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Chapter 13
Introduction to Inference
Essential Statistics
Chapter 13
1
Statistical Inference
 Provides
methods for drawing
conclusions about a population from
sample data
– Confidence Intervals
What
is the population mean?
– Tests of Significance
Is
the population mean larger than 66.5?
Essential Statistics
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Inference about a Mean
Simple Conditions
1.
2.
3.
SRS from the population of interest
Variable has a Normal distribution
N(m, s) in the population
Although the value of m is unknown,
the value of the population standard
deviation s is known
Essential Statistics
Chapter 13
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Confidence Interval
A level C confidence interval has two parts
1. An interval calculated from the data,
usually of the form:
estimate ± margin of error
2.
The confidence level C, which is the
probability that the interval will capture the
true parameter value in repeated samples;
that is, C is the success rate for the
method.
Essential Statistics
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Case Study
NAEP Quantitative Scores
(National Assessment of Educational Progress)
Rivera-Batiz, F. L., “Quantitative literacy and the likelihood of
employment among young adults,” Journal of Human
Resources, 27 (1992), pp. 313-328.
What is the average score for all young
adult males?
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Case Study
NAEP Quantitative Scores
The NAEP survey includes a short test of
quantitative skills, covering mainly basic
arithmetic and the ability to apply it to realistic
problems. Scores on the test range from 0 to
500, with higher scores indicating greater
numerical abilities. It is known that NAEP
scores have standard deviation s = 60.
Essential Statistics
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Case Study
NAEP Quantitative Scores
In a recent year, 840 men 21 to 25 years of
age were in the NAEP sample. Their mean
quantitative score was 272.
On the basis of this sample, estimate the
mean score m in the population of all 9.5
million young men of these ages.
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Case Study
NAEP Quantitative Scores
1.
2.
3.
To estimate the unknown population mean m,
use the sample mean x = 272.
The law of large numbers suggests that x
will be close to m, but there will be some error in
the estimate.
 distribution of x has the Normal
The sampling
distribution with mean m and 
standard deviation
s
60

 2.1
n
840

Essential Statistics
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Case Study
NAEP Quantitative Scores
Essential Statistics
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Case Study
NAEP Quantitative Scores
4.
The 68-95-99.7 rule
indicates that
x and m are within
two standard
deviations (4.2) of
each other in about
95% of all samples.
x  4.2 = 272  4.2 = 267.8
x + 4.2 = 272 + 4.2 = 276.2
Essential Statistics
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Case Study
NAEP Quantitative Scores
So, if we estimate that m lies within 4.2 of
we’ll be right about 95% of the time.
Essential Statistics
Chapter 13
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Confidence Interval
Mean of a Normal Population
Take an SRS of size n from a Normal population
with unknown mean m and known standard
deviation s. A level C confidence interval for m is:
σ
x z
n

z* is called the critical value, and z* and –z* mark off the
Central area C under a standard normal curve (next slide);
values of z* for many choices of C can be found at the
bottom of Table C in the back of the textbook, and the
most common values are on the next slide.
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Confidence Interval
Mean of a Normal Population
Confidence Level
C
Critical Value
z*
90%
1.645
95%
1.960
99%
2.576
Essential Statistics
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Case Study
NAEP Quantitative Scores
Using the 68-95-99.7 rule gave an approximate 95%
confidence interval. A more precise 95% confidence
interval can be found using the appropriate value of z*
(1.960) with the previous formula.
x  (1.960)(2. 1) = 272  4.116 = 267.884
x  (1.960)(2. 1) = 272  4.116 = 276.116
We are 95% confident that the average NAEP
quantitative score for all adult males is between
267.884 and 276.116.
Essential Statistics
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Careful Interpretation of a
Confidence Interval

“We are 95% confident that the mean NAEP score for
the population of all adult males is between 267.884
and 276.116.”
(We feel that plausible values for the population of males’ mean
NAEP score are between 267.884 and 276.116.)

