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Chapter 10
AP Statistics
St. Francis High School
Fr. Chris
Introduction to Inference
1. Estimating with Confidence
2. Tests of Significance
3. Using Significance Tests
4. Inference as a Decision.
Confidence Interval
• Two parts: the estimate and the confidence
level
• Estimate±Margin of Error
• If you have an SRS, you can assume that
the mean is normally distributed
• If you can estimate the spread, you can find
the boundaries where 90%, 95% or 99% of
the area under this normal curve lie.
A Pharmaceutical Manufacturer
To establish the concentration of an active ingredient of a drug,
the same specimen of a particular batch is measured three times:
0.8403, 0.8363, and 0.8447. The standard deviation is known to
be 0.0068 grams per liter. What is the 95% confidence interval?
.8403  .8363  .8447
x
 .8404
3
So .8404 is a good estimate of the actual amount, but how
big a margin of error do we need to be 95% confident the
actual measurement is in our interval?
The 95% Confidence Interval
On a standard normal curve (N[0,1]),
95% is between -1.96 and +1.96
standard deviations.
-1.96
95%
Since we know sigma, we convert
+1.96 the drug’s distribution to the
standard normal so the interval is
x z*

n
Where z = 1.96 (95% Confidence)
Substituting the numbers...

.0068
 .8404  1.96
 .8404  .0101
x z*
3
n
So we can be 95% sure that the actual measurement of the
sample is between .8303 and .8505
We would have a more narrow interval if we are willing to
drop our confidence interval to 90%. On a standard normal
curve, 90% of the area is between -1.645 and +1.65...
.0068
 .8404  1.645
 .8404 .0065
3
Or (.8339, .8469)
Caution!!!!!
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Data must be SRS from the population
Formulas for more complex designs are available
Fancy formulas cannot rescue badly produced data
Outliers can influence x-bar
If n is small (<15), population needs to be normal
Need to know the sigma of the population (If
sample is huge, you can use s, otherwise, wait for
more methods next chapter!)
10.2 Tests of Significance
•
•
•
•
•
H0 and Ha
One Sided vs. Two Sided Alternative
P-Value

Critical values and the z Statistic
Blood Pressure
Males age 33-44 have a systolic blood pressure of 128 with a =15.
A researcher looks at the records of 72 male ATC’s in this age
group and found their average systolic blood pressure to be
131.07. Obviously this is higher than the average. It is likely that
this is due to chance variation, or that ATC’s have a significantly
higher blood pressure? It is SIGNIFICANT if the result is unlikely
to be caused by chance.
The Null Hypothesis
Males age 33-44 have a systolic blood pressure of 128 with a =15.
72 of our ATC’s average at over 131. The null Hypothesis is to say
There is no difference between ATC’s and males in general.
H0 :  = 0
One way to state the alternative is to say that the average of
ATC’s is different, so the alternative hypothesis is
Ha :  ≠ 0
This is a “Two-tail” test, since the alternative includes two
possibilities: Either the ATC’s have a higher blood pressure or
they have a lower blood pressure.
Let’s get graphic
A two tailed test means
that the sample mean
falling on either
extreme will be
considered grounds to
reject the H0
P(Z  1.96)  .0250
P(Z  1.96)  .0250
Knowing what we know about the Normal
Distribution, we know the critical z score that
partitions the inner 95% is where the cumulative area
is at 2.5% on both ends (thus our =.05)
Are ATC’s average on the fringe?
If they belong to the same distribution as anyone else, we would
expect the  of the sample mean of 72 randomly selected males
to be
So to convert to the

15
x 

 1.7678 Standard Normal
n
72
Curve...
z
x
x
131.07  128

 1.7366
1.7678
Fail to reject the Null Hypothesis
Since 1.7366 is less than 1.96, we are in the 95% of sample
means, and we cannot reject the H0 …
But what if we only consider the alternative that ATC’s,
due to the stressful nature of their work, would have a
HIGHER blood pressure than other males...
The One Tailed Test
The one tailed test has the same
H0 :  =0 but a different
alternative hypothesis, viz.,
Ha :  >0
P(Z 1.645) 
1  P(Z 1.645) 1  .95  .05
The critical value is different, since 95% is on the left,
and 5% is on the upper end.
Rejecting the Null Hypothesis
Since our critical value is now 1.645, and the
sample mean of 131.07 has a z value of 1.7366, it
is in the extreme upper 5%, so we can reject the
null hypothesis in favor of the alternative
hypothesis: The mean of ATC’s systolic blood
pressure is higher than other males of this age
group.
Review of terms
Notice our  is still .05, but only when we consider the
one-tailed test (which is appropriate in this case), are we
able to reject the null hypothesis. Computers and
statistical calculators can compute a p-value (probability)
which allows to to see how unlikely the average is if the
null hypothesis is true.
Using your calculator
Select STAT, then TESTS
Select “1:Z-Test”
Select “Stats” since we don’t have the raw data (72 sbp’s)
First let’s select the two tailed alternative
0 :128
hypothesis   0 then press Calculate
 : 15
Notice that it shows the z-score, and the
x : 131.07
p-value (.08 is bigger than .05). So we
n : 72
cannot
 reject the null hypothesis.
Go back and instead of selecting
“Calculate”, select “Draw”
Using your calculator for
One Tail Test
Select STAT, then TESTS
Select “1:Z-Test”
Select “Stats” since we don’t have the raw data (72 sbp’s)
First let’s select the two tailed alternative
0 :128
hypothesis >0 then press Calculate
 : 15
Notice that it shows the z-score, and the
x : 131.07
p-value (.04 is less than .05). So we can
n : 72
reject the null hypothesis.
Go back and instead of selecting
“Calculate”, select “Draw”
10.3 Using Significance Tests
• Choosing a level of significance
• When is Statistical Inference Valid
• Examples
Choosing a level
• P-values are more informative than
Traditional  levels
• Small effects can be highly significant
• Lack of signifcance doesn’t imply H0 is true
Validity
• Faulty data collection
• Outliers
• Testing a hypothesis on the same data that
suggests a hypothesis can invalidate the test
• Many tests at once can produce significance
due to chance alone, even if the H0’s are true
Examples
Hawthorne Effect: Will background music increase
productivity? Any short term change can effect,
regard less of what it is (See Example 10.19, p 590)
A researcher looking for evidence of ESP tests 500 subjects.
Four do significantly better (P<.01) than random guessing.
Can we conclude these 4 have ESP? What should the
researcher do to test whether these 4 have ESP? (10.61 p.
592)
Kidney Failure
Patients with kidney failure are often treated by dialysis, and can
also cause the retention of phosphorous that must be corrected by
changes in the diet. One patient in a study had these 6
measurements over time: 5.6, 5.3, 4.6, 4.8, 5.7, 6.4 (These are
separated over time and can be considered an SRS of the
patient’s blood phosphorous level). If this level varies normally
with = 0.9 mg/dl, give a 90% confidence interval for the mean
blood phosphorus level.
(4.7956, 6.0044)
Conclusions
The normal range of phosphorous in the blood is considered
to be 2.6 to 4.8 mg/dl. Is there strong evidence that the
patient has a mean phosphorous level that exceeds 4.8?
What are the hypotheses?
Conclusion?
There is evidence that the patient’s
H0: =4.8 Ha: >4.8
What is the z-score/P-value? mean phosphorous level is higher
than 4.8 mg/dl, but it is not strong.
Z=1.633
Whether this is sufficient evidence is a
P-value is .051235
judgement call. .05 is not sacred
What if the inference is use to
determine a change in diet? Surgury?