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Chapter 18
Inferences About
Means
Copyright © 2014, 2012, 2009 Pearson Education, Inc.
1
Objectives
Perform a t-test for the population mean, to include:
writing appropriate hypotheses, checking the
necessary assumptions, drawing an appropriate
diagram, computing the P-value, making a decision,
and interpreting the results in the context of the
problem.
62. Compute and interpret in context a t-based confidence
interval for the population mean, checking the
necessary assumptions.
61.
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18.1
Getting Started:
The Central Limit
Theorem (Again)
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The Central Limit Theorem (CLT)
When a random sample is drawn from a population with
mean m and standard deviation s, the sampling
distribution has:
•
Mean: m
•
Standard deviation:
•
Approximately Normal distribution as long as the
sample size is large.
•
The larger the distribution, the closer to Normal.
s
n
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Weight of Angus Cows
The weight of Angus cows is Normally distributed with
m = 1309 pounds and s = 157 lbs. What does the CLT
say about the mean weight of a random sample of 100
Angus cows?
• Sample means y will average 1309 pounds.
•
SD( y ) 
s
n

157
100
 15.7 pounds
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Weight of Angus Cows (Continued)
•
The CLT says the sampling distribution will be
approximately Normal: N(1309, 15.7).
For the means of all random samples,
• 68% will be between 1293.3 and 1324.7 lb.
• 95% will be between 1277.6 and 1340.4 lb.
• 99.7% will be between 1261.9 and 1356.1 lb.
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The Challenge of the CLT
s
•
CLT tells us SD( y ) 
•
We would like to use this for Confidence Intervals
and Hypothesis Testing.
n
Unfortunately, we almost never know s.
s
• Using s almost works: SE ( y ) 
, but not quite.
n
• When using s, the Normal model has some error.
•
•
William Gosset came up with new models, one for
each n that works better.
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18.2
Gosset’s t
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Gosset the Brewer
At Guinness, Gosset experimented with beer.
• The Normal Model was not right, especially
for small samples.
• Still bell shaped, but details differed, depending on n
• Came up with the “Student’s t Distribution” as the
correct model
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Degrees of Freedom
•
For every sample size n there is a different
Student’s t distribution.
•
Degrees of freedom: df = n – 1.
•
Similar to the “n – 1” in the formula for sample
standard deviation
•
It is the number of independent quantities left after
we’ve estimated the parameters.
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Confidence Interval for Means
Sampling Distribution Model for Means
• With certain conditions (seen later), the standardized
sample mean follows the Student’s t model
with n – 1 degrees of freedom.
y m
t
SE ( y )
•
We estimate the standard deviation with
SE ( y ) 
s
n
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One Sample t-Interval for the Mean
•
When the assumptions are met (seen later), the
confidence interval for the mean is
y  t n* 1  SE ( y )
•
*
n 1
The critical value t
depends on the confidence
level, C, and the degrees of freedom n – 1.
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Contaminated Salmon
A study of mirex concentrations in salmon found
• n = 150, y = 0.0913 ppm, s = 0.0495 ppm
• Find a 95% confidence interval for mirex
concentrations in salmon.
• df = 150 – 1 = 149
• SE ( y ) 
•
0.0495
150
 0.0040
*
t149
 1.976
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Contaminated Salmon
Confidence Interval for m
•
*
y  t149
 SE ( y )  0.0913  1.976(0.0040)
= 0.0913 ± 0.0079
= (0.0834, 0.0992)
•
I’m 95% confident that the mean level of mirex
concentration in farm-raised salmon is between
0.0834 and 0.0992 parts per million.
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Thoughts about z and t
•
•
The Student’s t distribution:
• Is unimodal.
• Is symmetric about its mean.
• Has higher tails than Normal.
• Is very close to Normal for large df.
• Is needed because we are using s as an
estimate for s.
If you happen to know s, which almost never
happens, you could use the Normal model and not
Student’s t (but we won’t do this in our course).
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Assumptions and Conditions (know these!)
Independence Condition
• Randomization Condition: The data should arise
from a suitably randomized experiment.
•
Sample size < 10% of the population size.
Nearly Normal
• For large sample sizes (n > 40), not severely
skewed.
•
(15 ≤ n ≤ 40): Need unimodal and symmetric.
•
(n < 15): Need almost perfectly normal.
•
Check with a histogram.
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Assumptions for Salmon Contamination
Study
Researchers tested 150 salmon from 51 farms in eight
regions in six countries. Are the
assumptions and conditions for
inference satisfied?
 Independence Assumption:
Not a random sample, but likely
independently selected.
