Transcript [MSM04]

Hypothesis Testing II
Testing Hypothesis
Test of Significance of Mean for Large Sample (T)
• Examples: 1.: Assumes that the average annual
income for govt. employee in the nations is
reported by the census bureau to be Rs. 18750.
There was some doubt whether the average yearly
income of govt. employee in Pune was
representative of the national wage level.
A random sample of 100 govt. employee in Pune
was taken and found that their average salary was
Rs. 19240 with standard deviation of Rs. 2610. At
the  = 0.05 level of significance can we conclude
that the average salary of govt. employee in Pune
is representative of the national wage level.
( Standard value of z = 1.96)
Testing Hypothesis
Test of Significance of Mean for Large Sample (T)
• Examples:2.:
A sample of 400 male students is found to have a
mean height 67.47 inches. Can it be reasonably
regarded as a sample drawn from large
population with mean height 67.39 inches and
standard deviation (S.D.)1.30 inches?
• Test at a 5% level of significance.
• ( Standard value of z = 1.96)
Testing Hypothesis
Test of Significance of Mean for Large Sample (L)
• Examples: 3. The manufacturer of light bulbs claims
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that minimum average life of this bulbs is 1600
hours. We want to test his claim.
A sample of 100 light bulbs was taken at random
and the average bulbs life of this sample was
computed to be 1570 hours with standard deviation
of 120 hours. At  = 0.01, let us use test the validity of
the claim of this manufacturer.
( Standard value of z = - 2.33)
Testing Hypothesis
Test of Significance of Mean for Large Sample ®
• Examples:4.
An insurance company claims that it takes 14
days on an average to process an auto accident
claim with S.D. of 6 days. To test the validity of
this claim. An investigator randomly selected 36
people who recently field claims. This sample
reveled that it took the company an average of
16 days to process these claims. At 99%
confidence check if it takes the company more
than 14 days on an average to process a claim.
( Standard value of z = 2.33)
Testing Hypothesis
Test of Significance of Mean for Small Sample (T)
• Examples:5. A gas station repair shop claims
that it can do a lubrication job and oil change
in 30 minutes . The consumer protection
department wants to test this claim, a sample of
6 cars were sent to the station for oil change
and lubrication . The job took an average of 34
minutes with SD of 4 minutes . This claim is to
be tested at
 = 0.05.
( Standard value of t = 2.02)
t - distribution
Normal distribution
df = inf
t- distribution for sample
size n=15 (df = 15-1 = 14)
t- distribution for sample
size n=2 (df = 2-1 = 1)
Characteristics of the t - distribution
• Both t-distribution and Normal distribution
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are symmetrical.
t-distribution is flatter than the normal
distribution.
A t- distribution is lower at the mean and higher
at the tails than a normal distribution.
When sample size gets larger, the shape of the tdistribution loses its flatness & become
approximately equal to the normal distribution.
In fact sample size is more than 30, the tdistribution is so close to normal distribution
that we will use the normal to approximate the t.
t - distribution & Degree of Freedom
• There is separate t- distribution for each
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sample size.
There is different t- distribution for each of the
possible degree of freedom.
What is degree of freedom? Degree of Freedom
is the number of values we can choose freely.
(n-1)
When there are two elements in our sample we
have (2-1)= 1 degree of freedom, and with
seven elements in our sample we have (7-1)= 6
degree of freedom.
Testing Hypothesis
Test of Significance of Mean for Small Sample (T)
• Examples:6. A clam is maid that ICFAI College
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Students have an IQ OF 120. To test this claim
a random sample of 10students was taken and
their IQ score are recorded as follows.
105,110,120,125,100,130,120,115,125,130, at the
0.05 level of significance, test the validity of this
claim.
( Standard value of t = 2.26)
Testing Hypothesis
Test of Significance of Mean for Small Sample (T)
• Examples:7. A dispenser is set to dispense 500ml
of liquid. 15 samples were taken and it was
found that the mean of the sample is equal to
498ml and sample standard deviation is 5ml.
Determine whether the dispenser needs to be
readjusted at ά = 0.05.
• Ho: μ = 500 ml
• Ha: μ ≠ 500 ml
• As the sample size is n < 30 and σp is unknown,
we will use t-statistics
Testing Hypothesis
Test of Significance of Mean for Small Sample (T)
• Examples:7.
t=
=
= - 1.4966
• Assuming a 5% level of significance, the critical
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value of t for 14 degrees of freedom is ± 2.145.
As the calculated t value of -1.496 lies within
the acceptance region, so we accept the null
hypothesis i.e. the dispenser dispenses 500 ml of
liquid. Thus there is no need to readjust the
dispenser
Testing Hypothesis
Test of Significance of Difference between Two Means
(Large Sample) (T)
• Examples:8. The mean produce of wheat of a
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sample of 100 fields in 200 Qt per acre with SD
of 10 Qt. Another sample of 150 field gives the
mean of 220 Qt with SD of 12 Qt . Can the two
samples be considered to have been taken from
the same population whose SD is 11 Qt. Use  =
0.05.
