Download Unit 12

CHAPTER 12
Asking and Answering
Questions About A
Population Mean
Created by Kathy Fritz
LET'S REVIEW SOME STATISTICAL
NOTATION.
n
𝒙
s
m
s
When the purpose of a statistical study is to
learn about a population mean m, the sample
mean 𝑥 can be used as an estimate of m.
To understand statistical inference procedures
based on 𝑥, you must first study how sampling
variability causes 𝑥 to vary in value from one
sample to another.
The sample size n and characteristics of the
population (its shape, mean value m, and
standard deviation s) are important in
determining the sampling distribution of 𝑥.
THE SAMPLING DISTRIBUTION
OF THE SAMPLE MEAN
Properties of the Sampling
Distribution of 𝑥
Central Limit Theorem
The paper “Mean Platelet Volume in Patients with
Metabolic Syndrome and Its Relationship with Coronary
Artery Disease” (Thrombosis Research, 2007) includes
data that suggests that the distribution of
x = platelet volume
for patients who do not have metabolic syndrome is
approximately normal with mean m = 8.25 and standard
deviation s = 0.75.
Platelet Volume Continued . . .
The paper “Is the Overtime Period in an NHL Game Long
Enough?” (American Statistician, 2008) gave data on the
time (in minutes) from the start of the game to the first goal
scored for the 281 regular season games from the 20052006 season that went into overtime. The density histogram
for the data is shown below.
Let’s consider
these 281 values
as a population.
The distribution is
strongly positively
skewed with mean
m = 13 minutes
and with a
median of 10
minutes.
GENERAL PROPERTIES OF
SAMPLING DISTRIBUTIONS OF 𝑥
Rule 1:
Rule 2:
This rule is exact if the population is infinite, and is
approximately correct if the population is finite
and no more than 10% of the population is
included in the sample.
GENERAL PROPERTIES CONTINUED. . .
Rule 3:
GENERAL PROPERTIES CONTINUED . . .
Rule 4: Central Limit Theorem
n = 16
n=4
Population
In a study of the courtship of mating scorpion flies, one
variable of interest was x = courtship time, which was
defined as the time from the beginning of a femalemale interaction until mating. Data suggest that it is
reasonable to think that the population mean and
standard deviation of 𝑥 are m = 117.1 minutes and
s = 109.1 minutes.
The sampling distribution of 𝑥 = mean courtship time for
a random sample of 20 scorpion fly mating pairs would
have mean
𝜇𝑥 = 𝜇 = 117.1 minutes
The standard deviation of 𝑥 is
𝜎
109.1
𝜎𝑥 =
=
= 24.40
𝑛
20
It is not reasonable to assume that
the shape of the sampling
distribution of 𝑥 is normal.
A hot dog manufacturer claims that one of its brands of hot
dogs has a mean fat content of m = 18 grams per hot dog.
Consumers of this brand would probably not be disturbed if
the mean was less than 18 grams, but would be unhappy if it
exceeded 18 grams.
In this situation, the variable of interest is
x = fat content of a hot dog
For purposes of this example, suppose we know that s, the
standard deviation of the x distribution, is equal to 1 gram.
An independent testing organization is asked to analyze a
random sample of 36 hot dogs. The fat content for each of
the 36 hot dogs is measured and the sample mean is
calculated to be 𝑥 = 18.4 grams.
Does this result suggest that the manufacturer’s
claim that the population mean is 18 is
incorrect?
Hot Dogs Continued . . .
Let’s look at the sampling distribution of 𝑥 :
• The sample size, n = 36, is large enough to say that the
sampling distribution of 𝑥 will be approximately
normal.
• The standard deviation of the 𝑥 distribution is
𝜎
1
𝜎𝑥 =
=
= 0.1667
𝑛
36
• If the manufacturer’s claim is correct, you also know
that
𝜇𝑥 = 𝜇 = 18
Hot Dogs Continued . . .