 Nearly
Normal Condition:
The histogram is unimodal and not highly skewed.
OK since sample size is 51.
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Using TI-83+
Stat -> Tests -> #8 Tinterval
• A sample of 256 residents of a city indicated a monthly mean
electric bill of $90 with a standard deviation of $24. Construct a
95% confidence interval for the mean monthly electric bill of all
residents.
• Stat…Tests…#8 for T Interval…enter
• For inpt: choose Stats and fill in the rest of the screen with the
inputted stats.
• You should find the interval is $87.05 - $92.95. We are 95%
confident that the mean electric monthly bill for residents is
between $87.05 and $92.95.
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Using TI-83+
Stat -> Tests -> #8 Tinterval
Use this when the population (true) sigma is unknown. Relies on sample mean
and sample standard deviation
The survival times in weeks are given for 20 male rats which were exposed to a high
level of radiation.
152
152
115
109
137
88
94
77
160
165
20
128
123
136
101
62
152
83
69
125
Determine the 95% confidence interval on the mean survival time for rats.
First enter your data in L1 (if you have summary stats skip this)
Tinterval:
•
Inpt: Data
•
List: L1
•
Freq: 1
•
C-level .95
•
Result:94.57 weeks to 130.23 weeks.
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23. Suppose the elapsed time of airline itineraries between Washington, D.C. and
Boston is normally distributed with an unknown population mean and an
unknown population standard deviation. Further suppose that a sample of size
25 (therefore, n=25 and degrees of freedom=24) was taken and the following
statistics were gotten from the sample: sample mean (ybar) =135 and sample
standard deviation (s) = 40. Construct confidence intervals around the sample
mean corresponding to the following confidence levels (express the lower and
upper bounds of the intervals as INTEGERS):
Stat -> Tests -> Tinterval (here we’ll input stats instead of raw data)
Inpt: Stats: x(bar): 135, Sx: 40, n:25, C-level: try .80, .90, .95, .98, & .99
a. 80% (124, 146)
b. 90% (121, 149)
c. 95% (118, 152)
d. 98% (115, 155)
e. 99% (113, 157)
Hint: Your intervals should get wider and wider and should all be centered around
135.
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How Much Sleep do
College Students Get?
n = 25, µ= 6.64, s = 1.075
Build a 90% Confidence Interval
for the Mean.
Think →
• Plan: Data on 25 Students
Model →
• Randomization Condition
The data are from a random survey.
• Nearly Normal Condition
Unimodal and slightly skewed, so OK
• Use Student’s t-Model with df = 25 – 1 = 24.
• One-sample t-interval for the mean
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How Much Sleep?
Show →
• Mechanics: n = 25, y = 6.64, s = 1.075
• Stat->Tests-> TInterval
90% CI  6.64  0.368  (6.272, 7.008)
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How Much Sleep do
College Students Get?
Tell →
•
Conclusion: I’m 90 percent confident that the
interval from 6.272 and 7.008 hours contains the true
population mean number of hours that college
students sleep.
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Make a Picture, Make a Picture,
Make a Picture
Always Test the Normality Assumption
• Create a histogram to check for near normality.
• Good for seeing the nature of the deviations:
outliers, not symmetric, skewed
•
Also create a normal probability plot to see that it is
reasonably straight.
• Good for checking for normality
•
With technology at hand, there is no excuse not to
make these two plots.
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18.3
Interpreting
Confidence
Intervals
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What Not to Say
Don’t Say
• “90% of all students sleep between 6.272 and 7.008
hours each night.”
• The CI is for the mean sleep, not individual
students.
•
“We are 90% confident that a randomly selected
student will sleep between 6.272 and 7.008 hours
per night.”
• We are 90% confident about the mean sleep, not
an individual’s sleep.
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What Not to Say (Continued)
Don’t Say
• “The mean amount of sleep is 6.64 hours 90% of the
time.”
• The population mean never changes. Only sample
means vary from sample to sample.
•
“90% of all samples will have a mean sleep between
6.272 and 7.008 hours per night.”
• This interval does not set the standard for all other
intervals. This interval is no more likely to be
correct than any other.
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What You Should to Say
Do Say
• “90% of all possible samples will produce intervals
that actually do contain the true mean sleep.”
• This is the essence of the CI, but it may be too
formal for general readers.
•
“I am 90% confident that the true mean amount that
students sleep is between 6.272 and 7.008 hours
per night.”
• This is more personal and less technical.
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18.4
A Hypothesis Test
for the Mean
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One-Sample t-Test for the Mean
•
•
•
Assumptions are the same.
