Question

Clean water The Environmental Protection Agency (EPA) has determined that safe

drinking water should contain at most $1.3$mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of$30$randomly selected locations. The company performs a test at the α=0.05 significance level of$H_0:\mu =1.3$versus $H_a:\mu >1.3$, where μ is the

true mean copper content of the water from the new source.

a. Describe a Type I error and a Type II error in this setting.

b. Which type of error is more serious in this case? Justify your answer.

c. Based on your answer to part (b), do you agree with the company’s choice of $\alpha =0.05$? Why or why not?

Short answer

a. Type I error rejects null hypothesis $H_0$if it is true and type II error fails to reject null hypothesis $H_0$if it is false.

b. Type II error is more serious.

c.$\alpha =0.05$is better than$\alpha =0.10$

Step-by-step solution

Given Information

It is given that $H_0:\mu =1.3$

$H_1:\mu >1.3$

Explaining type I and type II error

Type I error: There is enough convincing evidence that true mean copper content from new source is $1.3$mg/l, when actual mean content of water from new source is $1.3$mg/l.

Type II error: Noconvincing evidence is present true mean copper content from new source is $1.3$mg/l when actual mean content of water from new source is over$1.3$ mg/l

Which error is worse

Here, type II error is worse as when water is not safe, people might drink it.

Choice of α

As type II error is more worse.

The chances of type II error decreases with increase in type I error and $\alpha$would be better.

Hence, $\alpha =0.10$is preferred over$0.05$