1. Chapter 5
STATISTICS (PART 3)

2. Hypothesis test
In statistics, a hypothesis is a claim or statement about a property of a population.
Hypothesis testing can be used to determine whether statement about the value of a population parameter should or should not be rejected.
The three methods used to test hypotheses are
The traditional method
The P-value method
The confidence interval method

3. Hypothesis & test procedures
A statistical test of hypothesis consist of : 1. The Null hypothesis, 2. The Alternative hypothesis, 3. The test statistic 4. The rejection region 5. The conclusion

4. Definition 1: Null hypothesis, H0 : is a tentative assumption about a
population parameter.
Alternative hypothesis, H1 : is the opposite of what is
stated in the null hypothesis.
Test Statistic is a function of the sample data on which the
decision is to be based.

5. 6 Steps to Perform a Hypothesis Testing
State the null hypothesis, H0 and the alternative hypothesis, H1
Choose the significance level,  and the sample size, n
Determine the appropriate test statistic
Determine the critical values that divide the rejection and non rejection regions
Collect data and compute the value of the test statistic
Make the statistical decision and state the managerial conclusion. If the test statistic falls into the non rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem.

6. Developing Null and Alternative Hypothesis
It is not always obvious how the null and alternative hypothesis should be formulated.
In some cases it is easier to identify the alternative hypothesis first. In other cases the null is easier.

7. Step 1:Statement of hypothesis
The null (H0) and alternative (H1) hypothesis are stated together as below:
Tail of a test:
Two-Tailed Test
Left-Tailed Test
Right-Tailed Test
Sign in H0 =
Sign in H1 ≠
<
>
Rejection Region
In both tail
In the left tail
In the right tail

8. HYPOTHESIS TESTING CoMMON PHRASES
>
Is greater than
Is above
Is longer than
Is bigger than
Is increased
<
Is bellow than
Is below
Is lower than
Is smaller than
Is decreased or reduced from =
Is equal to
Is the same as
Has not changed from ≠
Is not equal to
Is different from
Has changed from
Is not the same as

9. Example 6.6:
(a) Situation A: A chemist invents an additive to increase the life of an automobile battery. If the mean lifetime of the automobile battery without the additive is 36 months, then her hypotheses are, (b) Situation B: A contractor wishes to lower heating bills by using a special type of insulation in houses. If the average of the monthly heating bills is $78, her hypotheses about heating costs with the use of insulation are,

10. Solution:
(a)
In this situation, the chemist is interested only in increasing the lifetime of the batteries,
so her alternative hypothesis is that the mean is greater than 36 months. The null hypothesis is that the mean is equal to 36 months. This test is called right-tailed, since the interest is in an increase only.

11. Solution:
(b)
This test is a left-tailed test, since the contractor is interested only in lowering heating costs.

12. Exercise 6.7:
State the null and alternative hypotheses for each conjecture.
A researcher thinks that if expectant mothers use vitamin pills, the birth weight of the babies will increase. The average birth weight of the population is 8.6 pounds.
An engineer hypothesizes that the mean number of defects can be decreased in a manufacturing process of compact disks by using robots instead of humans for certain tasks. The mean number of defective disks per is 18.
A psychologist feels that playing soft music during a test will change the results of the test. The psychologist is not sure whether the grades will be higher or lower. In the past, the mean of the scores was 73.

13. Solution:

14. Population Proportion
TEST STATISTICS
Population Mean
σ known
Any size
σ Unknown
Population Proportion

15. How to decide whether to reject or fail to reject ?
The entire set of values that the test statistic may assume is divided into two regions. One set, consisting of values that support the H1 and lead to reject H0 , is called the rejection region. The other, consisting of values that support the H0 is called the acceptance region.

