Question

The FDA regulates that fresh Albacore tuna fish contains at most 0.82 ppm of mercury. A scientist at the FDA believes the mean amount of mercury in tuna fish for a new company exceeds the ppm of mercury. The hypotheses are: H0:µ = 0.82 H1:µ > 0.82 What is a type II error in the context of this problem? A. The fish is accepted by the FDA when in fact it had more than 0.82 ppm of mercury. B. The fish is accepted by the FDA when in fact it had less than 0.82 ppm of mercury. C. The fish is rejected by the FDA when in fact it had less than 0.82 ppm of mercury. D. The fish is rejected by the FDA when in fact it had more than 0.82 ppm of mercury.

Answer 1

A type II error in this context is when the fish is rejected by the FDA when it actually has less than 0.82 ppm of mercury.

Step-by-step explanation:

A type II error in this context refers to the situation where the FDA rejects a fish that actually has less than 0.82 ppm of mercury. In other words, it is when the fish is rejected by the FDA when in fact it had less than 0.82 ppm of mercury. This error occurs when the scientist at the FDA wrongly concludes that the mean amount of mercury in the tuna fish for the new company exceeds the limit set by the FDA, based on the actual data.

Answer 2

Option A

Step-by-step explanation:

A type II error occurs when the researcher fails to reject the bull hypothesis when it is false.

In this study, the

null hypothesis is H0:µ = 0.82 while the alternative hypothesis is H1:µ > 0.82

Thus, the type II error will be that the fish is accepted by the FDA when in fact it had more than 0.82 ppm of mercury.