1. Hypothesis and Hypothesis Testing
HYPOTHESIS A statement about the value of a population parameter developed for the purpose of testing.
HYPOTHESIS TESTING A procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.
TEST STATISTIC A value, determined from sample information, used to determine whether to reject the null hypothesis.
CRITICAL VALUE The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.

2. Important Things to Remember about H0 and H1
H0: null hypothesis and H1: alternate hypothesis
H0 and H1 are mutually exclusive and collectively exhaustive
H0 is always presumed to be true
H1 is the research hypothesis
A random sample (n) is used to “reject H0”
If we conclude 'do not reject H0', this does not necessarily mean that the null hypothesis is true, it only suggests that there is not sufficient evidence to reject H0; rejecting the null hypothesis then, suggests that the alternative hypothesis may be true.
Equality is always part of H0 (e.g. “=” , “≥” , “≤”).
“≠” “<” and “>” always part of H1
In actual practice, the status quo is set up as H0
In problem solving, look for key words and convert them into symbols. Some key words include: “improved, better than, as effective as, different from, has changed, etc.”
Keywords
Inequality
Symbol
Part of:
Larger (or more) than
> H1
Smaller (or less)
<
No more than  H0
At least ≥
Has increased
Is there difference? ≠
Has not changed =
Has “improved”, “is better than”. “is more effective”
See left text

3. Signs in the Tails of a Test

4. Two-tailed Test
Two-tailed tests - the rejection region is in both tails of the distribution
Rejection
Region
Rejection
Region
Acceptance
Region
One-tailed tests - the rejection region is in only on one tail of the distribution
One-tailed Test
Rejection
Region
Acceptance
Region

5. Types of Errors is true is false Type I error P(Type I)= Correct
Decision
Reject
Type II error
P(Type II)=
Correct
Decision
Do not reject
Type I Error -
Defined as the probability of rejecting the null hypothesis when it is actually true.
This is denoted by the Greek letter “”
Also known as the significance level of a test
Type II Error:
Defined as the probability of “accepting” the null hypothesis when it is actually false.
This is denoted by the Greek letter “β”

6. Hypothesis Setups for Testing a Mean () or a Proportion ()

7. Steps in hypothesis testing
- Define Null hypothesis
- Define Alternative hypothesis
- Calculate Test statistic
- Determine Rejection region
Compare Value of the test statistic with
Critical Value
- Conclusion

8. Testing for a Population Mean with a Known Population Standard Deviation- Example
Jamestown Steel Company manufactures and assembles desks and other office equipment . The weekly production of the Model A325 desk at the Fredonia Plant follows the normal probability distribution with a mean of 200 and a standard deviation of 16. Recently, new production methods have been introduced and new employees hired. The mean number of desks produced during last 50 weeks was The VP of manufacturing would like to investigate whether there has been a change in the weekly production of the Model A325 desk, at 1% level of significance.
Step 4: Formulate the decision rule.
Reject H0 if |Z| > Z/2
Step 5: Make a decision and interpret the result.
Because 1.55 does not fall in the rejection region, H0 is not rejected. We conclude that the population mean is not different from 200. So we would report to the vice president of manufacturing that the sample evidence does not show that the production rate at the plant has changed from 200 per week.
Step 1: State the null hypothesis and the alternate hypothesis.
H0:  = 200
H1:  ≠ 200
(note: This is a 2-tail test, as the keyword in the problem “has changed”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since σ is known

9. Testing for a Population Mean with a Known Population Standard Deviation- Another Example
Suppose in the previous problem the vice president wants to know whether there has been an increase in the number of units assembled. To put it another way, can we conclude, because of the improved production methods, that the mean number of desks assembled in the last 50 weeks was more than 200?
Recall: σ=16,  =200, α=.01
Step 1: State the null hypothesis and the alternate hypothesis.
H0:  ≤ 200
H1:  > 200
(note: This is a 1-tail test as the keyword in the problem “an increase”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since σ is known
Step 4: Formulate the decision rule.
Reject H0 if Z > Z
Step 5: Make a decision and interpret the result.
Because 1.55 does not fall in the rejection region, H0 is not rejected. We conclude that the average number of desks assembled in the last 50 weeks is not more than 200

10. p-value in Hypothesis Testing
EAMPLE p-Value
Recall the last problem where the hypothesis and decision rules were set up as:
H0:  ≤ 200
H1:  > 200
Reject H0 if Z > Z
where Z = 1.55 and Z =2.33
Reject H0 if p-value < 
is not < 0.01
Conclude: Fail to reject H0
p-VALUE is the probability of observing a sample value as extreme as, or more extreme than, the value observed, given that the null hypothesis is true.
In testing a hypothesis, we can also compare the p-value to the significance level ().
Decision rule using the p-value:
Reject null hypothesis, if p< α

11. Interpreting the p-value
Describing the p-value
If the p-value is less than 1%, there is overwhelming evidence that supports the alternative hypothesis.
If the p-value is between 1% and 5%, there is a strong evidence that supports the alternative hypothesis.
If the p-value is between 5% and 10% there is a weak evidence that supports the alternative hypothesis.
If the p-value exceeds 10%, there is no evidence that supports the alternative hypothesis.

