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.