Question

Refer to Exercise 9.15. Explain what each of the following would mean.

(a) Type I error

(b) Type II error

(c) Correct decision

Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean cadmium level in Boletus Pinicolamushrooms.

(d) equals the safety limit of $0.5$ppm.

(e) exceeds the safety limit of$0.5$ppm.

Short answer

(a) Rejecting a Null Hypothesis, when it is true.

(b) Rejecting a Null Hypothesis, when it is false.

(c) If the true null hypothesis is not rejected or the false null hypothesis is rejected it is a correct decision.

(d) A correct decision.

(e) Type II Error.

Step-by-step solution

Step 1. Given Information.

The Null Hypothesis is,

$H_0:\mu =0.5ppm$.

The Alternative Hypothesis is,

$H_0:\mu >0.5ppm$.

Step (a). Type I error.

According to the definition of the type I error it is to reject a null hypothesis when it is true. A type I error would occur in fact $\mu =0.5ppm$true, that is the mean cadmium level but the result of the sampling lead to the conclusion that the mean cadmium level is greater .

Part (b). Type II error.

According to the definition of the type II error, it is to not reject a null hypothesis when it is false. The results of the sampling fail to lead to the conclusion that $\mu >0.5ppm$.

Part (c). Correct Decision.

A correct decision would occur if the true null hypothesis is not rejected or a false null hypothesis is rejected. The mean cadmium level is $\mu =0.5ppm$ and the results of the sampling do not lead to rejection, so is a correct decision; or the mean cadmium level$\mu >0.5ppm$and the results of the sampling lead to that conclusion.

Part (d). Equals the safety limit of 0.5ppm.

Here the mean cadmium level equals the safety limit of $0.5ppm$, and the results of the hypothesis test lead to the non-rejection of the null hypothesis. We are not rejecting the null hypothesis, so we have taken the correct decision.

Part (e) Exceeds the safety limit of 0.5ppm.

Here the mean cadmium level exceeds the safety limits of $0.5ppm$, and the results of the hypothesis test lead to the non-rejection of the null hypothesis. We are not rejecting the false null hypothesis that is a result of the sampling the mean cadmium level is greater then $0.5ppm$, which is considered under the alternative hypothesis so we have committed Type II error.