** This does not mean that 95% of all males will have NAEP
scores between 267.884 and 276.116. **

Statistically: 95% of all samples of size 840 from the population
of males should yield a sample mean within two standard errors
of the population mean; i.e., in repeated samples, 95% of the
C.I.s should contain the true population mean.
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Reasoning of Tests of Significance
What
would happen if we repeated
the sample or experiment many
times?
How likely would it be to see the
results we saw if the claim of the test
were true?
Do the data give evidence against the
claim?
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Stating Hypotheses
Null Hypothesis, H0




The statement being tested in a statistical test
is called the null hypothesis.
The test is designed to assess the strength of
evidence against the null hypothesis.
Usually the null hypothesis is a statement of
“no effect” or “no difference”, or it is a statement
of equality.
When performing a hypothesis test, we
assume that the null hypothesis is true until
we have sufficient evidence against it.
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Stating Hypotheses
Alternative Hypothesis, Ha



The statement we are trying to find evidence for
is called the alternative hypothesis.
Usually the alternative hypothesis is a
statement of “there is an effect” or “there is a
difference”, or it is a statement of inequality.
The alternative hypothesis should express
the hopes or suspicions we bring to the
data. It is cheating to first look at the data
and then frame Ha to fit what the data show.
Essential Statistics
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Case Study I
Sweetening Colas
Diet colas use artificial sweeteners to avoid
sugar. These sweeteners gradually lose their
sweetness over time. Trained testers sip the
cola and assign a “sweetness score” of 1 to 10.
The cola is then retested after some time and the
two scores are compared to determine the
difference in sweetness after storage. Bigger
differences indicate bigger loss of sweetness.
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Case Study I
Sweetening Colas
Suppose we know that for any cola, the sweetness loss
scores vary from taster to taster according to a Normal
distribution with standard deviation s = 1.
The mean m for all tasters measures loss of sweetness.
The sweetness losses for a new cola, as measured by
10 trained testers, yields an average sweetness loss of
x = 1.02. Do the data provide sufficient evidence
that the new cola lost sweetness in storage?
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Case Study I
Sweetening Colas



If the claim that m = 0 is true (no loss of sweetness, on
average), the sampling distribution of x from 10 tasters
is Normal with m = 0 and standard deviation
σ
1

 0.316
n
10
The data yielded x = 1.02, which is more than three
standard deviations from
m = 0. This is strong evidence
that the new cola lost sweetness in storage.
If the data yielded x = 0.3, which is less than one
standard deviations from m = 0, there would be no
 that the new cola lost sweetness in storage.
evidence
Essential Statistics
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Case Study I
Sweetening Colas
Essential Statistics
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The Hypotheses for Means
Null:
H 0: m = m 0
One
sided alternatives
Ha: m >m0
Ha: m <m0
Two sided alternative
Ha: m m0
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Case Study I
Sweetening Colas
The null hypothesis is no average sweetness loss
occurs, while the alternative hypothesis (that which we
want to show is likely to be true) is that an average
sweetness loss does occur.
H0: m = 0
Ha: m > 0
This is considered a one-sided test because we are
interested only in determining if the cola lost sweetness
(gaining sweetness is of no consequence in this study).
Essential Statistics
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Case Study II
Studying Job Satisfaction
Does the job satisfaction of assembly workers
differ when their work is machine-paced rather
than self-paced? A matched pairs study was
performed on a sample of workers, and each
worker’s satisfaction was assessed after
working in each setting. The response variable
is the difference in satisfaction scores, selfpaced minus machine-paced.
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Case Study II
Studying Job Satisfaction
The null hypothesis is no average difference in scores in
the population of assembly workers, while the
alternative hypothesis (that which we want to show is
likely to be true) is there is an average difference in
scores in the population of assembly workers.
H0: m = 0
Ha: m ≠ 0
This is considered a two-sided test because we are
interested determining if a difference exists (the
direction of the difference is not of interest in this study).
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Test Statistic
Testing the Mean of a Normal Population
Take an SRS of size n from a Normal
population with unknown mean m and known
standard deviation s. The test statistic for
hypotheses about the mean (H0: m = m0) of a
Normal distribution is the standardized
version of x:
x  μ0
z
σ
n

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Case Study I
Sweetening Colas
If the null hypothesis of no average sweetness loss is
true, the test statistic would be:
x  μ0
1.02  0
z

 3.23
σ
1
10
n
Because the sample result is more than 3 standard
deviations above the hypothesized mean 0, it gives
strong evidence that the mean sweetness loss is not 0,
but positive.
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P-value
Assuming that the null hypothesis is true, the
probability that the test statistic would take a
value as extreme or more extreme than the
value actually observed is called the P-value
of the test.
The smaller the P-value, the stronger the
evidence the data provide against the null
hypothesis. That is, a small P-value indicates
a small likelihood of observing the sampled
results if the null hypothesis were true.
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P-value for Testing Means

