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

Paraphin [41]

Answer:

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!

3 0

3 years ago

sweet t has 2 orange picks for every 5 green. I f there are 21 picks in all, how many picks are orange?

weeeeeb [17]

If there are 21 picks, 6 are orange

8 0

3 years ago

Read 2 more answers

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

(c) What is Upper P ( $\overline{X}$ less than or equals 74.4 ) =

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$