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Chapter 9: Testing a Claim

Section 9.2

Tests About a Population Proportion

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Two-Sided Tests

According to the Centers for Disease Control and Prevention (CDC) Web site, 50%

of high school students have never smoked a cigarette. Taeyeon wonders whether

this national result holds true in his large, urban high school. For his AP Statistics

class project, Taeyeon surveys an SRS of 150 students from his school. He gets

responses from all 150 students, and 90 say that they have never smoked a

cigarette. What should Taeyeon conclude? Give appropriate evidence to support

your answer.

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State: We want to perform at test at the α = 0.05 significance level of

H0: p = 0.50

Ha: p ≠ 0.50

where p is the actual proportion of students in Taeyeon’s school who would say

they have never smoked cigarettes.

Plan: If conditions are met, we should do a one-sample z test for the population

proportion p.

Random Taeyeon surveyed an SRS of 150 students from his school.

Normal Assuming H0: p = 0.50 is true, the expected numbers of smokers and

nonsmokers in the sample are np0 = 150(0.50) = 75 and n(1 - p0) = 150(0.50) =

75. Because both of these values are at least 10, we should be safe doing

Normal calculations.

Independent We are sampling without replacement, we need to check the

10% condition. It seems reasonable to assume that there are at least 10(150) =

1500 students a large high school.

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Two-Sided TestsTe

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Conclude: Since our P-value, 0.0142, is less than the chosen significance

level of α = 0.05, we have sufficient evidence to reject H0 and conclude that

the proportion of students at Taeyeon’s school who say they have never

smoked differs from the national result of 0.50.

Do: The sample proportion is

ˆ p 60/150 0.60.

Test statistic z ˆ p p0

p0(1 p0)

n

0.600.50

0.50(0.50)

150

2.45

P-value To compute this P-value, we

find the area in one tail and double it.

Using Normalcdf(2.45, 100) yields P(z ≥

2.45) = 0.0071 (the right-tail area). So

the desired P-value is

2(0.0071) = 0.0142.

+ Checkpoint According to the National Institute for Occupational Safety and

Health, job stress poses a major threat to the health of workers.

In a national survey of restaurant employees, 75% said that

work stress had a negative impact on their personal lives.

Managers of a large restaurant claim wonder whether this

national result holds for their employees. A random sample of

100 employees finds that 68 answer “Yes” when asked, “Does

work stress have a negative impact on your personal life?” Is

this good reason to think that the proportion of all employees in

this claim who would say “Yes” differs from the national

proportion of 0.75? Support your answer with a significance

test.

State: We want to perform a test at the α = 0.05

significance level of Ho: p = 0.75 versus Ha: p ≠ 0.75,

where p is the actual proportion of restaurant employees

who say that work stress has a negative impact on their

personal lives.

Plan: If conditions are met, we should do a one-sample

z test for the population proportion p.

Random: The sample was randomly selected.

Normal: np = 75 and n(1 – p) = 25.

Independent: There were 100 in the sample; since it

is a “large restaurant chain” it is safe to assume there

are more than 1000 workers.

Do: The sample proportion is p-hat = 0.68. The

corresponding test statistic is z = -1.62. Since this is a

two sided test, the P-value is 0.1052.

Conclude: Since our P-value is greater than 0.05, we fail to

reject the null hypothesis, meaning we do not have enough

evidence to conclude that the proportion of restaurant

employees at this large restaurant chain who say that work

stress has a negative impact on their personal lives is

different from the national proportion of 0.75.

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Why Confidence Intervals Give More InformationTe

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The result of a significance test is basically a decision to reject H0 or fail

to reject H0. When we reject H0, we’re left wondering what the actual

proportion p might be. A confidence interval might shed some light on

this issue.

Taeyeon found that 90 of an SRS of 150 students said that they had never

smoked a cigarette. Before we construct a confidence interval for the

population proportion p, we should check that both the number of

successes and failures are at least 10.

The number of successes and the number of failures in the sample

are 90 and 60, respectively, so we can proceed with calculations.

Our 95% confidence interval is:

We are 95% confident that the interval from 0.522 to 0.678 captures the

true proportion of students at Taeyeon’s high school who would say

that they have never smoked a cigarette.

