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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

Short Answer

Expert verified

The required correct option is C.

The appropriate standardized test statistic isz=0.64-0.50.5(0.5)100

Step by step solution

01

Given information

n=100x=64p=50%=0.50

02

Concept used

The following expression is used to compute the test statistic:

z=p^-pp(1-p)n

Where sample proportion p^

Sample size:n

03

The objective is to find out the correct option which shows the appropriate standardized test statistic.

To begin, compute the sample proportion, which is the ratio of the number of successes to the sample size:

p^=xnp^=64100p^=0.64

The following expression is used to compute the test statistic:

z=p^-pp(1-p)n

Fill in the blanks with the values from the given data.

z=p^-pp(1-p)nz=0.64-0.50.5(1-0.5)100z=0.64-0.50.5(0.5)100

Therefore, the correct option is(c)

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Most popular questions from this chapter

Side effects A drug manufacturer claims that less than 10%of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300out of 5000Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25of the subjects experience nausea.

a. Describe a Type I error and a Type II error in this setting, and give a possible

consequence of each.

b. Do these data provide convincing evidence for the drug manufacturer’s claim?

Explaining confidence: Here is an explanation from a newspaper concerning one of its opinion polls. Explain what is wrong with the following statement.

For a poll of 1600 adults, the variation due to sampling error is no more than 3

percentage points either way. The error margin is said to be valid at the 95%

confidence level. This means that, if the same questions were repeated in 20 polls, the results of at least 19 surveys would be within 3 percentage points of the results of this survey.

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.

Restaurant power problems Refer to Exercises 86 and 88

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 50 people instead of 30 people.

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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