You know that even if m = 18, 𝑥 will not usually be exactly
18 due to sampling variability. But, is it likely that you
would see a sample mean at least as large as 18.4 when
the population mean is really 18?
Using the normal distribution, you can compute the
probability of observing a sample mean this large. If the
manufacturer’s claim is correct,
18.4 − 18
𝑃(𝑥 ≥ 18.4) ≈ 𝑃 𝑧 ≥
0.1667
= 𝑃 𝑧 ≥ 2.4 = 0.0082
Values of 𝑥 as large as 18.4 will be
observed only about 0.82% of the time
when a random sample of size 36 is taken
from a population with m = 18 and s = 1.
A CONFIDENCE INTERVAL FOR A
POPULATION MEAN
t-distributions
A One-Sample t Confidence
Interval for m
CONFIDENCE INTERVALS FOR M
WHEN S IS KNOWN
The general formula for a confidence interval estimate is
standard error
statistic ± (critical value)
of the statistic
From Section 12.1, you know that:
1. The sampling distribution of 𝑥 is centered at m.
𝜎
2. The standard deviation of 𝑥 is 𝜎𝑥 = .
𝑛
3. As long as n is large (n ≥ 30), the sampling distribution of
𝑥 is approximately normal.
This suggests that a confidence interval for a
population mean when the sample size is large and s
is known is . . .
Cosmic radiation levels rise with increasing
altitude, promoting researchers to consider how
pilots and flight crews might be affected by
increased exposure to cosmic radiation. A study
reported a mean annual cosmic radiation dose of 219
mrem for a sample of flight personnel of Xinjiang Airlines.
Suppose this mean is based on a random sample of 100
flight crew members and that s = 35 mrems. Let:
m = mean annual cosmic radiation exposure for all
Xinjiang Airlines flight crew members
A 95% confidence interval for m is . . .
𝑥 ± 𝑧 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝜎
35
= 219 ± (1.96)
𝑛
100
= (212.15, 225.86)
IMPORTANT PROPERTIES OF T
DISTRIBUTIONS
1) The t distribution corresponding to any particular
number of degrees of freedom is bell shaped
and centered at zero (just like the standard
normal (z) distribution).
2) Each t distribution is more spread out than the
standard normal distribution.
z curve
t curve for 2 df
0
IMPORTANT PROPERTIES OF T
DISTRIBUTIONS CONTINUED . . .
3) As the number of degrees of freedom increases,
the spread of the corresponding t distribution
decreases.
t curve for 8 df
t curve for 2 df
0
IMPORTANT PROPERTIES OF T
DISTRIBUTIONS CONTINUED . . .
4) As the number of degrees of freedom increases,
the corresponding sequence of t distributions
approaches the standard normal distribution.
z curve
t curve for 2 df
t curve for 5 df
0
FINDING T CRITICAL VALUES
Table 3 t Critical Values
Central area captured / Confidence level
df
.80
80%
.90
90%
.95
95%
.98
98%
.99
99%
.998
99.8%
.999
99.9%
1
3.08
6.31
12.71
31.82
63.66
318.31
636.62
2
1.89
2.92
4.30
6.97
9.93
23.53
31.60
3
1.64
2.35
3.18
4.54
5.84
10.21
12.92
4
1.53
2.13
2.78
3.75
4.60
7.17
8.61
5
1.48
2.02
2.57
3.37
4.03
5.89
6.86
6
1.44
4.94
2.45
3.14
3.71
5.21
5.96
7
1.42
4.90
2.37
3.00
3.50
4.79
5.41
8
1.40
1.86
2.31
2.90
3.36
4.50
5.04
9
1.38
1.83
2.26
2.82
3.25
4.30
4.78
10
1.37
1.81
2.23
2.76
3.17
4.14
4.59
ONE-SAMPLE T CONFIDENCE INTERVAL
FOR A POPULATION MEAN M
Appropriate when the following conditions are met:
• The sample is a random sample from the population of
interest or the sample is selected in a way that results in a
sample that is representative of the population.
