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Unformatted text preview: Chapter 10 Hypothesis Tests Developing Null and Alternative Hypotheses Type I and Type II Errors Hypothesis Tests for Population Mean: σ Known Hypothesis Tests for Population Mean: σ Unknown Slide 1 Summary of Forms for Null and Alternative Hypotheses about a Population Mean
s The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population mean µ must take one of the following
must take one of the following three forms (where µ0 is the hypothesized value of the population mean). H 0 : µ ≥ µ0
H a : µ < µ0 H 0 : µ ≤ µ0
H a : µ > µ0 H 0 : µ = µ0
H a : µ ≠ µ0 Onetailed
(lowertail) Onetailed
(uppertail) Twotailed Slide 2 Type I and Type II Errors
Population Condition Conclusion H0 True
(µ < µ0) Accept H0
(Conclude µ < µ0) Correct
Decision Type II Error Type I Error Correct
Decision Reject H0
(Conclude µ > µ0) H0 False (µ > µ0) Slide 3 Two Basic Approaches to Hypothesis Testing There are two basic approaches to conducting a hypothesis test: 1 pValue Approach, and 2 Critical Value Approach Slide 4 1 pValue Approach to
OneTailed Hypothesis Testing In order to accept or reject the null hypothesis the pvalue is
In
the computed using the test statistic Actual Z value. Reject H0 if the pvalue < α Do not reject (accept) H0 if the pvalue > α Slide 5 2 Critical Value Approach OneTailed Hypothesis Testing Use the Z table to find the critical Z value, and Use the equation to find the actual ZZ statistics. s The rejection rule is: • Lower tail: Reject H0 if Actual z < Critical zα
• Upper tail: Reject H0 if Actual z > Critical zα In other words, if the actual Z (Z statistics) is in the
In
actual
rejection region, then reject the null hypothesis.
rejection
Equation for finding the
Equation
actual Z value:
actual x −µ
z=
σ/ n Slide 6 Steps of Hypothesis Testing
Step 1. Develop the null and alternative hypotheses.
Step 2. Specify α and n.
and
Step 3. Compute critical Z and actual Z values.
Step 3
Step 4. Use either of the following approaches to make conclusion:
1 pValue Approach, or
2 Critical Approach Slide 7 OneTailed Tests About a Population Mean:
σ Known
s Example: Metro EMS
s The response times for a random sample of 40 medical emergencies
were tabulated. The sample mean
is 13.25 minutes. The population
standard deviation is believed to
be 3.2 minutes.
s The EMS director wants to
perform a hypothesis test, with a
0.05 level of significance, to determine
whether the service goal of the response time to be at most 12 minutes or less is being achieved. Slide 8 OneTailed Tests About a Population Mean:
σ Known: Solution p Value and Critical Value Approaches
1. Develop the hypotheses. H0: µ < 1 2
Ha: µ > 1 2 2. Level of significance and sample size are: α= .05
n = 40 3. Compute the value of the test statistic. x − µ 13.25 − 12
z=
=
= 2.47
σ / n 3.2 / 40
Actual z
Actual Slide 9 OneTailed Tests About a Population Mean:
σ Known: Solution Continued p –Value Approach
4. Compute the p –value.
From the Ztable the actual z = 2.47
using Z table, p–value = 0.5 .4932 = .0068 5. Make conclusion about H0
Because p–value = .0068 < α = .05, we reject H0.
s We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes. Slide 10 Solution Continued
Because p–value = .0068 < α = .05, we reject H0. α = .05 pvalue
= .0 0 6 8
z
0 Zc =
1.645 Za =
2.47 Slide 11 OneTailed Tests About a Population Mean:
σ Known: Solution Continued Critical Value Approach
4. Determine the critical value and rejection rule.
For α = .05, z.05 = 1.645
Reject H0 if actual z > 1.645
5. Make conclusion about H0 Finding critical z value
0.5 – 0.05 = 0.45
Then, from table
1.64 + 1.65
3.29 / 2 = 1.645 Because actual z = 2.47 > Critical z = 1.645 we reject H0.
