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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.
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.”
Inequality
Symbol
Part of:
Larger (or more) than
>
H1
Smaller (or less)
<
H1
No more than

H0
At least
≥
H0
Has increased
>
H1
Is there difference?
≠
H1
Has not changed
=
H0
Keywords
Has “improved”, “is better
than”. “is more effective”
See left
text
H1
Signs in the Tails of a Test
Two-tailed Test
Two-tailed tests
- the rejection
region is in both Rejection
tails of the
Region
distribution
One-tailed
tests - the
rejection
region is in
only on one
tail of the
distribution
Acceptance
Region
Rejection
Region
One-tailed Test
Rejection
Region
Acceptance
Region
Types of Errors
RejectH 0
Do not reject
H0
H 0 is true
H 0 is false
Type I error
P(Type I)= 
Correct
Decision
Correct
Decision
Type II error
P(Type II)= 
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”
Hypothesis Setups for Testing a Mean () or a
Proportion ()
MEAN
PROPORTION
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
Testing for a Population Mean with a
Known Population Standard Deviation- Example
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
203.5. 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 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
Step 4: Formulate the decision rule.
Reject H0 if |Z| > Z/2
Z  Z / 2
X 
 Z / 2
 / n
203.5  200
 Z .01/ 2
16 / 50
1.55 is not  2.58
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.
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
p-value in Hypothesis Testing

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:
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 < 
0.0606 is not < 0.01
Reject null hypothesis, if p< α
Conclude: Fail to reject H0
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.
The Power of Statistical Test
The power of a statistical test, given as 1 –  =
P (reject H0 when H0 is false), measures the
ability of the test to perform as required. This
1 –  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
Steps in Hypothesis Testing using SPSS
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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
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  is known we use the
following statistic to estimate and test a
x
population mean
z
 n
When  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
The t - Statistic
t
x
s n
The t distribution is mound-shaped, The “degrees of freedom”,
(a function of the sample size)
and symmetrical around zero.
determine how spread the
distribution is (compared to the
normal distribution)
d.f. = v2
d.f. = v1
v 1 < v2
0
Testing  when  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).
Testing  when  is unknown

Example – Solution
– The problem objective is to describe the
population of the number of packages
processed in one hour.
–
H0: = 450
H1: > 450
– The t statistic
t
x 
s
n
d.f. = n - 1 = 49
Testing  when  is unknown

Solution continued (solving by
hand)
From the data we have
– The rejection region is
t > t,n – 1
t,n - 1 = t.05,49
@ t.05,50 = 1.676.
 x i  23,019
2
x
 i  10,671,357, thus
23,019
x
 460.38, and
50

x

x 

n

2
s2
2
i
i
n 1
s  1507 .55  38.83
 1507 .55.
Testing  when  is unknown
Rejection region
• The test statistic is
t
x 
s
n
1.676

460.38  450
38.83
50
1.89
 1.89
• Since 1.89 > 1.676 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.
Solution using SPSS (use file PROD.sav)
One-Sample Statistics
N
Packages
Mean
50
Std. Deviation
460.38
Std. Error Mean
38.827
5.491
One-Sample Test
Test Value = 450
95% Confidence Interval of
the Difference
t
Packages
1.890
Sig. (2tailed)
df
49
.065
Mean
Difference
10.380
Lower
-.65
Upper
21.41
Inference About a Population
Proportion

Statistic and sampling distribution
– the statistic used when making inference
about p is:
x
p̂ 
where
n
x  the number of successes .
n  sample size .
– Under certain conditions, [np > 5 and n(1-p)
> 5], p̂ is approximately normally distributed,
with  = p and 2 = p(1 - p)/n.
Testing and Estimating the
Proportion

Test statistic
for p
p̂  p
Z
p(1  p) / n
where np  5 and n(1  p)  5
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.
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
Testing the Proportion
– Solution
• The rejection region is z < z = z.05 = -1.645.
• The sample proportion is pˆ  150 200  .75
• The value of the test statistic is
Z
pˆ  p
.75  .8

 1.786
p(1  p) / n
.8(1  .8) / 200
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.
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)
One-sample t-test

H1 : 
H1 : 
H1 : 
H0 :
0
 0
 0
 0

Paired sample t-test
H0 :
H1 :
H1 :
H1 :
d
d
d
d
 0
 0
 0
 0
Matched pairs
The mean of the population differences is
that is
1   2   D
Test statistic:
xD   D
t
s D nD
Degree of freedom =
nD  1
D
Independent sample t-test
H 0 : 1   2
H 1 : 1   2
H 1 : 1   2
H 1 : 1   2
The sampling process.
Population2
Population 1
Parameters:
Parameters:
 1and 1
 2 and 22
2
Statistics:
Statistics:
x2 ands
2
x1 ands1
Sample size:
n1
2
2
Sample size: n2
If the two population standard deviations are
unknown, then we can estimate the standard
error of the difference between two means.
ˆ x1  x2 
ˆ 1
2
n1

