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Hypothesis Tests Developing Null and Alternative Hypotheses Type I and Type II Errors
Hypothesis Tests for Population Mean: s Known Hypothesis Tests for Population Mean: s Unknown
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One-tailed(lower-tail)
One-tailed(upper-tail)
Two-tailed
0 0: H
0: aH 0 0: H
0: aH 0 0: H
0: aH
Summary of Forms for Null and Alternative Hypotheses about a
Population Mean The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population mean must take one of the following three forms (where 0 is the hypothesized value of the population mean).
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Type I and Type II Errors
CorrectDecision Type II Error
CorrectDecisionType I Error
Reject H0
(Conclude > µ0)
Accept H0
(Conclude < µ0)
H0 True( < µ0)
H0 False( > µ0)Conclusion
Population Condition
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Two Basic Approaches to Hypothesis Testing
There are two basic approaches to conducting a hypothesis test:
1- p-Value Approach, and 2- Critical Value Approach
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1- p-Value Approach toOne-Tailed Hypothesis Testing
Reject H0 if the p-value <
In order to accept or reject the null hypothesis the p-value is computed using the test statistic --Actual Z value.
Do not reject (accept) H0 if the p-value >
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2- Critical Value Approach One-Tailed Hypothesis Testing
Use the Z table to find the critical Z value, and
Use the equation to find the actual Z--Z . The rejection rule is:• Lower tail: Reject H0 if Actual z < Critical -z• Upper tail: Reject H0 if Actual z > Critical z
In other words, if the actual Z (Z ) is in the rejection region, then reject the null hypothesis.
s
/
xzn
Equation for finding the actual Z
value:
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Steps of Hypothesis Testing
Step 1. Develop the null and alternative hypotheses.Step 2. Specify and n.
Step 3. Compute critical Z and actual Z values.
Step 4. Use either of the following approaches to make conclusion:1- p-Value Approach, or
2- Critical Approach
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Example: Metro EMS
The EMS director wants toperform a hypothesis test, with a0.05 level of significance, to determinewhether the service goal of the response time to be at most 12 minutes or less is being achieved.
The response times for a randomsample of 40 medical emergencieswere tabulated. The sample meanis 13.25 minutes. The populationstandard deviation is believed tobe 3.2 minutes.
One-Tailed Tests About a Population Mean:s Known
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1. Develop the hypotheses.
2. Level of significance and sample size are:= .05n = 40
H0: Ha:
p -Value and Critical Value Approaches
One-Tailed Tests About a Population Mean:s Known: Solution
3. Compute the value of the test statistic.
s
13.25 12 2.47/ 3.2/ 40
xzn
Actual z
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5. Make conclusion about H0
We are at least 95% confident that Metro EMS is not meeting the response
goal of 12 minutes.
p –Value Approach
4. Compute the p –value.From the Z table the actual z = 2.47
p–value = 0.5 - .4932 = .0068
Because p–value = .0068 < = .05, we reject H0.
One-Tailed Tests About a Population Mean:s Known: Solution Continued
using Z table,
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p-value
0 Zc =1.645
= .05
z Za =2.47
Solution ContinuedBecause p–value = .0068 < = .05, we reject H0.
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5. Make conclusion about H0
We are at least 95% confident that Metro EMS is not meeting the response
goal of 12 minutes.
Because actual z = 2.47 > Critical z = 1.645
we reject H0.
Critical Value Approach
For = .05, z.05 = 1.6454. Determine the critical value and rejection rule.
Reject H0 if actual z > 1.645
Finding critical z value 0.5 – 0.05 = 0.45Then, from table
1.64 + 1.65 3.29 / 2 = 1.645
One-Tailed Tests About a Population Mean:s Known: Solution Continued
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Excel: SWStat
One-Tailed Tests About a Population Mean:s Known
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Excel: SWStat
One-Tailed Tests About a Population Mean:s Known
P Approach
Critical Approach
Because actual z = 2.47 > Critical z = 1.645 we reject H0, or
Because p–value = .0068 < α = .05, we reject H0
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Example: Glow Toothpaste
Two-Tailed Test for Population Mean: s Known
Quality assurance procedures call forthe continuation of the filling process if thesample results are consistent with the assumption thatthe mean filling weight for the population of toothpastetubes is 6 oz.; otherwise the process will be adjusted.
The production line for Glow toothpasteis designed to fill tubes with a mean weightof 6 oz. Periodically, a sample of 30 tubeswill be selected in order to check thefilling process.
oz.G
low
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Example Continued: Glow Toothpaste
Perform a hypothesis test, at the 0.03level of significance, to help determinewhether the filling process should continueoperating or be stopped and corrected.
Assume that a sample of 30 toothpastetubes provides a sample mean of 6.1 oz.The population standard deviation is believed to be 0.2 oz.
Two-Tailed Test for Population Mean: s Known
oz.G
low
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1. Determine the hypotheses.
2. Alpha and sample size are given
3. Compute the value of the test statistic.
= .03 and n=30
p –Value and Critical Value Approaches
H0: Ha: 6
Two-Tailed Tests About a Population Mean:s Known: Solution
s
0 6.1 6 2.74/ .2/ 30
xzn
Actual z
Glow
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5. Determine whether to reject or to accept H0.
p –Value Approach
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
Because p–value = .0062 < = .03, we reject H0. We are at least 97% confident that the
mean filling weight of the toothpaste tubes is not 6 oz.