ˆ p z *ˆ p (1 ˆ p )

n 0.60 1.96

0.60(0.40)

150 0.60 0.078 (0.522,0.678)

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Confidence Intervals and Two-Sided TestsTe

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There is a link between confidence intervals and two-sided tests. The 95%

confidence interval gives an approximate range of p0’s that would not be rejected

by a two-sided test at the α = 0.05 significance level. The link isn’t perfect

because the standard error used for the confidence interval is based on the

sample proportion, while the denominator of the test statistic is based on the

value p0 from the null hypothesis. A two-sided test at significance

level α (say, α = 0.05) and a 100(1 –

α)% confidence interval (a 95%

confidence interval if α = 0.05) give

similar information about the

population parameter.

If the sample proportion falls in the

“fail to reject H0” region, like the

green value in the figure, the

resulting 95% confidence interval

would include p0. In that case, both

the significance test and the

confidence interval would be unable

to rule out p0 as a plausible

parameter value.

However, if the sample proportion

falls in the “reject H0” region, the

resulting 95% confidence interval

would not include p0. In that case,

both the significance test and the

confidence interval would provide

evidence that p0 is not the

parameter value.

+ Checkpoint

1) James, a starting player for a major college basketball team,

made only 40% of his free throws last season. During the

summer, he worked on developing a softer shot in hopes of

improving his free throw accuracy. In the first 8 games of this

season, James made 25 free throws in 40 attempts. You want

to investigate whether James’ work over the summer will result

in a higher proportion of free-throw successes this season.

What conclusion would you draw at the α = 0.01 level about

James’ free throw shooting? Justify your answer with a

complete significance test.

State: We wish to test Ho: p = 0.4 versus Ha: p > 0.4,

where p = James’ new free throw accuracy, expressed

as the true proportion of all shots he will take this season

that he’ll make. We will use a α = 0.01 significance

level.

Plan: The procedure is a one-sample z test for a

proportion.

Random: We will have to assume that James first 40

attempts this season are a SRS.

Normal: np = 16 and n(1 – p) = 24

Independent: We are assuming that the success of

each free throw is independent from all free throw

attempts.

Do:

0019.0

29.0

40

)6.0)(4.0(

4.0625.0

625.040

25ˆ

valueP

z

pConclude: A P-value of

0.0018 is less than α = 0.01,

so we reject Ho and conclude

that there is convincing

evidence that James free-

throw shooting percentage

has improved.

+ Checkpoint

Is the ratio of male births to female births even? A simple random

sample of births in a major metropolitan area found 1345 boys

among 2546 firstborn children. A 99% confidence interval for p

= the proportion of male births in this population is given by

(0.5028, 0.5538).

2) Use the confidence interval to draw a conclusion about the

hypothesis Ho: p = 0.5 against & Ha: p ≠ 0.5. Be sure to

indicate the appropriate significance level.

Since the 99% confidence interval does not contain the

null value of 0.5, we can reject Ho at the α = 0.01 level.

We have convincing evidence that the proportion of males

births is different from the proportion of female births.

+ 3) What information is provided by the confidence interval that

would not be provided by a test of significance alone?

The confidence interval gives a range of plausible

values for the proportion of male births; the test of

significance does not.

+Section 9.2

Tests About a Population Proportion

In this section, we learned that…

As with confidence intervals, you should verify that the three conditions—

Random, Normal, and Independent—are met before you carry out a

significance test.

Significance tests for H0 : p = p0 are based on the test statistic

with P-values calculated from the standard Normal distribution.

The one-sample z test for a proportion is approximately correct when

(1) the data were produced by random sampling or random assignment;

(2) the population is at least 10 times as large as the sample; and

(3) the sample is large enough to satisfy np0 ≥ 10 and n(1 - p0) ≥ 10 (that is,

the expected numbers of successes and failures are both at least 10).

Summary

z ˆ p p0

p0(1 p0)

n

+Section 9.2

Tests About a Population Proportion

In this section, we learned that…

Follow the four-step process when you carry out a significance test:

STATE: What hypotheses do you want to test, and at what significance level?

Define any parameters you use.

PLAN: Choose the appropriate inference method. Check conditions.

DO: If the conditions are met, perform calculations.

• Compute the test statistic.

• Find the P-value.

CONCLUDE: Interpret the results of your test in the context of the problem.

Confidence intervals provide additional information that significance tests do

not—namely, a range of plausible values for the true population parameter p. A

two-sided test of H0 : p = p0 at significance level α gives roughly the same

conclusion as a 100(1 – α)% confidence interval.

Summary