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Chapter 9: Hypothesis Tests for One Population
Mean
9.3 Hypothesis Testing for a Population Mean
When σ is Known
I.
Critical Values:
• We have previously defined critical values. Now we must see how we
can obtain them. It is important to remember that all critical values are
based on α and must be looked up on a tables in the appendix of the
text (the z – table and t – tables being only two of these types of tables;
there are others).
• Once we know how to obtain the critical values, we will be ready to
conduct a full hypothesis test.
• If a hypothesis test is to be performed at a specified significance level, α,
then the critical value(s) must be chosen so that, if the null hypothesis is
true, the probability is α that the test statistic will fall in the rejection
region.
• Here are three examples that will help make this last statement more
understandable:
Example: Determine the critical value(s) for a hypothesis test at the 5%
significance level (α = 0.05) if the test is (a) two-tailed, (b) left-tailed, and
(c) right-tailed.
Solution:
(a) Since α = 0.05, we choose critical values so that the area under the
SNC that lies above the rejection region equals 0.05. As seen in the
following figure, for a two-tailed test, the rejection region is in both tails,
i.e. the total area is 0.05, or each area is 0.05/2 or 0.025. The area is
located on the z – table and the corresponding z – score is found in the
manner of Chapter 6. Here they are ± zα = ± z0.05 = ± z0.025 = ±1.96 :
2
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2
Critical Values
(b) Likewise for a left-tailed test, the rejection region is on the left and
the critical value (from Table II) is − z0.05 = −1.645 . Note that the alpha
value is not divided in half as for a two-tailed test:
Critical Values
(c) An similarly for the right-tailed test, z0.05 = 1.645 :
Critical Values
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• More generally, in summary, for any specified significance level, we
obtain the critical value(s) as follows:
Critical Values
• There are some significance levels which in practice are more common
than others. These are listed here, and are found just below the t – table
in the appendix (they have no relationship to the t – table, that’s just
where they were put by the publisher):
Values of zα
z 0.10
1.28
II.
z 0.05
1.645
z 0.025
1.96
z 0.01
2.33
z 0.005
2.575
One-Sample Z – Test for a Population Mean, σ known:
• Now that we have defined our terms and learned how to look up the
critical values, we are ready to look at the first type of hypothesis test, a
test that, like the z – interval procedure of Chapter 8, is not a commonly
done test, but is given first because it is easier from a teaching
standpoint. Here’s how it is done—and we will always follow exactly the
procedure as outlined in your text:
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One-Sample z-test for a
Population Mean
(Critical-Value Approach)
Assumptions
1. Normal population or large sample
2. σ known
Step 1 The null hypothesis is Ho: µ = µo, and the alternative
hypothesis is
Ha: µ = µo
(two-tailed)
or
Ha: µ < µo
(left-tailed)
or
Ha: µ > µo
(right-tailed)
(Continued)
One-Sample z-test for a
Population Mean
(Critical-Value Approach)
Step 5 If the value of the test statistic falls in the rejection
region, reject Ho; otherwise, do not reject Ho.
Step 6 Interpret the results of the hypothesis test.
The hypothesis test is exact for normal populations and is
approximately correct for large samples from nonnormal
populations.
One-Sample z-test for a
Population Mean
(Critical-Value Approach)
Step 5 If the value of the test statistic falls in the rejection
region, reject Ho; otherwise, do not reject Ho.
Step 6 Interpret the results of the hypothesis test.
The hypothesis test is exact for normal populations and is
approximately correct for large samples from nonnormal
populations.
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• And just as we did with CIs, here is when you can use the z – test and
expect good answers:
When to Use z-Test
1. If n < 15: use only if variable is normally distributed
or close to it.
2. If 15 < n < 30: use unless (a) outliers present or (b)
variable is far from normally distributed.
3. If n > 30: use essentially without restrictions.
4. Use if outliers present but they can be justifiably
removed.
• We are now ready to do some examples. We will do three. Here’s the
first:
Body Temperature
Researchers at the University of Maryland
obtained body temperatures from 93 healthy
Humans. At the 1% significance level, do the
data provide sufficient evidence to conclude
that the mean body temperature of healthy
humans differs from 98.6o F. Assume σ = 0.63oF
o
and the sum of the 93 temperatures is 9125.5 F.
