For each of the following situations, indicate what the general impact on the type ii error probability will be:

a. nbsp the alpha level is increased.
b. nbsp the "true" population mean is moved farther from the hypothesized popoulation mean.
c. nbsp the alpha level is decreased.
d. nbsp the sample size is increased.

Step-by-step explanation:

Hello!

The type II error is committed when you fail to reject a false null hypothesis.

The probability associated with this type of error is symbolized as β.

a. The alpha level is increased.

α and β are two probabilities that are closely related in such a way that if alpha is decreased, beta automatically increases and vice versa when the sample size n is predetermined.

If you increase α then the probability of committing type II error (β) decreases.

b. The "true" population mean is moved farther from the hypothesized population mean.

If the true population mean is moved away from its hypothesized value, then the probabilities of "keeping" a false null hypothesis are bigger.

c. The alpha level is decreased.

This is the inverse situation from a. If you decrease α then, β is automatically increased.

d. The sample size is increased.

If the sample size "n" is left free, it can be shown that by increasing the sample size, both α and β decrease. In practice, α is fixed and β is calculated with several possible sizes of n. Finally choosing as sample size the one that minimizes or gives a reasonable value of β.

I hope it helps!

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116