s We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes. Slide 12 OneTailed Tests About a Population Mean:
σ Known
s Excel: SWStat Slide 13 OneTailed Tests About a Population Mean:
σ Known
s Excel: SWStat P Approach
Critical Approach
Because actual z = 2.47 >
Because actual
Critical z = 1.645 we
Critical
reject H0, or
or
Because p–value = .0068
Because
< α = .05, we reject
we
H0 Slide 14 Example: Glow Toothpaste
s TwoTailed Test for Population Mean: σ Known The production line for Glow toothpaste
is designed to fill tubes with a mean weight
of 6 oz. Periodically, a sample of 30 tubes
o
will be selected in order to check the
z.Glow
filling process. Quality assurance procedures call for
the continuation of the filling process if the
sample results are consistent with the assumption that
the mean filling weight for the population of toothpaste
tubes is 6 oz.; otherwise the process will be adjusted. Slide 15 Example Continued: Glow Toothpaste
s TwoTailed Test for Population Mean: σ Known Assume that a sample of 30 toothpaste
tubes provides a sample mean of 6.1 oz.
The population standard deviation is believed to be 0.2 oz. Perform a hypothesis test, at the 0.03
level of significance, to help determine
whether the filling process should continue
operating or be stopped and corrected. o
z.Glow Slide 16 TwoTailed Tests About a Population Mean:
σ Known: Solution
Glow p –Value and Critical Value Approaches
1. Determine the hypotheses. H0: µ = 6
Ha: µ ≠ 6 2. Alpha and sample size are given α = .03 and n=30 3. Compute the value of the test statistic. x − µ0
6.1 − 6
z=
=
= 2.74
σ / n .2 / 30
Actual z Slide 17 TwoTailed Tests About a Population Mean:
σ Known: Solution Continued p –Value Approach Glow 4. Compute the p –value.
For actual z = 2.74, the probability = 0.4969, thus p–value = 2(0.5 – 0.4969) = 2 (0.0031) = 0.0062
5. Determine whether to reject or to accept H0.
Because p–value = .0062 < α = .03, we reject H0.
s We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz. Slide 18 Solution Continued Glow Because p–value = .0062 < α = .03, we reject H0. 1/2
p value
= .0031 1/2
p value
= .0031 α/2 = .015 α/2 = .015 z
z = 2.74 zα/2 = 2.17 0 zα/2 = 2.17 za = 2.74 Slide 19 TwoTailed Tests About a Population Mean:
σ Known: Solution Continued
Critical Value Approach To Find the Critical Z Value: G lo w 0.5
Given that α = 0.03, thus α/2 = .015 and 0.5
– 0.015 = 0.485 Then from the table we need to find the z
value of 0.485. Critical zα/2 Locate 0.485 in the Z Table. Thus, the critical z value for 0.485 is 2.17
0.485 Critical zα/2 = 2.17 α/2 = .015 Slide 20 TwoTailed Tests About a Population Mean:
σ Known: Solution Continued Critical Value Approach
G lo w Conclusion: Because actual z of 2.74 > critical z of 2.17, we reject H 0 s We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.
Actual Z z= x − µ0
6.1 − 6
=
= 2.74
σ / n .2 / 30 Slide 21 Critical Approach: Solution Continued Glow Because actual z of 2.74 > critical z of 2.17, we reject H0. Actual Z Value
z= x − µ0
σ / n = 2.74 Reject H0 Reject H0 α/2 = .015 2.17 α/2 = .015
0
Critical Z values 2.17 z Slide 22 TwoTailed Tests About a Population Mean:
σ Known
s Glow Excel: SWStat Slide 23 TwoTailed Tests About a Population Mean:
σ Known
s Excel: SWStat P Approach
Critical Approach Glow Slide 24 THOUGHT
THOUGHT Confidence Intervals Versus
Hypothesis Tests A standard confidence interval is equivalent to a
twotail hypothesis test. All two tails tests can be handled either as
hypothesis tests or as confidence intervals. The confidence interval has the appeal of
providing a graphic feeling for how close the
hypothesized value lies to the ends of confidence
interval. Rejection Rule: If the confidence interval
does not contain H0 , we reject H0. Thinking Challenge Example 32 males between the ages of 40 and 69 years with
32
40
69
moderate carotid disease were tested at the Henry
Hospital over 39,months period. Their mean systolic
pressure was 146.6 mmHg with a standard deviation of
146.6
17.3 mmHg. At a = 0.05, is this sample consistent with a
17.3
0.05
population mean of 140 mmHg, which is considered a
140
borderline for dangerously high blood pressure (note:
borderline
recent medical evidence suggests 130 as a borderline, but
we will use the older benchmark)?