•
The sample size is large (n ≥ 30) or the population
distribution is normal.
When these conditions are met, a confidence interval
for the population mean is
𝑠
𝑥 ± (𝑡 critical value)
𝑛
The t critical value is based on df = n – 1 and the desired
confidence level.
ONE-SAMPLE T CONFIDENCE INTERVAL
FOR A POPULATION MEAN M
CONTINUED …
Interpretation of Confidence Interval
You can be confident that the actual value of the
population mean is included in the computed interval.
In a given problem, this statement should be worded in
context.
Interpretation of Confidence Level
The confidence level specifies the long-run proportion
of the time that this method is expected to be
successful in capturing the actual population mean.
During a flu outbreak, many people visit emergency rooms.
Before being treated, they often spend time in crowded
waiting rooms where other patients may be exposed. A
study was performed investigating a drive-through model
where flu patients are evaluated while they remain in their
cars.
In the study, 38 people were each given a scenario for a flu
case that was selected at random from the set of all flu cases
actually seen in the emergency room. The scenarios
provided the “patient” with a medical history and a
description of symptoms that would allow the patient to
respond to questions from the examining physician.
The patients were processed using a drive-through procedure
that was implemented in the parking structure of Stanford
University Hospital. The time to process each case from
admission to discharge was recorded.
The following sample statistics were computed from
the data:
n = 38
𝒙 = 26 minutes
s = 1.57 minutes
Drive-through Model Continued . . .
The following sample statistics were computed from the data:
n = 38
𝑥 = 26 minutes
s = 1.57 minutes
Step 1 (Estimate):
You want to estimate the value of m, the mean time to
process a flu case using the new drive-through model.
Step 2 (Method):
Because the answers to the four key questions are: 1)
estimation, 2) sample data, 3) one numerical variable,
and 4) one sample, consider using a one-sample t
confidence interval for a population mean. A
confidence level of 95% was specified for this
example.
Drive-through Model Continued . . .
The following sample statistics were computed from the data:
n = 38
𝑥 = 26 minutes
s = 1.57 minutes
Step 3 (Check):
• Because the sample size is large (38 > 30), you do not
need to worry about whether the population
distribution is approximately normal.
• The sample is a random sample because the 38 flu
cases were randomly selected from the population
of all flu cases seen at the emergency room
Step 4 (Calculate):
t critical value (from Table 3) = 2.02
𝑥 ± 𝑡 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑠
1.57
= 26 ± 2.02
𝑛
38
= (25.486, 26.514)
Drive-through Model Continued . . .
The following sample statistics were computed from the data:
n = 38
𝑥 = 26 minutes
s = 1.57 minutes
Step 5 (Communicate Results):
Confidence Interval
You can be 95% confident that the actual mean
processing time for emergency room flu cases using the
new drive-through model is between 25.486 minutes
and 26.514 minutes
Confidence Level
The method used to construct this interval
estimate is successful in capturing the actual
value of the population mean about 95% of the
time.
In a study, seven chimpanzees learned to use an
apparatus that dispensed food when either of two
ropes was pulled.
When one of the ropes was pulled, only the chimp
controlling the apparatus received food. When the other
rope was pulled, food was dispensed both to the chimp
controlling the apparatus and also to a chimp in the
adjoining cage.
The accompanying data represent the number of times
out of 36 trials that each of seven chimps chose the
option that would provide food to both chimps.
23 22 21 24 19 20 20
Construct a 99% confidence interval for the mean
number of times out of 36 trials chimpanzees would
choose the option that would provide food to both
chimps.
Chimp Problem Continued . . .
Step 1 (Estimate):
The mean number of times out of 36 chimps choose the
charitable response, m, will be estimated.
Step 2 (Method):
Because the answers to the four key questions are
estimation, sample data, one numerical variable, and
one sample, a one-sample t confidence interval for a
population mean will be considered. A confidence
level of 99% was specified.
