AP Statistics Section 11.2 B
A 95% confidence interval captures the
true value of  in 95% of all samples. If we
are 95% confident that the true lies in our
interval, we are also confident that values
of  that fall outside our interval are
incompatible with the data. That sounds
like the conclusion of a significance test.
Confidence Intervals and Two-Sided Tests
A level  two-sided significance test rejects
H 0 :    0 whenever the value  0 falls
1
outside a level ______
confidence interval
for  .
The link between two-sided
significance tests and confidence
intervals is called _______.
duality The
following example illustrates this
link.
Example 1: The Deely Laboratory analyzes specimens of a drug
to determine the concentration of the active ingredient. Such
chemical analyses are not perfectly precise. Repeated
measurements on the same specimen will give slightly different
results. The results of repeated measurements follow a Normal
distribution quite closely. The analysis procedure has no bias, so
the mean  of the population of all measurements is the true
concentration of the specimen. The standard deviation of this
distribution is a property of the analysis method and is known to
be   0 . 0068 . The laboratory analyzes each specimen three
times and reports the mean result.
A client sends a specimen for which the concentration of active
ingredient is supposed to be 0.86%. Deely’s three analyses give
concentrations
0.8403
0.8363
0.8447
Is there significant evidence at the 1% level that the true
concentration is not 0.86%?
Let’s do both a significance test and a 99% confidence interval.
Hypothesis: The population of interest is all specimens
for this type of drug. We wish to test the claim that the
mean concentration of the active ingredient is not
.86%.
H 0 :   . 86 %
H a :   . 86 %
Conditions:
SRS: I will assume the sample is an SRS, but if it is not,
the results may not generalize to the population.
Normality of x : Population is Normal so the distribution
of x will be Normal.
Independence: Must assume N  (10)(3) or 30.
Calculations:
z
. 8404  . 86
. 0068
  4 . 99
3
P - value  2(.0002)  .0004
Note : .0002 is from the table
Interpretation: Since the p-value of .0004 < the
significance level of .01, we reject the H0. We
conclude that the mean concentration of the
active ingredient in this drug type is not .86%.
Parameter: The population of interest is all specimens
for this type of drug. We wish to estimate the true
mean concentration level of the active ingredient in
this drug.
Conditions:
Calculations:
xt

s
n
 . 8404  2 . 576
. 0068
3
 (. 8303 ,. 8505 )
Interpretation: We are 99% confident that the
true mean concentration level of the active
ingredient in this drug is between .8303% and
.8505%.
What if the null hypothesis in Example 1
had been H :   . 85 ? Would we still reject
H :   . 85 in favor of the two-sided
alternative H :   . 85 ?
0
0
a
NO , since .85 now lies in the 99% confidence
interval, we would
not reject H 0 :   . 85 in our   .01 significan ce test.
. 8303
. 8505
  . 85
  . 86