Ha: m> m0


Ha: m< m0


P-value is the probability of getting a value as large or
larger than the observed test statistic (z) value.
P-value is the probability of getting a value as small or
smaller than the observed test statistic (z) value.
Ha: mm0

P-value is two times the probability of getting a value as
large or larger than the absolute value of the observed test
statistic (z) value.
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Case Study I
Sweetening Colas
For test statistic z = 3.23 and alternative hypothesis
Ha: m > 0, the P-value would be:
P-value = P(Z > 3.23) = 1 – 0.9994 = 0.0006
If H0 is true, there is only a 0.0006 (0.06%) chance that
we would see results at least as extreme as those in the
sample; thus, since we saw results that are unlikely if H0
is true, we therefore have evidence against H0 and in
favor of Ha.
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Case Study I
Sweetening Colas
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Case Study II
Studying Job Satisfaction
Suppose job satisfaction scores follow a Normal
distribution with standard deviation s = 60. Data from
18 workers gave a sample mean score of 17. If the null
hypothesis of no average difference in job satisfaction is
true, the test statistic would be:
x  μ0
17  0
z

 1.20
σ
60
n
18
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Case Study II
Studying Job Satisfaction
For test statistic z = 1.20 and alternative hypothesis
Ha: m ≠ 0, the P-value would be:
P-value = P(Z < -1.20 or Z > 1.20)
= 2 P(Z < -1.20) = 2 P(Z > 1.20)
= (2)(0.1151) = 0.2302
If H0 is true, there is a 0.2302 (23.02%) chance that we
would see results at least as extreme as those in the
sample; thus, since we saw results that are likely if H0 is
true, we therefore do not have good evidence against H0
and in favor of Ha.
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Case Study II
Studying Job Satisfaction
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Statistical Significance



If the P-value is as small as or smaller than the
significance level a (i.e., P-value ≤ a), then we say
that the data give results that are statistically
significant at level a.
If we choose a = 0.05, we are requiring that the data
give evidence against H0 so strong that it would occur
no more than 5% of the time when H0 is true.
If we choose a = 0.01, we are insisting on stronger
evidence against H0, evidence so strong that it would
occur only 1% of the time when H0 is true.
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Tests for a Population Mean
The four steps in carrying out a significance test:
1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Find the P-value.
4. State your conclusion in the context of the
specific setting of the test.
The procedure for Steps 2 and 3 is on the next page.
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Essential Statistics
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Case Study I
Sweetening Colas
1.
Hypotheses:
2.
Test Statistic:
H 0: m = 0
H a: m > 0
z
x  μ0
σ

1.02  0
1
n
3.
4.
 3.23
10
P-value: P-value = P(Z > 3.23) = 1 – 0.9994 = 0.0006
Conclusion:
Since the P-value is smaller than a = 0.01, there is very strong
evidence that the new cola loses sweetness on average during
storage at room temperature.
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Case Study II
Studying Job Satisfaction
1.
Hypotheses:
2.
Test Statistic:
H 0: m = 0
H a: m ≠ 0
z
x  μ0
σ

17  0
60
n
3.
4.
 1.20
18
P-value: P-value = 2P(Z > 1.20) = (2)(1 – 0.8849) = 0.2302
Conclusion:
Since the P-value is larger than a = 0.10, there is not sufficient
evidence that mean job satisfaction of assembly workers differs
when their work is machine-paced rather than self-paced.
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Confidence Intervals & Two-Sided Tests
A level a two-sided significance test
rejects the null hypothesis H0: m = m0
exactly when the value m0 falls outside a
level (1 – a) confidence interval for m.
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Case Study II
Studying Job Satisfaction
A 90% confidence interval for m is:
xz
 σ
n
 17  1.645
60
 17  23.26
18
 6.26 to 40.26
Since m0 = 0 is in this confidence interval, it is plausible that
the true value of m is 0; thus, there is not sufficient evidence
(at a = 0.10) that the mean job satisfaction of assembly
workers differs when their work is machine-paced rather
than self-paced.
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