16. Critical Value Approach to One-Tailed Hypothesis Testing
The test statistic z has a standard normal probability distribution.
We can use the standard normal probability distribution table to find the z-value with an area of a in the left/lower (or right/upper) tail of the distribution.
The value of the test statistic that established the boundary of the rejection region is called the critical value, Za for the test.
The rejection rule is:
Left tail (H1: µ<µ0) Reject H0 if Z < -Za
Right tail (H1: µ>µ0) Reject H0 if Z > Za

17. left-Tailed Test About a Population Mean:s Known
Critical Value Approach
Sampling
Distribution of
Reject H0
a 1
Do Not Reject H0 z
-Za = -1.28

18. Right-Tailed Test About a Population Mean:s Known
Critical Value Approach
Sampling
Distribution of
Reject H0

Do Not Reject H0 z
Za = 1.645

19. Example 6.7:

20. Solution

21. Exercise 6.8: The response times for a random sample of 40
medical emergencies were tabulated. The sample mean is minutes. The population standard deviation is believed to be 3.2 minutes. The EMS director wants to perform a hypothesis test, with a .05 level of significance, to determine whether the service goal is greater than 12 minutes is being achieved.
(Answer: reject H0)

22. Critical Value Approaches
1. Develop the hypotheses.
H0: =
H1:>
2. Specify the level of significance.
a = 0.05
3. Compute the value of the test statistic.

23. 4. Determine the critical value and rejection rule.
For a = 0.05, Z0.05 = 1.645
Reject H0 if Z > 1.645
5. Determine whether to reject H0.
Because 2.47> 1.645, we reject H0 at α=0.05. we conclude that claim is true.

24. Right-Tailed Tests About a Population Mean:
s Known
Critical Value Approach
Sampling
Distribution of
Reject H0

Do Not Reject H0 z
Za = 1.645
Z =2.47

25. Exercise 6.9:
A researcher claims that the average cost of men’s athletic shoes is less than $80. He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in dollars). (The costs have been rounded to the nearest dollar.) Is there enough evidence to support the researcher’s claim at a = 0.10? Assume that . (Answer: reject H0)

26. Critical Value Approaches
1. Develop the hypotheses.
H0:  = 80
H1:<80
2. Specify the level of significance.
a = 0.10
3. Compute the value of the test statistic.

27. 4. Determine the critical value and rejection rule.
For a = 0.10, Z0.10 =
Reject H0 if Z <
5. Determine whether to reject H0.
Because < , we reject H0.
We conclude that there is enough evidence to support the claim that the average cost of men’s athletic shoes is less than $80.

28. Left-Tailed Tests About a Population Mean:
s Known
Critical Value Approach
Sampling
Distribution of
Reject H0
a 1
Do Not Reject H0 z
-Za =
Z =

29. Critical Value Approach to two-Tailed Hypothesis Testing
The critical values will occur in both the left and right tails of the standard normal curve.
Use the standard normal probability distribution table to find Z/2 (the Z-value with an area of a/2 in the right tail of the distribution).
The rejection rule is:
Two tail (H1: µ≠µ0) Reject H0 if Z < -Z  /2 or Z > Z  /2

30. two-Tailed Test About a Population Mean:s Known
Critical Value Approach
Sampling
distribution of
Reject H0
Do Not Reject H0
Reject H0
a/2 = .015
a/2 = .015 z
-2.17
2.17

31. Two-Tailed Tests About a Population Mean:
s Known
Example 6.8: Glow Toothpaste
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. 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. 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 .03 level of significance, to help determine whether the filling
process should continue operating or be stopped and
corrected.

32. Two-Tailed Tests About a Population Mean:
s Known
Critical Value Approaches
H0: = 6
H1: ≠ 6
1. Determine the hypotheses.
2. Specify the level of significance.
a =0.03
3. Compute the value of the test statistic.

33. 4. Determine the critical value and rejection rule.
For a/2 =0 .03/2 =0 .015, Z0.015 = 2.17
Reject H0 if Z < or Z > 2.17
5. Determine whether to reject H0.
Because 2.74 > 2.17, we reject H0.
There is sufficient statistical evidence to
infer that the alternative hypothesis is true
(i.e. the mean filling weight is not 6 ounces).

34. Two-Tailed Tests About a Population Mean:
s Known
Critical Value Approach
Sampling
distribution of
Reject H0
Do Not Reject H0
Reject H0
a/2 = .015
a/2 = .015 z
-2.17
2.17

35. Exercise 6.10:
The Medical Rehabilitation Education Foundation reports that the average cost of rehabilitation for stroke victims is $24,672. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $26,343. The standard deviation of the population is $3251. At a 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24,672? (Answer: reject H0)

36. Test About a Population Mean:s unKnown
Test Statistic
This test statistic has a t distribution
with n - 1 degrees of freedom.