ˆ 2
2
n2
Test statistic:
z
 x1  x2 
ˆ
2
1
n1

ˆ
2
2
n2
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
2
“ pooled estimate” of 
 x1  x2
1 1
 sp   
 n1 n2 
2
Where:

n  1s  n  1s
s 
2
p
1
2
1
2
n1  n2  2
2
2
Test statistic:
t
x1  x2 
1 1
s   
 n1 n2 
2
p
Degree of freedom =
n1  n2  2
One way
Analysis of Variance ( ANOVA )
ANOVA is a technique used to test a
hypothesis concerning the means
of three or more populations.
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
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.
H 0 : 1   2  3  ........
H1 : Not all means are same
ANOVA – Example (File Airlines.sav)
EXAMPLE
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.
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.
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 = {(94-75.64)2 + (90-75.64)2 + ……+ (6575.64)2 }
= 1485.10
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 = {(94-87.25)2 + (90-87.25)2 + ……+ (80-87.25)2 } + {(75-78.20)2 + (68-78.20)2 +
……+ (88-78.20)2 } + {(70-72.86)2 + (73-72.86)2 + ……+ (65-72.86)2 } + {(68-69)2 + (7069)2 + ……+ (65-69)2 } = 594.41
Finally, determine SST = SS total – SSE.
SST = 1485.10 – 594.41 = 890.69
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.
ANOVA Example – SPSS Output
Test of Homogeneity of Variances
Satisfaction
Levene Statistic
df1
df2
.962
3
18
Sig.
.432
ANOVA
Satisfaction
Between
Groups
Within Groups
Total
Sum of
Squares
890.684
df
3
Mean Square
296.895
594.407
18
33.023
1485.091
21
F
8.991
Sig.
.001
ANOVA Example – SPSS Output
Multiple Comparisons
Satisfaction
Tukey HSD
(I) Carrier
Eastern
TWA
Allegheny
Ozark
(J) Carrier
95% Confidence Interval
Mean
Difference (ILower
Upper
Std. Error
Sig.
J)
Bound
Bound
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
Eastern
-9.050
3.855
.124
-19.95
1.85
Allegheny
5.343
3.365
.410
-4.17
14.85
Ozark
9.200
3.480
.071
-.63
19.03
*
Eastern
-14.393
3.602
.004
-24.57
-4.21
TWA
-5.343
3.365
.410
-14.85
4.17
Ozark
3.857
3.197
.631
-5.18
12.89
*
Eastern
-18.250
3.709
.001
-28.73
-7.77
TWA
-9.200
3.480
.071
-19.03
.63
Allegheny
-3.857
3.197
.631
-12.89
5.18
*. The mean difference is significant at the 0.05 level.
ANOVA Example – SPSS Output
Homogeneous Subsets
Satisfaction
Tukey HSDa,b
Carrier
Subset for alpha = 0.05
Ozark
N
6
1
69.00
Allegheny
7
72.86
TWA
5
78.20
Eastern
4
Sig.
2
78.20
87.25
.078
.085
Means for groups in homogeneous subsets are displayed.
a. Uses Harmonic Mean Sample Size = 5.266.
b. The group sizes are unequal. The harmonic mean of the group
sizes is used. Type I error levels are not guaranteed.
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
The Chi-square Distribution
At the outset, we should know that the chisquare 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.
Properties of Chi-square Distribution
1. Chi-square is non-negative in value; it is
either zero or positively valued.
2. It is not symmetrical; it is skewed to the
right.
3. There are many chi-square distributions.
As with the t-distribution, there is a
different chi-square distribution for each
degree-of-freedom value.
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
O  E 

2
k
i
i 1
i
Ei
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.
Contingency table c2 test –
Example
Degree
BA
BENG
BBA
Other
Accounting
31
8
12
10
61
Finance
13
16
10
5
44
The observed values
Marketing
16
7
17
7
47
60
31
60
39
152
Contingency table c2 test –
Example

Solution
– The hypotheses are:
H0: The two variables are independent
H1: The two variables are dependent
– The test statistic
c
(Oi  Ei )

Ei
i 1
k
2
k is the number of cells in
the contingency table.
– The rejection region
2
c2  c2,(r 1)( c 1)
Estimating the expected
frequencies
Undergraduate
Degree
Accounting
BA
BENG
BBA
Other
6161
Probability
61/152
MBA Major
Finance Marketing
44
44
44/152
6060
31
3939
22
47
47/152
Probability
60/152
31/152
39/152
22/152
152
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
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
(Column j total)(Row i total)
Eij =
Sample size
(Oi  Ei )
c 
Ei
i 1
k
2
2
Calculation of the c2 statistic
• Solution – continued
Undergraduate
Degree
Accounting
31 (24.08)
24.08
BA
k
BENG
2 8 (12.44)
BBA 31 24.08
12 (15.65)
Other
10 (8.83)
i61
1
31 24.08
c 
31
24.08
31
c2=
24.08

MBA Major
Finance
Marketing
13 (17.37) 2 16 (18.55)
16
(8.97)
7 (9.58)
i
i
10 (11.29) 17 (12.06)
(6.39) 77 6.80
(6.80)
55 6.39
i
44
47
(f  e )
e
5 6.39
The expected frequency
5 6.39
60
31
39
22
152
7 6.80
7 6.80
7 6.80
5 6.39
(31 - 24.08)2
(5 - 6.39)2
(7 - 6.80)2
=
+….+
+….+
24.08
6.39
6.80
14.70
Contingency table c2 test –
Example
• Solution – continued
– The critical value in our example is:
c 2 ,( r 1)( c 1)  c.205,( 4 1)( 31)  12.5916
• Conclusion:
Since c2 = 14.70 > 12.5916, there
is sufficient evidence to infer at 5% significance
level that students’ undergraduate degree
and MBA students courses selection
are dependent.
SPSS Output
Chi-Square Tests
Value
Pearson Chi-Square
Likelihood Ratio
Linear-by-Linear Association
N of Valid Cases
Asymp. Sig. (2sided)
df
14.702a
6
.023
13.781
6
.032
2.003
1
.157
152
a. 0 cells (.0%) have expected count less than 5. The minimum expected count is
6.37.
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
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  5).
If the expected frequency in a cell is
less than 5, combine it with other cells.