12. The Power of Statistical Test
The power of a statistical test, given as 1 – b  = P (reject H0 when H0 is false), measures the ability of the test to perform as required. This 1 – b  is called the power of the function. This means that greater the power of the function the better would be the decision rule.
There are two types of tail test
1. One-tailed tests - the rejection region is in only
one tail of the distribution
2. Two-tailed tests - the rejection region is in both
tails of the distribution

13. Steps in Hypothesis Testing using SPSS
State the null and alternative hypotheses
Define the level of significance (α)
Calculate the actual significance : p-value
Make decision : Reject null hypothesis, if p≤ α, for 2-tail test; and
if p*≤ α, for 1-tail test.(p* is p/2 when p is obtained from 2-tail test)
Conclusion

14. Inference About a Population Mean When the Population Standard Deviation Is Unknown or When the Sample Size is Small
In practice, the population standard deviation will be unknown.
Recall that when s is known we use the following statistic to estimate and test a population mean
When s is unknown or when the sample size is small, we use its point estimator s, and the z-statistic is replaced then by the t-statistic

15. The t - Statistic t s d.f. = v2 d.f. = v1 v1 < v2
The “degrees of freedom”,
(a function of the sample size)
determine how spread the
distribution is (compared to the
normal distribution)
The t distribution is mound-shaped,
and symmetrical around zero.
d.f. = v2
d.f. = v1
v1 < v2

16. Testing m when s is unknown
Example
In order to determine the number of workers required to meet demand, the productivity of newly hired trainees is studied.
It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring.
Can we conclude that this belief is correct, based on productivity observation of 50 trainees (see file PROD.sav).

17. Testing m when s is unknown
Example – Solution
The problem objective is to describe the population of the number of packages processed in one hour.
H0:m = H1:m > 450
The t statistic d.f. = n - 1 = 49

18. Testing m when s is unknown
Solution continued (solving by hand)
The rejection region is t > ta,n – 1 ta,n - 1 = t.05,49 @ t.05,50 =

19. Testing m when s is unknown
The test statistic is
Rejection region
1.676
1.89
Since 1.89 > we reject the null hypothesis in favor of the alternative.
There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level.

20. Solution using SPSS (use file PROD.sav)
One-Sample Statistics N
Mean
Std. Deviation
Std. Error Mean
Packages 50
460.38
38.827
5.491
One-Sample Test
Test Value = 450 t df
Sig. (2-tailed)
Mean Difference
95% Confidence Interval of the Difference
Lower
Upper
Packages
1.890 49
.065
10.380
-.65
21.41

21. Inference About a Population Proportion
Statistic and sampling distribution
the statistic used when making inference about p is:
Under certain conditions, [np > 5 and n(1-p) > 5], is approximately normally distributed, with m = p and s2 = p(1 - p)/n.

22. Testing and Estimating the Proportion
Test statistic for p

23. Testing the Proportion
Example 12.6
A pharmaceutical company claimed that its medicine was 80% effective in relieving allergy. In a sample of 200 persons, who were given medicine only 150 persons had relief. Do you thank that the effectiveness is below 80%? Use 0.05 level of significance.

24. Testing the Proportion
Solution
The problem objective is to test the effectiveness of medicine.
The data are nominal.
The parameter to be tested is ‘p’.
Success is defined as “having relief”.
The hypotheses are:
H0: p = .8
H1: p < .8

25. Testing the Proportion
Solution
The rejection region is z < za = z.05 =
The sample proportion is
The value of the test statistic is
Since calculated z is less than critical value, we reject null hypothesis and conclude that the claim of the company that its medicine is 80% effective is not justified.

26. T-Tests : When sample size is small (<30) or When the Population Standard Deviation Is Unknown
Variable : Normal
Types of t-tests: One-sample t-test Paired or dependent sample t-test Independent samples t-test (Equal and
Unequal Variance)

27. One-sample t-test

28. Paired sample t-test

29. Matched pairs
The mean of the population differences is
that is
Test statistic:
Degree of freedom =

30. Independent sample t-test

31. The sampling process.
Population 1
Population2
Parameters:
Parameters:
Statistics:
Statistics:
Sample size:
Sample size:

32. If the two population standard deviations are
unknown, then we can estimate the standard
error of the difference between two means.

