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The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset Vaia Explanations$ Math

6th Edition · 837 pages

ISBN: 9781319113339

Chapter 10: Q 31. (page 643)

Let $p_{M}, p_{F}$ be the proportions of all college males and
females who worked last summer. The hypotheses to be tested are
a. $H_{0} : p_{M} - p_{F} = 0 v e r s e s H_{a} : p_{M} - p_{F} \neq 0$
b. $H_{0} : p_{M} - p_{F} = 0 v e r s e s H_{a} : p_{M} - p_{F} > 0$
c. $H_{0} : p_{M} - p_{F} = 0 v e r s e s H_{a} : p_{M} - p_{F} < 0$
d. $H_{0} : p_{M} - p_{F} > 0 v e r s e s H_{a} : p_{M} - p_{F} = 0$
e. $H_{0} : p_{M} - p_{F} \neq 0 v e r s e s H_{a} : p_{M} - p_{F} = 0$

Short Answer

Expert verified

Option (a) is correct.

Step by step solution

01

Given Information

It is given that $p_{M}$ is the proportion of all college males who worked last summer.

$p_{F}$ is the proportion of all college females who worked last summer.

Claim: Difference in proportion of males and females.

02

Explanation

As $p_{M} \neq p_{F}$

According to given data, appropriate hypothesis is:

Null: $H_{0} : p_{M} = p_{F}$

Alternate: $H_{a} : p_{M} \neq p_{F}$

Solving, we get

$H_{0} : p_{M} - p_{F} = 0$

$H_{a} : p_{M} - p_{F} \neq 0$

Option (a) is correct.

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Most popular questions from this chapter

Which of the following will increase the power of a significance test?

a. Increase the Type II error probability.

b. Decrease the sample size.

c. Reject the null hypothesis only if the P-value is less than the significance level.

d. Increase the significance level $α$ .

e. Select a value for the alternative hypothesis closer to the value of the null hypothesis.

There are two common methods for measuring the concentration of a pollutant in fish tissue. Do the two methods differ, on average? You apply both methods to each fish in a random sample of $18$ carp and use

a. the paired t test for $μ d i f f \frac{305}{1526} = 0.200 = 20.0 % μ d i f f$ .

b. the one-sample z test for p.

c. the two-sample t test for $μ 1 - μ 2 \frac{305}{1526} = 0.200 = 20.0 % μ 1 - μ 2$ .

d. the two-sample z test for $p 1 - p 2 \frac{305}{1526} = 0.200 = 20.0 % p 1 - p 2$ .

e. none of these.

A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained an SRS of $60$ high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be $6$ hours with a standard deviation of $3$ hours. The researcher also obtained an independent SRS of $40$ high school students in a large city school district and found the mean time spent in extracurricular activities per week to be $5$ hours with a standard deviation of $2$ hours. Suppose that the researcher decides to carry out a significance test of $H_{0}$ : μsuburban=μcity versus a two-sided alternativ

The P-value for the test is $0.048$ . A correct conclusion is to

a. fail to reject $H_{0}$ because $0.048 < α = 0.05$ . There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts.

b. fail to reject $H_{0}$ because $0.048 < α = 0.05$ . There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts.

c. fail to reject $H_{0}$ because $0.048 < α = 0.05$ . There is convincing evidence that the average time spent on extracurricular activities by students in the suburban and city school districts is the same.

d. reject $H_{0}$ because $0.048 < α = 0.05$ . There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts.

e. reject $H_{0}$ because $0.048 < α = 0.05$ . There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts.

An SRS of size $100$ is taken from Population A with proportion $0.8$ of successes. An independent SRS of size $400$ is taken from Population B with proportion $0.5$ of successes. The sampling distribution of the difference (A − B) in sample proportions has what mean and standard deviation?

a. mean $= 0.3$ ; standard deviation $= 1.3$

b. mean $= 0.3$ ; standard deviation $= 0.40$

c. mean $= 0.3$ ; standard deviation $= 0.047$

d. mean $= 0.3$ ; standard deviation $= 0.0022$

e. mean $= 0.3$ ; standard deviation $= 0.0002$

Where’s Egypt? In a Pew Research poll, $287$ out of $522$ randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When $520$ randomly selected U.S. women were asked, $233$ were able to do so.

a. Construct and interpret a $95 %$ confidence interval for the difference in the true

proportion of U.S. men and U.S. women who can identify Egypt on a map.

b. Based on your interval, is there convincing evidence of a difference in the true

proportions of U.S. men and women who can identify Egypt on a map? Justify your

answer.

See all solutions

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