Chimp Problem Continued . . .
Step 3 (Check):
1. It was stated that it is reasonable to regard the
sample as representative of the population
2. The normal probability
plot is reasonably
straight, so it seems
plausible that the
population distribution
is approximately
normal.
Chimp Problem Continued . . .
Step 4
(Calculation):
𝑥 = 21.29 𝑠 = 1.80
𝑥 ± 𝑡 critical value
𝑠
= 21.29 ± 3.71
𝑛
= (18.77, 23.81)
1.80
7
Step 5 (Communicate Results):
Based on this sample, you can be 99% confident that the
population mean number of charitable responses (out of
36 trials) is between 18.76 and 23.81.
CHOOSING A SAMPLE SIZE
The sample size required to estimate a
population mean m with a specified margin of
error M is
If the value of s is unknown, it may be estimated
based on previous information or, for a population
range
that is not skewed, by using
.
4
A college financial advisor wants to estimate the
mean cost of textbooks per quarter for students
at the college. For the estimate to be useful, it
should have a margin of error of $20 or less.
How large a sample should be used to be
confident of achieving this level of accuracy?
Suppose the financial advisor thinks that the
amount spent on books varies widely, but that
most values are between $150 to $550.
A reasonable estimate of s is :
𝑟𝑎𝑛𝑔𝑒 550 − 150
=
= 100
4
4
Using this estimate of the population standard
deviation, the required sample size is:
2
2
1.96𝜎
1.96(100)
𝑛=
=
= 96.04
𝑀
4
TESTING HYPOTHESES ABOUT A
POPULATION MEAN
HYPOTHESES:
When testing hypotheses about a population
mean, the null hypothesis will have the form:
where m0 is a particular hypothesized value.
The alternative hypothesis has one of the
following three forms, depending on the
research question being addressed:
TEST STATISTIC
If n is large (n ≥ 30) or if the population
distribution is approximately normal, the
appropriate test statistic is
If the null hypothesis is true, this test statistic has a
t distribution with df = n – 1.
This means that the P-value for a hypothesis test about a
population mean will be based on a t distribution and
not the standard normal distribution.
COMPUTING P-VALUES
1. Upper-tailed test:
Ha: m > m0
t curve
2. Lower-tailed test:
Ha: m < m0
P-value = area
in lower tail
P-value = area
in upper tail
Calculated t
3. Two-tailed test:
Ha: m ≠ m0
t curve
Calculated -t
t curve
P-value = sum
of area in
two tails
Calculated –t and t
THE ONE-SAMPLE T-TEST FOR A
POPULATION MEAN
Appropriate when the following conditions are met:
1. The sample is a random sample from the population of
interest or the sample is selected in a way that results in a
sample that is representative of the population.
2. The sample size is large (n ≥ 30) or the population
distribution is normal.
When these conditions are met, the following test statistic
can be used:
𝑥 − 𝜇0
𝑡= 𝑠
𝑛
Where m0 is the hypothesized value from the null hypothesis.
THE ONE-SAMPLE T-TEST FOR A
POPULATION MEAN CONTINUED . .
.
Null hypothesis:
H0 : m = m 0
When the conditions are met and the null hypothesis is true, the t test
statistic has a t distribution with df = n – 1.
When the Alternative
Hypothesis Is . . .
The P-value Is . . .
Area under the t curve to the right of
Ha: m > m0
the calculated value of the test statistic
Ha: m < m0
Area under the t curve to the left of the
calculated value of the test statistic
Ha: m ≠ m0
2·(area to the right of t) if t is positive
Or
2·(area to the left of t) if t is negative
A study conducted by researchers at Pennsylvania State
University investigated whether time perception, an
indication of a person’s ability to concentrate, is impaired
during nicotine withdrawal. After a 24-hour smoking
abstinence, 20 smokers were asked to estimate how
much time had elapsed during a 45-second period.
