## 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 , 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 , 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.