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Chapter 9: Q T9.5 (page 616)

A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the standardized test statistic for performing a test of H0:p=0.5H0:p=0.5versus Ha:p>0.5Ha:p>0.5

is which of the following?

a)z=0.59-0.50.59(0.41)100

b)z=0.59-0.50.5(0.5)100

c)z=0.5-0.590.59(0.41)100

d)z=0.5-0.590.5(0.5)100

e)z=0.59-0.5100

Short Answer

Expert verified

The test statistic is0.59-0.500.30(1-0.300100.

Step by step solution

01

Step 1:Given information

Number of trials (n)=100

Number of events(x)=59

Level of significance(α)=0.05

The null and alternative hypotheses are:

H0:p=0.5

Ha:p>0.5

02

Step 2:Explaination

The formula to compute the test statistic is:

z=p^-p0P01-p0n

Where,

p0 is the population proportion

p^ is the sample proportion

Now,

The sample proportion can be computed as:

Sample proportion(p^)=xn

=59100=0.59

Now, test statistic can be calculated as:

z=p^-p0p01-p0n

=0.59-0.500.50(1-0.50)100

Thus, the test statistic is0.59-0.500S5(1-0.000)100.

Hence, the correct option is (d).

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Most popular questions from this chapter

Error probabilities and power You read that a significance test at the α=0.01

significance level has probability 0.14of making a Type II error when a specific alternative is true.

a. What is the power of the test against this alternative?

b. What’s the probability of making a Type I error?

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?

Which of the following has the greatest probability?

a.P(t>2)if t has 5 degrees of freedom.

b. P(t>2) if t has 2 degrees of freedom.

c. P(z>2) if z is a standard Normal random variable.

d.P(t<2)if t has 5 degrees of freedom.

e.P(z<2) if z is a standard Normal random variable.

pg559¯

No homework Refer to Exercises 1 and 9. What conclusion would you make at theα=0.05α=0.05level?

Jump around Refer to Exercise 78.

a. Construct and interpret a 90% confidence interval for the true mean vertical jump μ(in inches) of the students at Haley, Jeff, and Nathan’s school. Assume that the conditions for inference are met.

b. Explain why the interval in part (a) is consistent with the result of the test in Exercise 78
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