frequencies of \( \frac{1}{2} \) for dark stems and fruit, \( \frac{1}{4} \) for light stems and light fruit, and \( \frac{1}{4} \) for light stems and plain fruit.

Calculate the expected frequencies

First, determine the total number of observed plants, which is the sum of the three observed categories: \(220 + 129 + 105 = 454\text{ plants}\).Next, calculate the expected frequency for each category using the genetic model: \(454 \times \frac{1}{2} = 227\), \(454 \times \frac{1}{4} = 113.5\), and \(454 \times \frac{1}{4} = 113.5\).

Compute the chi-square test statistic

Use the formula \[ \text{Chi-square (}\text{χ}^2\text{)} = \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each category. The chi-square test statistic is the sum of these values for all categories. \[ \text{χ}^2 = \frac{(220 - 227)^2}{227} + \frac{(129 - 113.5)^2}{113.5} + \frac{(105 - 113.5)^2}{113.5} = \frac{(-7)^2}{227} + \frac{(15.5)^2}{113.5} + \frac{(-8.5)^2}{113.5} \] \[ = \frac{49}{227} + \frac{240.25}{113.5} + \frac{72.25}{113.5} \approx 2.39 \]

Decision Rule

Compare the computed chi-square value with the critical value from the chi-square distribution table based on the degrees of freedom (df) and the significance level \(\alpha\). For this test, df is the number of categories minus 1, which is 2. With \(\alpha = 0.10\), we look up the critical value for 2 df to determine if we can reject the null hypothesis. However, since the p-value of 0.23 is given and it's higher than the significance level, we do not need to refer to the table and can directly state the conclusion based on the p-value.

Draw the conclusion

Because the p-value (0.23) is greater than the significance level (0.10), we do not reject the null hypothesis. Therefore, there is no statistically significant evidence at the 0.10 level to conclude that the data does not follow the proposed genetic model of \(1/2, 1/4, 1/4\) for the different types of squash.

Key concepts

Null Hypothesis

When delving into the realm of statistics, the null hypothesis is akin to a starting point for comparison. It is a statement used in statistical analysis that suggests there is no effect or no difference, serving as the baseline assumption for the test. In the chi-square test context of our exercise, the null hypothesis (\(H_0\)) posits that the observed number of acorn squash with various characteristics—dark stems with dark fruit, light stems with light fruit, and light stems with plain fruit—does not significantly differ from what would be expected based on genetic predictions. This hypothesis is essentially what we aim to verify: whether or not the observed data align with the expected frequencies of the genetic model.
To articulate it simply, if at the end of the test the null hypothesis stands unrefuted, it indicates that our observations could indeed be a result of the predicted genetic proportions of acorn squashes.

Expected Frequencies

A fundamental aspect of the chi-square test is the expected frequencies. These are the numbers we would expect to see in each category if the null hypothesis were true. They're estimated based on a specific model or distribution that we believe represents the real-world scenario. In the case of our exercise, the expected frequencies of acorn squash types were calculated from a genetic model predicting half of the population to have dark characteristics and a quarter each for the other two types.
These frequencies serve as the target numbers against which we measure the observed counts. We determine them by multiplying the total count of observations by the probability of each outcome as per the genetic model. Once these expected frequencies are in hand, they become pivotal in computing the chi-square test statistic, which gauges the discrepancies between what we see in reality and what our initial hypothesis would have us expect.

Degrees of Freedom

The phrase 'degrees of freedom' in statistics often flusters students, but its concept is quite straightforward. Consider it as the number of values in a calculation that are free to vary. In a chi-square test, the degrees of freedom are calculated as the number of categories of data minus one. For the acorn squash experiment, with three categories of squash, the degrees of freedom is two.
Why is this crucial? Because the degrees of freedom define the shape of the chi-square distribution—this is the curve we use to determine the probability of observing our test statistic under the null hypothesis. Having the correct degrees of freedom is paramount for proper analysis. They literally 'free' us to compare our calculated chi-square value against the correct segment of the chi-square distribution curve to make inferences about our hypothesis.

p-Value

Next is the p-value, which often mystifies, yet is one of the most pertinent outcomes of hypothesis testing. Put plainly, the p-value measures the strength of evidence against the null hypothesis provided by our data. It's the probability of observing a test statistic as extreme as, or more extreme than, the value computed from the data, assuming that the null hypothesis is true.
The p-value helps us make a decision on whether to reject the null hypothesis. In the context of the chi-square test in our exercise, a p-value of 0.23 indicates that there's a 23% chance of observing a chi-square as large as 2.39 (or larger) if the null hypothesis is accurate. It's a threshold; if the p-value is below a certain significance level (say, 0.05 or 5%), we would reject the null hypothesis, deeming our results statistically significant. However, if it's above that level—as it is in this case—we maintain the status quo and do not reject the null.

Significance Level

Lastly, we touch upon the significance level, denoted as \( \alpha \). This is a critical component of hypothesis testing, acting as the deciding factor that determines the 'rarity' of data required to reject the null hypothesis. The significance level is the probability at which we're willing to risk rejecting a true null hypothesis. Commonly used values for \( \alpha \) are 0.05 (5%), 0.01 (1%), and as in our exercise, 0.10 (10%).
Think of the significance level as a strictness gauge. Setting a low \( \alpha \) (making it 'strict') means demanding stronger evidence to reject the null hypothesis, whereas a higher \( \alpha \) ('lenient') requires less strong evidence. In our squash example, the significance level is 0.10, so as the p-value 0.23 exceeds this, we hold onto the null hypothesis, concluding that there is no statistical reason to believe the actual distribution of squash types deviates from the genetic model's predictions.