we Apply confidence interval approach to test the hypothesis Thinking Challenge Example
Solution Confidence Interval Approach: For this problem, the twosided hypothesis would be:
H0: µ = 1 4 0
Ha: µ ≠ 1 4 0 The 95% confidence interval (α=0.05) for µ is: for s
x ± tα / 2
Margin of Error
Margin
n
146 – + (2.040) 17.3 /5.657 Since interval 140.36 < µ < 152.84 does not contain µ =140 we =140 would reject the hypothesis H0: µ = 1 4 0 in favor of Ha: µ ≠ 1 4 0 . Slide 28 Hypothesis Tests About a Population Mean:
σ Unknown
s Test Statistic
Actual t Value x − µ0
t=
s/ n This test statistic has a t distribution with n 1 degrees of freedom. Slide 29 Example: Highway Patrol
s OneTailed Test About a Population Mean: σ Unknown A State Highway Patrol periodically samples
vehicle speeds at various locations
on a particular roadway. The sample of vehicle speeds
is used to test the hypothesis
H0: µ < 65 The locations where H0 is rejected are deemed
the best locations for radar traps. Slide 30 Example Continued: Highway Patrol At Location F on I75, a sample of 64 vehicles shows a
mean speed of 66.2 mph with a
sample standard deviation of
4.2 mph. Use α = .05 to
test the hypothesis. Use Excel Slide 31 Using SWStat Slide 32 Solution Using SWStat P Approach
Critical Approach H0: µ < 65
Since p=0.0128 < α=0.05
Since
=0.05
we reject H0
we The locations where H0
is rejected are deemed
rejected
the best locations for radar
the best
traps.
traps.
s Slide 33 OneTailed Test About a Population Mean:
σ Unknown: Solution Continued Reject H0
Do Not Reject H0 0 α = .0 5 critical tα = 1.669 t t Statistic = Actual t = 2.286 Slide 34 Thinking Challenge The current rate for producing 5
amp fuses at Ariana Electric Co. is
250 per hour. A new machine has
been purchased and installed that,
according to the supplier, will
increase the production rate. The
production hours are normally
distributed. A sample of 10
randomly selected hours from last
month revealed that the mean hourly
production on the new machine was
256 units, with a sample standard
deviation of 6 per hour. and
Solution At the .05 significance level
can Ariana Electric Co.
conclude that the new
machine is faster? Slide 35 The null hypothesis is rejected if t > 1.833 or, using the
pvalue, the null hypothesis is rejected if p ≤ 0.05
Step 4
State the decision rule.
There are 10 – 1 = 9
degrees of freedom.
Step 1
State the null and
alternate hypotheses.
H0: µ < 250
H1: µ > 250 Step 3
Find a test statistic. Use
the t distribution since σ
is not known and n < 30.
Step 2
Select the level of
significance. It is .05. Slide 36 Step 5
Make a decision
and interpret the
results. t= X −µ
s Actual t
Computed t (or actual t) of 3.162 >
critical t of 1.833 and From Excel, p of .0058 < α = .0 5
So we reject Ho n = 256 − 250
6 10 = 3.162 The p(t >3.162) is .0058
for a onetailed test. Conclusion
The mean number of
amps produced by the
new machine is more
than 250 per hour. Slide 37 Solution Using SWStat Slide 38 Solution Continued
H0: µ < 250 Since computed t (or actual t)
of 3.162 > critical t of 1.833
and since p of .0058 <
α = .0 5
thus, we reject Ho Hence, we conclude that the
mean number of amps produced by the new machine
is more than 250 per hour. Slide 39 and
Solution Thinking Challenge s A group of young businesswomen wish to open a high fashion boutique in a vacant store, but only if the average income of households in the area is more than $45,000. A random sample of 9 households showed the following results.
$48,000 $44,000 $46,000 $43,000 $47,000 $46,000 $44,000 $42,000 $45,000 Slide 40 Thinking Challenge (Continued)
s Use the statistical techniques in Excel (SWStat) to advise the group on whether or not they should locate the boutique in this store. Use a 0.05 level of significance. (Assume the population is normally distributed). Slide 41 Thinking Challenge 4 (Solution) Slide 42 Summary of Selecting an Appropriate Test Statistic for a Test about a Population Mean Slide 43 Cartoon Classics
Cartoon End of Chapter 10 The End
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 Fall '09

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