Download SECTION 9.3 – SIGNIFICANCE TESTS ABOUT MEANS

SECTION 9.3 – SIGNIFICANCE
TESTS ABOUT MEANS
Overview
—  For
the most part this is done similarly to
significance tests for proportions.
—  We also need to check the assumption that
the population is normally distributed.
◦  This is important when n is small and we are
computing a one-sided test.
◦  Two-sided tests are robust concerning the
normality assumption.
First Example
A job placement director claims that the average starting
salary for nurses is $24,000. A researcher questions this
claim – she thinks the true mean is lower than that figure.
She takes a random sample of 10 starting nurses, and gets the
following data:
23,400 19,000 20,200 22,000 22,145
25,150 24,530 25,000 21,000 25,500
Is there enough evidence to reject the director’s claim at α =
0.05?
Note: Using my TI-83, I found a mean of $22,792.50 and a
standard deviation of $2273.37.
Hypothesis Testing Steps
1. 
2. 
3. 
What is the population mean to be tested?
µ = the average starting salary among all
nurses.
State the null hypothesis.
H0 is “µ = 24000”
State the alternative hypothesis.
Ha is “µ < 24000”
Hypothesis Testing Steps
5. 
5. 
6. 
Calculate the t test statistic and the P-value.
t = (22792.50 – 24000)/718.90 = -1.68
P-value
= Pr( t < -1.68)
= tcdf(-10000, -1.68, 9)
= 0.064
Do we reject the null hypothesis? Since the Pvalue is not small, we do not reject H0.
English conclusion: “There is not statistically
significant evidence that the average starting
salary among all nurses is less than $24K.”
Example 1
The diastolic blood pressure for American women ages 18 to
44 has approximately a normal distribution with mean 75
millimeters of mercury (mm Hg) and standard deviation 10 mm
Hg. We suspect that regular exercise will lower the blood
pressure. A sample of 25 women who jog at least 5 miles per week
gives a sample mean blood pressure of 71 mm Hg. Is this good evidence (at the 5% level) that the mean diastolic
blood pressure for the population of regular exercisers is lower
than 75 mm Hg?
Hypothesis Testing Steps
1. 
2. 
3. 
What is the population mean to be tested?
µ = the average blood pressure among all
nurses.
State the null hypothesis.
H0 is “µ = 24000”
State the alternative hypothesis.
Ha is “µ < 24000”
Hypothesis Testing Steps
5. 
Calculate the t test statistic and the P-value.
6. 
Do we reject the null hypothesis?
7. 
State the conclusion in plain English.
Example 1
The diastolic blood pressure for American women ages 18 to
44 has approximately a Normal distribution with mean 75
millimeters of mercury (mm Hg) and standard deviation 10 mm
Hg. We suspect that regular exercise will lower the blood
pressure. A sample of 25 women who jog at least 5 miles per
week gives a sample mean blood pressure of 71 mm Hg. Is
this good evidence (at the 5% level) that the mean diastolic
blood pressure for the population of regular exercisers is lower
than 75 mm Hg?
Step 1: Assumptions
—  The
variable of interest is quantitative – What
is your diastolic blood pressure?
◦  This is a test with means.
—  The
data is produced using randomized
methods.
—  The sample size is smaller than 30, but the
population distribution is said to be normal.
Step 2: Hypotheses
—  µ
= the mean diastolic blood pressure (mm
Hg) among all women who jog at least 5
miles per week.
—  H0: µ = 75 mm Hg
—  Ha: µ < 75 mm Hg
Step 3: Test Statistic
10
s.e. =
=
= 2 mm Hg
n
25
σ
x = 71 mm Hg
71 − 75
z=
= −2
2
Step 4: P-value
—  normalcdf(-1000,
71, 75, 2) = 0.023
—  If the true mean diastolic blood pressure
for female joggers was 75 mm Hg (like
women in general), we would still get a
sample at least this far from the true mean
2.3% of the time.
Step 5: Conclusion
—  This
evidence is strong against the null
hypothesis.
—  Thus we are led to believe that the mean
diastolic blood pressure for female joggers
is less than 75 mm Hg.
Example 2
The following figures refer to the actual weights (in ounces)
of a simple random sample of ten "one-pound" cans of
peaches distributed by a company. It is known that weights
of this type vary with a Normal distribution.
16.3
15.5
15.4
15.7
16.0
16.8
16.8
16.2
16.5
15.8
At the 5% level of significance, decide whether or not the
true mean of all of the cans distributed by the company
is different from 16 ounces.
Step 1: Assumptions
—  The
variable of interest is quantitative –
What is the weight of each can?
◦  This is a test with means.
—  The
data is produced using randomized
methods.
—  The sample size is smaller than 30, but the
population distribution is said to be
Normal.
Step 2: Hypotheses
—  µ
= the average weight (oz) of all cans of
peaches distributed by the company
—  H0: µ = 16 ounces
—  Ha: µ ≠ 16 ounces
Step 3: Test Statistic
s
0.5055
s.e. ≈
=
≈ 0.16 ounces
n
10
x = 16.1 ounces
16.1 − 16
t=
= 0.625
0.16
Step 4: P-value
—  2*tcdf(0.625,
1000, 9) = 0.547
—  If the mean weight of cans was 16 ounces
(as the label claims), we would still get a
sample off by this much or more 54.7% of
the time.
Step 5: Conclusion
—  The
P-value is not small.
—  This evidence is not strong enough to reject
the null hypothesis.
—  This evidence is not strong enough to
conclude that the cans weigh significantly
different from the posted weight of 16
ounces.
SECTION 9.5 – LIMITATIONS OF
SIGNIFICANCE TESTS
Common Misconceptions
—  Small
P-values correspond to statistically
significant results. It does not mean that the
results are practically significant.
—  The sample value can be extremely close to
the target parameter value, but really large
sample sizes can make the result appear
statistically significant.
—  “Not rejecting” the null hypothesis is not the
same as “accepting” the null hypothesis.
—  The P-value is NOT the probability that the
null hypothesis is true.
Common Misinterpretations
—  Some
tests may be statistically significant by
chance.
◦  Repeated testing might find significant results just
because one test is bound to appear significant
by chance.
—  The
only reported or publicized data might
be the extreme data.
Significance Tests are Less Useful that
Confidence Intervals
—  A
significance test merely indicates
whether a particular parameter is
plausible.
◦  If our significance test fails to reject the null
hypothesis, we have no idea what the plausible
parameters are.
—  A
confidence interval gives a whole range
of plausible values for the parameter.