33. Test statistic:

34. If population variance unknown and the sample size
is small and the population variances are equal
Then we will use the weighted average called a
“ pooled estimate” of
Where:

35. Test statistic:
Degree of freedom =

36. Analysis of Variance ( ANOVA )
One way
Analysis of Variance ( ANOVA )
ANOVA is a technique used to test a hypothesis concerning the means of three or more populations.

37. Comparing Means of Three or More Populations
The F distribution is used for testing whether two or more sample means came from the same or equal populations.
Assumptions:
The sampled populations follow the normal distribution.
The populations have equal standard deviations.
The samples are randomly selected and are independent.
The Null Hypothesis is that the population means are the same. The Alternative Hypothesis is that at least one of the means is different.
H0: µ1 = µ2 =…= µk
H1: The means are not all equal
Reject H0 if F > F,k-1,n-k

38. The test statistic used to test the hypothesis is
F statistic
Assumptions:
1. The random variable is normally
distributed.
2. The population variances are equal.

39. ANOVA – Example (File Airlines.sav)
Recently a group of four major carriers joined in hiring Brunner Marketing Research, Inc., to survey recent passengers regarding their level of satisfaction with a recent flight. The survey included questions on ticketing, boarding, in-flight service, baggage handling, pilot communication, and so forth.
Twenty-five questions offered a range of possible answers: excellent, good, fair, or poor. A response of excellent was given a score of 4, good a 3, fair a 2, and poor a 1. These responses were then totaled, so the total score was an indication of the satisfaction with the flight. Brunner Marketing Research, Inc., randomly selected and surveyed passengers from the four airlines.
Is there a difference in the mean satisfaction level among the four airlines?
Use the .01 significance level.
Step 1: State the null and alternate hypotheses.
H0: µE = µA = µT = µO
H1: The means are not all equal
Reject H0 if F > F,k-1,n-k
Step 2: State the level of significance.
The .01 significance level is stated in the problem.

40. ANOVA – Example
Step 3: Find the appropriate test statistic. Use the F statistic
Calculations: It is convenient to summarize the calculations of F statistic in an ANOVA Table.

41. ANOVA – Example
Compute the value of F and make a decision
We find deviation of each observation from the grand mean, square the deviations, and sum this result for all 22 observations.
SS total = {( )2 + ( )2 + ……+ ( )2 } =
To compute SSE, find deviation between each observation and its treatment mean. Each of these values is squared and then summed for all 22 observations.
SSE = {( )2 + ( )2 + ……+ ( )2 } + {( )2 + ( )2 + ……+ ( )2 } + {( )2 + ( )2 + ……+ ( )2 } + {(68-69)2 + (70-69)2 + ……+ (65-69)2 } =
Finally, determine SST = SS total – SSE.
SST = – =

42. ANOVA – Example
Step 3: Find the appropriate test statistic. Use the F statistic
Calculations: It is convenient to summarize the calculations of F statistic in an ANOVA Table.
Step 4: State the decision rule.
Reject H0 if: F > F,k-1,n-k
F > F.01,4-1,22-4
F > F.01,3,18
F > 5.09
Step 5: Make a decision.
The computed value of F is 8.99, which is greater than the critical value of 5.09, so the null hypothesis is rejected.
Conclusion: The mean scores are not the same for the four airlines; at this point we can only conclude there is a difference in the treatment means. We cannot determine which treatment groups differ or how many treatment groups differ.

43. ANOVA Example – SPSS Output
Test of Homogeneity of Variances
Satisfaction
Levene Statistic
df1
df2
Sig.
.962 3 18
.432
ANOVA
Satisfaction
Sum of Squares df
Mean Square F
Sig.
Between Groups 3
8.991
.001
Within Groups 18
33.023
Total 21

44. ANOVA Example – SPSS Output
Multiple Comparisons
Satisfaction
Tukey HSD
(I) Carrier
(J) Carrier
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
Eastern
TWA
9.050
3.855
.124
-1.85
19.95
Allegheny
14.393*
3.602
.004
4.21
24.57
Ozark
18.250*
3.709
.001
7.77
28.73
-9.050
-19.95
1.85
5.343
3.365
.410
-4.17
14.85
9.200
3.480
.071
-.63
19.03 *
-24.57
-4.21
-5.343
-14.85
4.17
3.857
3.197
.631
-5.18
12.89 *
-28.73
-7.77
-9.200
-19.03
.63
-3.857
-12.89
5.18
*. The mean difference is significant at the 0.05 level.