H0: m = m0
y  m0
t n 1 
SE ( y )
s
•
Standard Error of y : SE ( y ) 
•
When the conditions are met and H0 is true, the
statistic follows the Student’s t Model.
•
Use this model to find the P-value.
n
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Are the Salmon Unsafe?
EPA recommended mirex screening is 0.08 ppm.
• Are farmed salmon contaminated beyond the
permitted EPA level?
•
Recap: Sampled 150 salmon. Mean 0.0913 ppm,
Standard Deviation 0.0495 ppm.
•
H0: m = 0.08
•
HA: m > 0.08
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Are the Salmon Unsafe?
One-Sample t-Test for the Mean
•
n = 150, df = 149, y = 0.0913, s = 0.0495
• SE ( y ) 
0.0495
150
 0.0040, t149
0.0913  0.08

 2.825
0.0040
• P(t149 2.825)  0.0027
•
Since the P-value is so low, reject H0 and conclude
that the population mean mirex level does exceed
the EPA screening value.
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One sample T-test for the mean - TI-83+
Judy is a ad designer who designs the newspaper ads for the Giant grocery store. Electronic
counters at the entrance total the number of people entering the store. Before Judy was hired,
the mean number of people entering every day was 3018. Since she has started working at the
Giant the management thinks that this average has increased. A random sample of 42
business days gave an average of 3333 people entering the store daily with a standard
deviation of 287. Does this indicate that the average number of people entering the store every
day has increased? Use an alpha of 0.01.
Stat…. Tests…T-Test
•
Inpt: choose data or stats according to what is available (usually you use stats)
•
μ0 : stands for the hypothecated mean which is in your hypothesis In this case we are testing
against the previous 3018.
•
X bar is the sample mean of 3333
•
Sx is the standard deviation of 287
•
n: the number in your sample 42 days
•
μ : choose the notation used in the alternative hypothesis In this case we are looking for an
increase so choose >
•
Calculate
We get a p value of .000000006 which is very, very slight. Therefore, we reject the null hypothesis
and conclude that the average number of people entering the store has certainly increased.
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24. Suppose the elapsed time of airline itineraries between Washington, D.C. and
Boston is normally distributed with an unknown population mean and unknown
population standard deviation. Suppose we randomly sample 25 itineraries and the
sample average is calculated to be 135 minutes and the sample standard deviation is
calculated to be 40. Further suppose that we want to test the hypothesis that the true
population mean (μ) equals 150 minutes. Conduct 2-sided hypothesis tests with the
following alpha levels:
a. 0.20
b. 0.10 c. 0.05
d. 0.02
e. 0.01
H0: µ=150
HA: µ≠150
Test statistic (t) = -1.875
P-value = 0.0730
Conclusions:
• a. 0.20
• b. 0.10
• c. 0.05
• d. 0.02
• e. 0.01
Reject
Reject
Don’t Reject
Don’t Reject
Don’t Reject
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Do Students Get at Least Seven
Hours of Sleep? 25 Surveyed.
Think →
• Plan: Does the mean amount of sleep
exceed 7 hours?
• Hypotheses: H0: m = 7
HA: m > 7
• Model
 Randomization Condition: The
students were randomly and
independently selected .
 Nearly Normal Condition: unimodal and symmetric
• Use the Student’s t-model, df = 24
• One-sample t-test for the mean
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Do Students Get at Least Seven
Hours of Sleep? 25 Surveyed.
Show →
• Mechanics: n = 25, y = 6.64, s = 1.075
• Use StatTestsT-Test
P-value  P(t25  1.67)  0.054
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Do Students Get at Least Seven
Hours of Sleep? 25 Surveyed.
Tell →
• Conclusion: P-value = 0.054 says that
if students do sleep an average of 7 hrs., samples of
25 students can be expected to have an observed
mean of 6.64 hrs. or less about 54 times in 1000.
• With 0.05 cutoff, there is not quite enough evidence
to conclude that the mean sleep is less than 7.
• The 90% CI: (6.272, 7.008) contains 7.
• Collecting a larger sample would reduce the ME and
give us a better handle on the true mean hours of
sleep.
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The Relationship between Intervals and
Tests
Confidence Intervals
• Start with data and find plausible values for the
parameter.
•
Always 2-sided
Hypothesis Tests
• Start with a proposed parameter value and then use
the data to see if that value is not plausible.
•
2-sided test: Within the confidence interval means
fail to reject H0. P-value = 1 – C is the cutoff.
