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5. 5Slide
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

6. 6Slide
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).

7. 7Slide
Type I and Type II Errors
Correct
Decision
Type II Error
Correct
Decision
Type I Error
Reject H0
(Conclude  > µ0)
Accept H0
(Conclude  < µ0)
H0 True
( < µ0)
H0 False
( > µ0)Conclusion
Population Condition

8. 8Slide
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

9. 9Slide
1- p-Value Approach to
One-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 > 

10. 10Slide
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


/
x
z
n
Equation for finding the
actual Z value:

11. 11Slide
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

12. 12Slide
 Example: Metro EMS
 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.
 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.
One-Tailed Tests About a Population Mean:
s Known

13. 13Slide
1. Develop the hypotheses.
2. Level of significance and sample size are: = .05
n = 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
x
z
n
Actual z

14. 14Slide
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,

15. 15Slide
p-value

0 Zc =
1.645
 = .05
z
Za =
2.47
Solution Continued
Because p–value = .0068 <  = .05, we reject H0.

16. 16Slide
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 = 1.645
4. Determine the critical value and rejection rule.
Reject H0 if actual z > 1.645
Finding critical z value
0.5 – 0.05 = 0.45
Then, from table
1.64 + 1.65
3.29 / 2 = 1.645
One-Tailed Tests About a Population Mean:
s Known: Solution Continued

17. 17Slide
 Excel: SWStat
One-Tailed Tests About a Population Mean:
s Known

18. 18Slide
 Excel: SWStat
One-Tailed Tests About a Population Mean:
s Known
PApproach
Critical Approach
Because actual z = 2.47 >
Critical z = 1.645 we
reject H0, or
Because p–value = .0068
< α = .05, we reject
H0

19. 19Slide
Example: Glow Toothpaste
 Two-Tailed Test for Population Mean: s Known
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.
The production line for Glow toothpaste
is designed to fill tubes with a mean weight
of 6 oz. Periodically, a sample of 30 tubes
will be selected in order to check the
filling process.

20. 20Slide
Example Continued: Glow Toothpaste
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.
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.
 Two-Tailed Test for Population Mean: s Known

21. 21Slide
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
x
z
n
Actual z

22. 22Slide
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

23. 23Slide
/2 =
.015
0
z/2 = 2.17
z
/2 =
.015
-z/2 = -2.17
za = 2.74z = -2.74
1/2
p -value
= .0031
1/2
p -value
= .0031
Solution Continued
Because p–value = .0062 <  = .03, we reject H0.

24. 24Slide
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

25. 25Slide
 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

s
 
  0 6.1 6
2.74
/ .2/ 30
x
z
n
Actual Z

26. 26Slide
/2 = .015
0 2.17
Reject H0
z
Reject H0
-2.17
Actual Z Value
z
x
n


s
0
/
/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.

27. 27Slide
 Excel: SWStat
Two-Tailed Tests About a Population Mean:
s Known

28. 28Slide
 Excel: SWStat
Two-Tailed Tests About a Population Mean:
s Known
PApproach
Critical Approach

29. THOUGHT

30. 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 interval
does not contain H0 , we reject H0.

31.  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

32. 32Slide
 Confidence Interval Approach:
 For this problem, the two-sided hypothesis would be:
H0:  4
Ha:  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 Example
Solution
146 – + (2.040) 17.3 /5.657

33. 33Slide
 Test Statistic
Hypothesis Tests About a Population Mean:
s Unknown
t
x
s n

 0
/
This test statistic has a t distribution
with n - 1 degrees of freedom.
Actual t Value

34. 34Slide
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
Example: Highway Patrol
 One-Tailed Test About a Population Mean: s Unknown
The locations where H0 is rejected are deemed
the best locations for radar traps.
H0:  < 65

35. 35Slide
Example Continued: Highway Patrol
At Location F on I-75, a sample of 64 vehicles shows a
mean speed of 66.2 mph with a
sample standard deviation of
4.2 mph. Use  = .05 to
test the hypothesis.
Use Excel

36. 36Slide
Using SWStat

37. 37Slide
Solution Using SWStat
PApproach
Critical Approach
Since p=0.0128 < α=0.05
we reject H0
 The locations where H0
is rejected are deemed
the best locations for radar
traps.
H0:  < 65

38. 38Slide

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

39. 39Slide
 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

40. 40Slide
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 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

41. 41Slide
162.3
106
250256





ns
X
t

 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.
Conclusion
The mean number of
amps produced by the
new machine is more
than 250 per hour.
Actual t

42. 42Slide
Solution Using SWStat

43. 43Slide
Solution Continued
 Since computed t (or actual t)
of 3.162 > critical t of 1.833
and since p of .0058 < 
thus, we reject Ho
Hence, we conclude that the
mean number of amps
produced by the new machine
is more than 250 per hour.
H0: µ < 250

44. 44Slide
 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
and
Solution

45. 45Slide
 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)

46. 46Slide
Thinking Challenge 4 (Solution)

47. 47Slide
Summary of
Selecting an Appropriate Test Statistic for a
Test about a Population Mean

48. End of Chapter 10

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