Two-Tailed Tests About a Population Mean:s Known: Solution Continued
Glow
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Glow
/2 = .015
0z/2 = 2.17
z
/2 = .015
-z/2 = -2.17za = 2.74z = -2.74
1/2p -value= .0031
1/2p -value= .0031
Solution ContinuedBecause p–value = .0062 < = .03, we reject H0.
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Two-Tailed Tests About a Population Mean:s Known: Solution Continued
Critical Value Approach To Find the Critical Z Value:
/2 = .015
Given that = 0.03, thus /2 = .015 and 0.5 – 0.015 = 0.485
Then from the table we need to find the z value of 0.485.
Locate 0.485 in the Z Table. Thus, the critical z value for 0.485
is 2.17
Critical z/2 = 2.17
Critical z/2
0.485
0.5
Glow
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Critical Value Approach
Conclusion:
We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.
Because actual z of 2.74 > critical z of 2.17, we reject H0
Two-Tailed Tests About a Population Mean:s Known: Solution Continued
Glow
s
0 6.1 6 2.74/ .2/ 30
xzn
Actual Z
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/2 = .015
0 2.17
Reject H0
z
Reject H0
-2.17
Actual Z Value
z xn
s0
/
/2 = .015
Critical Approach: Solution Continued
= 2.74
Critical Z values
Because actual z of 2.74 > critical z of 2.17, we reject H0.
Glow
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Excel: SWStatGlo
w
Two-Tailed Tests About a Population Mean:s Known
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Excel: SWStat
Two-Tailed Tests About a Population Mean:s Known
P Approach
Critical Approach
Glow
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THOUGHT
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Confidence Intervals Versus Hypothesis Tests
A standard confidence interval is equivalent to a two-tail 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 intervaldoes not contain H0 , we reject H0.
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32 males between the ages of 40 and 69 years with 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 17.3 mmHg. At a = 0.05, is this sample consistent with a population mean of 140 mmHg, which is considered a borderline for dangerously high blood pressure (note: recent medical evidence suggests 130 as a borderline, but we will use the older benchmark)?
Thinking Challenge Example
Apply confidence interval approach to test the hypothesis
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Confidence Interval Approach: For this problem, the two-sided hypothesis would
be:H0: 4Ha:4
The 95% confidence interval (α=0.05) for is: x t s
n /2
Since interval 140.36 < < 152.84 does not contain =140 we would reject the hypothesis H0: 4 in favor of Ha:4
Margin of Error
Thinking Challenge ExampleSolution
146 – + (2.040) 17.3 /5.657
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Test Statistic
Hypothesis Tests About a Population Mean:
s Unknown
t xs n
0
/
This test statistic has a t distribution with n - 1 degrees of freedom.
Actual t Value
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A State Highway Patrol periodically samplesvehicle speeds at various locationson a particular roadway. The sample of vehicle speedsis used to test the hypothesis
Example: Highway Patrol One-Tailed Test About a Population Mean: s Unknown
The locations where H0 is rejected are deemedthe best locations for radar traps.
H0: < 65
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Example Continued: Highway Patrol
At Location F on I-75, a sample of 64 vehicles shows amean speed of 66.2 mph with asample standard deviation of4.2 mph. Use = .05 totest the hypothesis.
Use Excel
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36 Slide
Using SWStat
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Solution Using SWStat
P Approach
Critical Approach
Since p=0.0128 < α=0.05 we reject
H0 The locations where H0
is rejected are deemedthe best locations for radar traps.
H0: < 65
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38 Slide
0 critical t = 1.669
Reject H0
Do Not Reject H0
t
One-Tailed Test About a Population Mean:s Unknown: Solution Continued
t Statistic = Actual t = 2.286
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39 Slide
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.
At the .05 significance level can Ariana Electric Co. conclude that the new
machine is faster?
Thinking Challenge and
Solution
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Step 4 State the decision rule.There are 10 – 1 = 9 degrees of freedom.
Step 1 State the null and
alternate hypotheses.H0: µ < 250H1: µ > 250
Step 2 Select the level of
significance. It is .05.
Step 3 Find a test statistic. Use the t distribution since s is not known and n < 30.
The null hypothesis is rejected if t > 1.833 or, using the p-value, the null hypothesis is rejected if p ≤ 0.05
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41 Slide
162.3106250256
ns
Xt
Computed t (or actual t) of 3.162 > critical t of 1.833 and
From Excel, p of .0058 < So we reject Ho
The p(t >3.162) is .0058 for a one-tailed test.
Step 5 Make a decision and interpret the
results.
ConclusionThe mean number of
amps produced by the new machine is more
than 250 per hour.
Actual t
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42 Slide
Solution Using SWStat
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43 Slide
Solution Continued
Since computed t (or actual t)
of 3.162 > critical t of 1.833 and since p of .0058 < thus, we reject HoHence, we conclude that the
mean number of amps
produced by the new machine
is more than 250 per hour.
H0: µ < 250
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44 Slide
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.
Thinking Challenge
$48,000 $44,000 $46,000$43,000 $47,000 $46,000$44,000 $42,000 $45,000
andSolution
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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).
Thinking Challenge (Continued)
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46 Slide
Thinking Challenge 4 (Solution)
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47 Slide
Summary of Selecting an Appropriate Test Statistic for
a Test about a Population Mean
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End of Chapter 10
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