Researchers wanted to see whether smoking abstinence
had a negative impact on time perception, causing
elapsed time to be overestimated. Suppose the resulting
data on perceived elapsed time (in seconds) were as
follows:
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
n = 20
𝑥 = 59.30
s = 9.84
Smoking Abstinence Continued . . .
Step 1 (Hypotheses):
The population mean is
m = mean perceived elapsed time for smokers who
have abstained from smoking for 24 hours
Null hypothesis:
Alternative hypothesis:
H0: m = 45
Ha: m > 45
Step 2 (Method):
Because the answers to the four key questions are:
1) hypothesis test, 2) sample data, 3) one numerical
variable, and 4) one sample, consider a one-sample t
test for a population mean. When the null hypothesis is
true, this statistic will have a t distribution with df = 20 – 1 =
19.
Significance level: a = 0.05
Smoking Abstinence Continued . . .
Step 3 (Check):
• The researchers conducting the study indicated that
they believed that the sample was selected in a way
that would result in a sample that was representative of
all smokers in general.
• Because n is only 20 in this example, you need to verify
that the normality condition is reasonable.
A boxplot of the sample data is
shown here. Although the
boxplot is not perfectly
symmetric, it is not too skewed
and there are no
outliers. It is reasonable to think that the population
distribution is at least approximately normal.
Smoking Abstinence Continued . . .
Step 4 (Calculate):
n = 20
Test statistic:
𝑥 = 59.30
𝑡=
s = 9.84
59.30−45
9.84
20
= 6.50
This is an upper-tailed test, so the P-value is the area under the t curve
with df = 19 to the right side of the computed t value.
Associated P-value:
P-value = area under t curve to the right of 6.50
= P(t > 6.50) ≈ 0
Smoking Abstinence Continued . . .
Step 5 (Communicate Results):
Decision:
0 < 0.05, Reject H0
Conclusion: There is convincing evidence that the mean
perceived time elapsed is greater than the actual time
elapsed of 45 seconds. It would be very unlikely to see a
sample mean this extreme just by chance when H0 is
true.
STATISTICAL VERSUS PRACTICAL
SIGNIFICANCE
Carrying out a hypothesis test amounts to deciding
whether the value obtained for the test statistic could
plausibly have resulted when H0 is true.
When the value of the test statistic leads to rejection of
H0, it is customary to say that the result is statistically
significant a the chosen significance level a.
However, statistical significance does NOT mean that
the true situation differs from what the null states in any
practical sense.
See the following example.
Let m denote the actual mean score on a standardized
test for children in a large school district. The mean
score for all children in the United States is known to be
100. District administrators are interested in testing
H0: m = 100 versus Ha: m >100 using a significance level of
a = 0.001. Data from a random sample of 2500 children
in the district resulted in 𝑥 = 101.0 and s = 15.0. Minitab
output from a one-sample t test is shown here:
One-Sample T
Test of mu = 100 vs > 100
n
Mean
StDev
SEMean
95%
Lower
Bound
2500
101.000
15.000
0.300
100.506
T
P
3.33
0.000
From the Minitab output, P-value ≈ 0, so H0 is rejected.
But, there is only a difference of 1 between the sample
mean of 101 and the population mean of 100.
AVOID THESE COMMON
MISTAKES
AVOID THESE COMMON MISTAKES
1. You will need to think about the distinction
between proportions and means when
choosing an appropriate method.
The best way to distinguish which method to use is to
focus on the type of data -
Think
Numerical
Think
Categorical
Proportions
Means
AVOID THESE COMMON MISTAKES
2. Be sure to keep in mind that conditions are
important. Use of the one-sample t
confidence interval and hypothesis test
REQUIRE that conditions are met.
Be sure to check these conditions are met
before using these methods.
AVOID THESE COMMON MISTAKES
3. Remember that the results of a hypothesis test
can never provide strong support for the null
hypothesis. Make sure that you don’t confuse
“I am not convinced that the null is false” with
the statement “I am convinced that the null
hypothesis is true”. These are not the same!