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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^=xn⇒p^=64100⇒p^=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)n⇒z=0.64-0.50.5(1-0.5)100⇒z=0.64-0.50.5(0.5)100

Therefore, the correct option is(c)

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Most popular questions from this chapter

A government report says that the average amount of money spent per U.S. household per week on food is about \(158. A random sample of50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of \)165 and a standard deviation of \(20. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of \)158?

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.

A95%confidence interval for the proportion of viewers of a certain reality television

show who are over 30 years old is (0.26,0.35). Suppose the show's producers want to est the hypothesis \H0:p=0.25against Ha: Ha:p≠0.25. Which of the following is an appropriate conclusion for them to draw at the α=0.05

a. Fail to reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old equals 0.25

b. Fail to reject H0there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from0.25.

c. Reject H0; there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25

. d. Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old is greater than 0.25.

e. Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from0.25.

Significance tests A test of H0:p=0.65 against Ha:p<0.65

based on a sample of size 400 yields the standardized test statistic z=−1.78 .

a. Find and interpret the P-value.

b. What conclusion would you make at the α=0.10 significance level? Would

your conclusion change if you used α=0.05 instead? Explain your reasoning.

c. Determine the value of pÁåœ= the sample proportion of successes.

A researcher claims to have found a drug that causes people to grow taller. The coach of the basketball team at Brandon University has expressed interest but demands evidence. Over 1000 Brandon students volunteer to participate in an experiment to test this new drug. Fifty of the volunteers are randomly selected, their heights are measured, and they are given the drug. Two weeks later, their heights are measured again. The power of the test to detect an average increase in height of 1 inch could be increased by

a. using only volunteers from the basketball team in the experiment.

b. usingα=0.05 instead of α=0.05

c. using α=0.05instead of α=0.01

d. giving the drug to 25 randomly selected students instead of 50.

e. using a two-sided test instead of a one-sided test.

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