37. Tests About a Population Mean: s Unknown
Rejection Rule: Critical Value Approach
H1: 
Reject H0 if t > t
H1: 
Reject H0 if t < - t
H1: ≠
Reject H0 if t < - t or t > t

38. One-Tailed Tests About a Population Mean:
s Unknown
Exercise 6.8:
An educator claims that the average salary of substitute teachers in school districts in Allegheny County, Pennsylvania, is less than $60 per day. A random sample of eight school districts is selected, and the daily salaries (in dollars) are shown. Is there enough evidence to support the educator’s claim at a a = 0.10?

39. Critical Value Approaches
H1: m < 65
1. Determine the hypotheses.
2. Specify the level of significance.
a = 0.10
3. Compute the mean and s.t.d of sample.

40. 4. Compute the value of the test statistic.
4. Determine the critical value and rejection rule.
For a = 0.10 , d.f.=7 t0.10 =
Reject H0 if t <
5. Determine whether to reject H0.
Because > , we fail to reject H0.
We conclude that there is not enough evidence to support the educator’s claim that the average salary of substitute teachers in Allegheny County is less than $60 per day.

41. One-Tailed Tests About a Population Mean:
s Unknown
Critical Value Approach
Sampling
Distribution of
Reject H0
a 1
Do Not Reject H0 z
-ta =
Z =2.286

42. Steps of Hypothesis Testing
Step 1. Develop the null and alternative hypotheses.
Step 2. Specify the level of significance, .
Step 3. Collect the sample data and compute the value of the test statistic.
Step 4. Use the level of significance to determine
the critical value and the rejection rule.
Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H0.

43. A Summary of Forms for Null and Alternative Hypotheses About a Population Proportion
The equality part of the hypotheses always appears
in the null hypothesis.
In general, a hypothesis test about the value of a
population proportion p must take one of the
following three forms (where p0 is the hypothesized
value of the population proportion).
H0: p = p0
H1: p < p0
H0: p = p0
H1: p > p0
H0: p = p0
H1: p ≠ p0
One-tailed
(left tail)
One-tailed
(right tail)
Two-tailed

44. Tests About a Population Proportion
Null hypothesis: H0: p = p0
Test Statistic

45. Rejection Rule: Critical Value Approach
H1: p>p
Reject H0 if Z > Z
Reject H0 if Z < -Z
Reject H0 if Z < -Z or Z > Z
H1: p<p
H1: p≠p

46. Two-Tailed Test About a Population Proportion
Example 6.9:
For a Christmas and New Year’s week, the National Safety Council estimated that 500 people would be killed and 25,000 injured on the nation’s roads. The NSC claimed that 50% of the accidents would be caused by drunk driving. A sample of 120 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC’s claim with a = .05.

47. Two-Tailed Test About a Population Proportion
Critical Value Approaches
1. Determine the hypotheses.
2. Specify the level of significance.
a =0 .05
3. Compute the value of the test statistic.

48. 4. Determine the critical value and rejection rule.
For a/2 = .05/2 = .025, z.025 = 1.96
Reject H0 if z < or z > 1.96
5. Determine whether to reject H0.
Because > and < 1.96,
we cannot reject H0.
We conclude that there is not enough evidence to support that the accidents would be caused by drunk driving.

49. Two-Tailed Tests About a Population Mean:
s Known
Critical Value Approach
Sampling
distribution of
Reject H0
Do Not Reject H0
Reject H0
a/2 = .015
a/2 = .015 z
-1.96
1.96
Z =1.278

50. Exercise 6.11:
A statistician read that at least 77% of the population oppose replacing $1 bills with $1 coins. To see if this claim is valid, the statistician selected a sample of 80 people and found that 55 were opposed to replacing the $1 bills. At a a = 0.01, test the claim that at least 77% of the population are opposed to the change. (Answer: fail to reject H0)

51. Two-Tailed Test About a Population Proportion
Critical Value Approaches
1. Determine the hypotheses.
2. Specify the level of significance.
a =0 .01
3. Compute the value of the test statistic.

52. 4. Determine the critical value and rejection rule.
For a =0.01, Z0.01 =
Reject H0 if Z <
5. Determine whether to reject H0.
Because > -1.96, we cannot reject H0.
We conclude that there is not enough evidence to reject the claim that at least 77% of the population oppose replacing $1 bills with $1 coins.