21.5 Exercises

Selected answers are available in Sect. D.20.

Exercise 21.1 A researcher was computing a 95% CI for a single proportion to estimate the proportion of trees with apple scabe (Hirst and Stedman 1962), and found that \(\hat{p} = 0.314\) and \(\text{s.e.}(\hat{p}) = 0.091\).

What is wrong with the following conclusion that the researcher made?

The approximate 95% CI for the sample proportion is between 0.223 and 0.405.

Exercise 21.2 A researcher was computing a 95% CI for a single proportion to estimate the proportion of trees with apple scabe (Hirst and Stedman 1962), and found that \(\hat{p} = 0.314\) and \(\text{s.e.}(\hat{p}) = 0.091\).

What is wrong with the following conclusion that the researcher made?

This CI means we are 95% confident that between 22.3 and 40.5 trees are infected with apple scab.

References

Hirst JM, Stedman OJ. The epidemiology of apple scab (Venturia inaequalis (Cke.) Wint.) III. The supply of ascospores. Annals of Applied Biology. Wiley Online Library; 1962;50(3):551–67.