Problem 1

Consider an experiment on chlorophyll inheritance in maize. A genetic theory predicts the ratio of green to yellow to be 3:1. In a sample of [\(n = 1103\)]{.math .inline} seedlings, [\(n_1 = 854\)]{.math .inline} were green and [\(n_2 = 249\)]{.math .inline} were yellow.

Part (a)

The statistic is defined as [\[X^2 = \sum_{j=1}^{c} \frac{(n_j - n \pi_{j 0})^2}{n \pi_{j 0}}.\]]{.math .display}

Under the null model, the odds are 3:1, or [\(\pi_{1 0} = 0.75\)]{.math .inline} and [\(\pi_{1 1} = 0.25\)]{.math .inline}. We are given [\(n = 1103\)]{.math .inline}, [\(n_1 = 854\)]{.math .inline}, and [\(n_2 = 249\)]{.math .inline}. Under the null model, these values have expectations given respectively by [\(n \pi_{1 0} = 827.25\)]{.math .inline} and [\(275.75\)]{.math .inline}.

The observed statistic is thus given by [\[X_0^2 = \frac{(854 - 827.25)^2}{827.25} + \frac{(249 - 275.75)^2}{275.75} = 3.46.\]]{.math .display}

Part (b)

The reference distribution is the chi-squared distribution with [\(1\)]{.math .inline} degree of freedom, denoted by [\(\chi^2(1)\)]{.math .inline}.

The upper 10th percentile given [\(1\)]{.math .inline} degree of freedom, denoted by [\(\chi_{0.10}^2(1)\)]{.math .inline}, is found by solving for [\(\chi_{0.10}^2(1)\)]{.math .inline} in the equation [\(\Pr(\chi^2(1) \geq \chi_{0.10}^2(1)) = 1-0.10 = 0.9\)]{.math .inline}, which yields the result [\[\chi_{0.10}^2(1) = 2.71.\]]{.math .display}

Part (c)

We see that any observed statistic [\(X_0^2\)]{.math .inline} with [\(\rm{df}=1\)]{.math .inline} greater than [\(\chi^2_{0.10}(1) = 2.71\)]{.math .inline} is not compatible with the null model at significance level [\(\alpha = 0.10\)]{.math .inline}.

Since the observed statistic [\(X_0^2 = 3.46 > 2.71\)]{.math .inline}, the null model, which is the genetic theory where the ratio of green to yellow is 3:1, is not compatible with the data.

Part (d)

Hypothesis testing, as a dichotomous measure of evidence, does not provide as much information as a more quantitative evidence measure. For instance, it does not provide information about effect size.

Problem 2

Part (a)

The MLE of [\(\pi_1\)]{.math .inline} is given by [\(\hat{\pi}_1 = \frac{n_1}{n} = \frac{854}{1103} = 0.774\)]{.math .inline}. Letting [\(\alpha = 0.10\)]{.math .inline} and inverting the Wald test statistic, we get the [\(90\%\)]{.math .inline} confidence interval for [\(\pi_1\)]{.math .inline}, [\[\hat{\pi}_1 \pm z_{1-alpha/2} \sigma_{\hat\pi} = 0.744 \pm 1.645 \sqrt{\frac{0.774(1-0.774)}{1103}},\]]{.math .display} which may be rewritten as [\[[0.754, 0.795].\]]{.math .display}

Part (b)

Based on the observed data, we estimate that the probability [\(\pi_1\)]{.math .inline} of a green strain is between [\(0.754\)]{.math .inline} and [\(0.795\)]{.math .inline}.

As expected, the null model specifies a value for [\(\pi_1\)]{.math .inline} ([\(0.75\)]{.math .inline}) that is not contained in this confidence interval.