Assume the temperatures are normally
distributed.
Solution: (I like to put my data down at the beginning)
¾ n = 91, σ = 0.63, x =
9125.5
= 98.12° F
93
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¾ ASSUMPTIONS: Normal distribution is given and σ is given.
¾ STEP 1:
H 0 : µ = 98.6° F
H a : µ ≠ 98.6° F
¾ STEP 2:
α = 0.01
¾ STEP 3:
z=
x − µ 98.12 − 98.6
=
= −7.35
σ n
0.63 93
¾ STEP 4: Critical values from z – table:
zα = z0.01 = z0.005 = ±2.575
2
2
¾ STEP 5: −7.35 < −2.575 ∴ reject H 0 because the test statistic
(- 7.35) falls in the rejection region.
¾ Step 6: At the 1% significance level, the data provide sufficient
evidence to conclude that the mean body temperature µ of all
healthy humans differs from the widely accepted value of 98.6ºF.
• Example #2:
Iron Deficiency
The Food and Nutrition Board of the National
Academy of Science states that the
recommended daily allowance (RDA) of iron in
adult females under the age of 51 is 18 mg. Iron
intakes for 45 randomly selected adult females
under the age of 51 were obtained. At the 1%
significance level, do the data suggest that adult
females under age 51, on average, get less than
18 mg of iron daily? Assume σ = 4.2 mg and
x = 14.68 mg and the iron intakes are ND.
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Solution:
¾ n = 45, σ = 4.2, x = 14.68
¾ ASSUMPTIONS: Normal distribution is given and σ is given.
¾ STEP 1:
H 0 : µ = 18 mg
H a : µ < 18 mg
¾ STEP 2:
α = 0.01
¾ STEP 3:
z=
x − µ 14.68 − 18.00
=
= −5.30
σ n
4.2 45
¾ STEP 4: Critical value from z – table:
zα = −2.325
¾ STEP 5: −5.30 < −2.326 ∴ reject H 0 because the test statistic
(- 5.30) falls in the rejection region.
¾ Step 6: At the 1% significance level, the data provide sufficient
evidence to conclude that adult females under the age of 51, on
average, are getting less than the RDA of 18 mg of iron.
• Example #3:
Hospital Costs
The American Hospital Association reports in
Hospital Stat that the mean cost to community
hospitals per patient day in U.S. hospitals was $1033
in 2002. In that same year, a random sample of 30
daily costs in Ohio hospitals had a mean of $1123.
Assuming a population
σ = $350 for Ohio hospitals, do the data provide
sufficient evidence, at the 5% significance level, to
conclude that, in 1997, the mean cost in Ohio
hospitals exceeded the national mean of $1033?
Assume a ND for hospital costs.
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Solution:
¾ n = 30, σ = $350, x = $1123
¾ ASSUMPTIONS: Normal distribution is given and σ is given.
¾ STEP 1:
H 0 : µ = $1033
H a : µ > $1033
¾ STEP 2:
α = 0.05
¾ STEP 3:
z=
x − µ 1123 − 1033
=
= 1.41
σ n
350 30
¾ STEP 4: Critical value from z – table:
zα = z0.05 = 1.645
¾ STEP 5: 1.41 < 1.645 ∴ do not reject H 0 because the test
statistic (1.41) does not fall in the rejection; it falls in the
nonrejection region.
¾ Step 6: At the 5% significance level, the data do not provide
sufficient evidence to conclude that, in 2002, the mean daily costs
per patient in community hospital in Ohio exceeded the national
mean of $1033.
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9.4 Type II Error Probabilities and Power
• I just want to bring two concepts here to your attention to keep you
informed of the terminology. We will not calculate the power of a HT.
• β is the probability of making a Type II error, i.e. the probability of not
rejecting a false null hypothesis.
• The Power of a Hypothesis Test is defined as the probability of not
making a Type II error, i.e.
Power = 1 − P ( Type II Error )
Power = 1 − β
• Power measures the ability of a hypothesis test to detect a false null
hypothesis. If power is near 0, the hypothesis test is not good at
detecting a false null hypothesis. If the power is near 1, the test is
extremely good at detecting a false null hypothesis.
• There is an important relation between power and sample size: for a
fixed significance level, increasing the sample size increases the power.
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