Transcript Inference Toolbox for Significance Tests

AP Statistics Section 11.2 A
Inference Toolbox for Significance
Tests
Although the reasoning of
significance testing isn’t simple,
carrying out a test is. The four-step
Inference Toolbox will once again
guide us through the inference
procedure.
Inference Toolbox for Significance Tests
Step 1: Hypothesis: Identify the population of interest and the
parameter you want to draw conclusions about. State
hypotheses.
Step 2: Conditions: Choose the appropriate inference procedure.
Verify the conditions for using it.
Step 3: Calculations: If the conditions are met, carry out the
inference procedure by calculating the _______________and
test statistic
find the ____________.
p - value
Step 4: Interpretation: Interpret your results in the context of
the problem by interpreting the __________
p - value or make a decision
about _____
H 0 using statistical significance. Don’t forget the three
C’s: ____________,
connection and __________.
conclusion ____________
context
z-Test for a Population Mean
To test the hypothesis based on an SRS of size n from a
population with unknown mean  and known standard deviation  ,
compute the one-sample z-statistic.
To determine the p-value, compute the probability of getting a value at
least as extreme as the value of our test statistic. The alternative
hypothesis ( H a) tells us if we are right-tailed, left-tailed or two-tailed.
These P-values are exact if the population is Normal and are
approximately correct for __________
large n in other cases.
Example 1: The medical director of a large company is
concerned about the effects of stress on the company’s
younger executives. According to the National Center
for Health Statistics, the mean systolic blood pressure
for males 35 to 44 years of age is 128, and the standard
deviation in this population is 15. The medical director
examines the medical records of 72 male executives in
this age group and finds that their mean systolic blood
pressure is x  129.93. Is this evidence that the mean
blood pressure for all the company’s younger male
executives is different from the national average?
(Assume that the executives have the same  as the
general population.)
Hypothesis : The population of interest is middle - age executives at this company.
We want to test the claim the claim that the mean systolic blood pressure for this
population is different from the national average of 128.
H 0 :   128
H a :   128
Conditions :
SRS : Seems safe to assume the 72 were randomly chosen. If not
an SRS, results may not generalize to the population.
Normality of x : n  72 means the CLT will give x a distribution that is
approximately Normal
Independence : Since we are sampling without replacement,
we must asume that N  10(72) or 720.
Calculations :
129.93 - 128
z
 1.09
15
72
p  value  2(.1379)  .2758
Interpretation : A p - value of .2758 indicates that there is a 27.58% chance
of a random sample having a mean systolic blood pressure of at least 129.93
when the true population mean is assumed to be 128. There is no good evidence
that the mean systolic blood pressure of young execs at this company differ from
the national average.
Note: The data in Example 1 do not
establish that the mean blood
pressure for this company’s
middle-aged male executives is
128. We simply failed to find
convincing evidence that the mean
differed from 128.
Failing to find evidence against H0
means only that the data are
consistent with H0, not that we
have clear evidence that H0 is
true.
Hypothesis : The population of interest is all employees at this company. We want
to test a claim that the health promotion campaign caused a drop in the mean blood
pressure (before minus after) of employees.
H0 :   0
Ha :   0
Conditions :
SRS : Director chose a random sample. If not an SRS, results
may not generalize to the population.
Normality of x : n  50 means the CLT gives a distribution for x
that is approx. Normal
Independence : Since we are sampling without replacement, we must
assume that the population of all employees  10(50) or 500.
Calculations :
60
z
 2.12
20
50
p  value  .017
Interpretation: Our p-value of .017 is less than the
significance level of .05. We reject the null hypothesis
and conclude that the mean difference in blood
pressure from before and after the campaign is less
than 0. This suggests that the mean blood pressure did
decrease in employees at this company.