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.

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.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)