45. ANOVA Example – SPSS Output
Homogeneous Subsets
Satisfaction
Tukey HSDa,b
Carrier N
Subset for alpha = 0.05 1 2
Ozark 6
69.00
Allegheny 7
72.86
TWA 5
78.20
Eastern 4
87.25
Sig.
.078
.085
Means for groups in homogeneous subsets are displayed.
a. Uses Harmonic Mean Sample Size =
b. The group sizes are unequal. The harmonic mean of the group sizes is used. Type I error levels are not guaranteed.

46. Chi-squared Test of a Contingency Table
Test of Independence : Test on association between two nominal variables regarding contingency tables.
Null Hypothesis : Two variables are independent
Alternative Hypothesis : The two variables are dependent

47. The Chi-square Distribution
At the outset, we should know that the chi-square distribution has only one parameter called the ‘degrees of freedom’ (df ) as is the case with the t-distribution. The shape of a particular chi-square distribution depends on the number of degrees of freedom.

48. Properties of Chi-square Distribution
Chi-square is non-negative in value; it is either zero or positively valued.
It is not symmetrical; it is skewed to the right.
There are many chi-square distributions. As with the t-distribution, there is a different chi-square distribution for each degree-of-freedom value.

49. The chi-squared statistic measures the difference
between the actual counts and the expected
counts ( assuming validity of the null hypothesis)
( Observed count - Expected count )2
The sum
Expected count

50. Contingency table c2 test – Example
In an effort to better predict the demand for courses offered by a certain MBA program, it was hypothesized that students’ academic background affect their choice of MBA major, thus, their courses selection.
A random sample of last year’s MBA students was selected. The data is given in the file Chi-Sq_MBA.sav. The following contingency table summarizes relevant data.
The file Chi_Sq_MBA_Table.sav gives the data as per the contingency table.

51. Contingency table c2 test – Example
The observed values

52. Contingency table c2 test – Example
Solution
The hypotheses are:
H0: The two variables are independent
H1: The two variables are dependent
k is the number of cells in
the contingency table.
The test statistic
The rejection region

53. Estimating the expected frequencies
Undergraduate
MBA Major
Degree
Accounting
Finance
Marketing
Probability 60 BA 60
60/152
BENG 31
31/152
BBA 39 39
39/152
Other 22
22/152 61 44
152
152 61 44 47
152
Probability
61/152
44/152
47/152
Under the null hypothesis the two variables are independent:
P(Accounting and BA) = P(Accounting)*P(BA)
= [61/152][60/152].
The number of students expected to fall in the cell “Accounting - BA” is
eAcct-BA = n(pAcct-BA) = 152(61/152)(60/152) = [61*60]/152 = 24.08
The number of students expected to fall in the cell “Finance - BBA” is
eFinance-BBA = npFinance-BBA = 152(44/152)(39/152) = [44*39]/152 = 11.29

54. The expected frequencies for a contingency table
The expected frequency of cell of raw i and column j in the contingency table is calculated by
Eij =
(Column j total)(Row i total)
Sample size

55. Calculation of the c2 statistic
Solution – continued
Undergraduate
MBA Major
Degree
Accounting
Finance
Marketing BA
31 (24.08)
13 (17.37)
16 (18.55) 60
BENG
8 (12.44)
(8.97)
7 (9.58) 31
BBA
12 (15.65)
10 (11.29)
17 (12.06) 39
Other
(8.83)
5 (6.39)
7 (6.80) 22 61 44 47
152
The expected frequency
c2=
( )2
24.08
( )2
6.39
( )2
6.80
= 14.70
+….+
+….+

56. Contingency table c2 test – Example
Solution – continued
The critical value in our example is:
Conclusion:
Since c2 = > , there is sufficient evidence to infer at 5% significance level that students’ undergraduate degree and MBA students courses selection are dependent.

57. SPSS Output Chi-Square Tests Value df Asymp. Sig. (2-sided)
Pearson Chi-Square
14.702a 6
.023
Likelihood Ratio
13.781
.032
Linear-by-Linear Association
2.003 1
.157
N of Valid Cases
152
a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 6.37.

58. Yates’ Correction for Continuity
Chi-square distribution is a continuous distribution. Whenever the degrees of freedom (in case of a 2x2 table), certain corrections for continuity can be made

59. Required conditions – the rule of five
The test statistic used to perform the test is only approximately Chi-squared distributed.
For the approximation to apply, the expected cell frequency has to be at least 5 for all the cells (np ³ 5).
If the expected frequency in a cell is less than 5, combine it with other cells.