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A researcher plans to conduct a significance test at the α=0.01 significance level. She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter she used is

a.0.01.b.0.10.c.0.89.d.0.90.e.0.99.

Short Answer

Expert verified

The correct option is (b) 0.10

Step by step solution

01

Given information

α=0.01

02

Explanation

The most appropriate response to the given statement "A researcher intends to conduct a significance test with a significance level of α=0.01. She plans for a power of 0.90for a certain alternative value of the parameter of interest in her investigation. She computed the power at a certain alternative value of the parameter, the risk of the researcher making a Type II error is 0.10"

Power is calculated in this case as,

1P(Type2error)P(Type2error)+Power=1P(Type2error)=1PowerP(Type2error)=10.90P(Type2error)=0.10

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Most popular questions from this chapter

How much juice? Refer to Exercises 3 and 11 .

a. What conclusion would you make at the α=0.10α=0.10level?

b. Would your conclusion from part (a) change if a 5 \% significance level was used instead? Explain your reasoning.

Don’t argue Refer to Exercises 2 and 12.

a. What conclusion would you make at the α=0.01 level?

b. Would your conclusion from part (a) change if a 5% significance level was used

instead? Explain your reasoning.

Potato power problems Refer to Exercises 85 and 87

a. Explain one disadvantage of using α=0.10 instead of α=0.05 when

performing the test.

b. Explain one disadvantage of taking a random sample of 500 potatoes instead of 250 potatoes from the shipment.

After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of all living alumni is in favor of firing the coach. The appropriate standardized test statistic is

(a)z=0.64-0.50.64(0.36)100z=0.64-0.50.64(0.36)100

role="math" localid="1654432946823" (b)t=0.64-0.50.64(0.36)100t=0.64-0.50.64(0.36)100

(c)z=0.64-0.50.5(0.5)100z=0.64-0.50.5(0.5)100

(d)z=0.64-0.50.64(0.36)64z=0.64-0.50.64(0.36)64

(e)z=0.5-0.640.5(0.5)100z=0.5-0.640.5(0.5)100

The reason we use t procedures instead of z procedures when carrying out a test about a population mean is that

a. z requires that the sample size be large.

b. z requires that you know the population standard deviation σ

c. z requires that the data come from a random sample.

d. z requires that the population distribution be Normal.

e. z can only be used for proportions.

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