Chapter 9: Testing a Claim - Interboro School District · 2017. 3. 23. · Chapter 9: Testing a...
Transcript of Chapter 9: Testing a Claim - Interboro School District · 2017. 3. 23. · Chapter 9: Testing a...
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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