•
1-sided test: P-value = (1 – C)/2 is the cutoff.
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Sleep, Confidence Intervals and
Hypothesis Tests
90% Confidence interval: (6.272, 7.008)
• For a 2-tailed test with a 10% cutoff, any
6.272 ≤ m0 ≤ 7.008 would result in failing to reject H0.
•
For a 1-tailed test (“<”) with a 5% cutoff, any
m0 ≤ 7.008 would result in failing to reject H0.
•
For a 1-tailed test (“>”) with a 5% cutoff, any
6.272 ≤ m0 would result in failing to reject H0.
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The Special Case with Proportions
Confidence Intervals
• Use p̂ to calculate SE  pˆ  .
Hypothesis Tests
• Use p to calculate SD(p).
If SE  pˆ  and SD(p) are far from each other, the
relationship between the confidence interval and the
hypothesis test breaks down.
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25. Hoping to lure more shoppers downtown, a city builds a new
public parking garage. The city plans to pay for the structure
through parking fees. During a two month period (44 week
days) daily fees collected averaged $126 with a standard
deviation of $15. If a consultant claimed that the average daily
income would be $130, should we reject her claim using
alpha=0.10 (perform a 2-sided test)?
H0: µ=130
HA: µ≠130
Test statistic (t): -1.77
P-value: 0.08
Conclusion: Reject the null. It is likely that the consultant’s
claim is false.
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26. In 1998, the Nabisco Company announced a “1000 Chips
Challenge” claiming that every 18 ounce bag of Chips Ahoy
contained at least 1000 chocolate chips. Below are the counts
of chips in selected bags.
1219
1214 1087 1200 1419 1121 1325 1345 1244
1258 1356 1132 1191 1270 1295 1135
Perform a one-sided test (HA: µ>1000). What does this evidence
say about Nabisco’s claim (let alpha=0.05)?
H0: µ=1000
HA: µ>1000
Test statistic (t) = 10.1
P-Value = tiny #
Conclusion: We can reject the null hypothesis. It appears that
Nabisco’s claim is valid.
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27. When consumers apply for credit, their credit is rated using
FICO scores. A random sample of credit ratings is obtained,
and the FICO scores are summarized with these statistics:
n=18, ybar=660.3, s=95.9. Use an alpha of 0.05 and do a 2sided hypothesis test to test the claim that the mean credit
score (of the general population) is equal to 700 (Triola 2008).
H0: µ=700
HA: µ≠700
Test statistic (t) = -1.76
P-Value = 0.097
Conclusion: We cannot reject the null hypothesis. There is
not enough statistical evidence to conclude that the mean
is different than 700.
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28. Different cereals are randomly selected, and the sugar content is
obtained for each cereal, with the results given below for Cheerios,
Harmony, Smart Start, Cocoa Puffs, Lucky Charms, Corn Flakes,
Fruit Loops, Wheaties, Cap’n Crunch, Frosted Flakes, Apple Jacks,
Bran Flakes, Special K, Rice Krispies, Corn Pops, and Trix. Use an
alpha of 0.05 to test the claim of a cereal lobbyist that the mean of all
cereals is LESS than 0.3 g (Triola 2008).
0.03
0.24 0.30 0.47 0.43 0.07 0.47 0.13
0.44
0.39 0.48 0.17 0.13 0.09 0.45 0.43
H0: µ=0.3
HA: µ<0.3
Test statistic (t): -0.12
P-value: 0.45
Conclusion: Fail to reject the null. The statistical analysis does not
back up the lobbyist’s claim.
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What Can Go Wrong?
Don’t confuse proportions and means.
• When counting successes with a proportion, use the
Normal model.
• With quantitative data, use Student’s t.
Beware of multimodality.
• If the histogram is not unimodal, consider separating
into groups and analyzing each group separately.
Beware of skewed data.
• Look at the normal probability plot and histogram.
• Consider re-expressing if the data is skewed.
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What Can Go Wrong?
Set outliers aside.
• Outliers violate the Nearly Normal Condition.
• If removing a data value substantially changes the
conclusions, then you have an outlier.
• Analyze without the outliers, but be sure to include a
separate discussion about the outliers.
Watch out for bias.
• With bias, even a large sample size will not save
you.
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What Can Go Wrong?
Make sure cases are independent.
• Look for violations of the independence assumption.
Make sure the data are from an appropriately randomized
sample.
• Without randomization, both the confidence interval
and the p-value are suspect.
Interpret your confidence interval correctly.
• The CI is about the mean of the population, not the
mean of the sample, individuals in samples, or
individuals in the population.
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