(Standard value of z = 1.96)
Testing Hypothesis
Test of Significance of Difference between Two Means
(Large Sample) (T)
• Examples:9. Given the SD of each of the two
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population is 2.5 and the means of two large
samples of sized 1000 & 2000 are 67.5 & 68
respectively. Test equality of means at ά= 0.05.
n1=1000, n2= 2000, x1= 67.5, x2= 68,
•σ
p1 =2.5
& σp1 =2.5
• (Standard value of z = 1.96)
Testing Hypothesis
Test of Significance of Difference between Two Means
(Large Sample) (T)
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Examples:10. INC Student is conducting a survey in
solapur and pune on the hourly wages of landless labours.
Results of the survey are as follows. Find is there
significance difference between wages of landless labours
in solapur and pune at  = 0.05
Town
Mean Hourly wages
SD
Sample
Solapur
Rs 8.95
Rs 0.40
200
Pune
Rs 9.10
Rs 0.60
175
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n1=200, n2= 175, x1= 8.95, x2= 9.10,
• σ = 0.40 & σ
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p1
p1 =0.60
(Standard value of z = 1.96)
Testing Hypothesis
Test of Significance of Difference between Two Means
(Large Sample)
• Examples: 11. A company manufacturing
chocolates want to compare the average
manufacturing time at its two facilities. At
facility A, on the basis of 500 samples it was
observed that, the manufacturing time was 5
hours with standard deviation of 0.5 hours. At
facility B, on the basis of 600 samples it was
observed that, the manufacturing time was 5.5
hours with standard deviation of 0.5 hours.
Test the equality of means at 0.05 significance
level.
Testing Hypothesis
Test of Significance of Difference between Two Means
(Large Sample)
• Examples:11. solution
• H0 : µA = µB
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There is no difference in the means of two populations
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There is difference in the means of two populations
• Ha : µA≠ µB
• na=500, nb=600, σa =0.5, σb =0.5,
• = 5, = 5.5
Testing Hypothesis
Test of Significance of Difference between Two Means
(Small Sample) (T)
• Examples:12.
• It is desired to find out if there is any significant
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difference in the average amount of money
carried by male and female students at Ferguson
college pune, on valentine day. A random sample
of 8 male & 10 female students was selected and
the amount of money each had was found from
each sample as follows.
n1= 8, n2= 10, x1= $20.50, x2= $17.00,
σS1= $2.00 & σS1= $1.5
(Standard value of t for df =16 is 2.12)
Testing Hypothesis
Test of Significance of Difference between Two Means
(Small Sample) (T)
• Examples:13.
• The number of sales, average size of the sales
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and SD of two sales person A & B are as
follows.
A
B
Number of sales
10
17
Average size (in Rs) - 6200
5600
Standard deviation (Rs)- 690
600
Examine whether the figure s of average sales
size differ significantly (at  = 0.05)
(Standard value of t for df = 25 is 2.06)
Testing Hypothesis
Test of Significance of Difference between Two Means
(Small Sample)
• Example: 14. A supervisor observed the output of
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two machines. Machine 1 produces 25 units in an
hour and it was found that 5 units are defective
with standard deviation of 3 units. Machine 2
produces 22 units in an hour and it was found that
4 units are defective with a standard deviation of 2
units. Determine whether the output of two
machines differ significantly
= 5,
= 4, n1 = 25, n2 = 22, σS1 =3, σS1 =2
Testing Hypothesis
Test of Significance for single Proportions (Large Sample) (T)
• Example:15. The sponsor of a television show
believes that his studio audience is divided
equally between men and women. Out of 400
persons attending the show one day, there
were 230 men (57.5%) at ά= 0.05 test the belief
of the sponsor is correct.
Testing Hypothesis
Test of Significance for single Proportions (Large Sample) (T)
• Example:16. A company
contemplating to give
VRS to its employee, The human resources
director has given report to the president that
roughly 60% of the employee are eligible for VRS.
The president forms a special committee conduct
the eligible candidates. The committee conducts in
depth interviews with 120 employee and find that
in its judgment only 70% of the sample are
qualified for VRS. The president wants to know
whether the finding of the human resources
director are corrected or not. (ά= 0.05)
Testing Hypothesis
Test of Significance for single Proportions (Large Sample)
• Example:16.
• PHO = 0.6
•q
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= 0.4
n = 120
= 0.7
=0.3
HO
Testing Hypothesis
Test of Significance for single Proportions (Large Sample) (L)
• Example:17. The mayor of the city claims
that 60% the people of the city follows him,
and support his policies. we want to test
whether his claim is valid or not. A random
sample of 400 persons was taken & it was
found that 220 of these people supported the
mayor at the ά = 0.01 what can be conclude
about mayors claim.
Testing Hypothesis
Test of Significance for single Proportions (Large Sample) (L)
• Example:17.
• PHO = 0.6
•q
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= 0.4
n= 400
= 220/400 = 0.55
=180/ 400 = 0.45
HO
Testing Hypothesis
Test of Significance for single Proportions (Large Sample) (L)
• Example:18. An organization wants to
introduce flexible work timings, if more than
60% of the people are in favor of the policy.
For this purpose it conducted survey among
500 employees and found that 450 employees
were in the favor of flexible work timings. On
the basis of sample survey, what the company
should do? Assume significance level as 5%.
Testing Hypothesis
Test of Significance for single Proportions (Large Sample)
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Example:18.
PHO = 0.60 (Hypothesized value of the population
proportion in the favor of flexible work timings)
qHO = 0.40 (Hypothesized value of the population
proportion not in the favor of flexible work timings)
n= 500
= 450/500= 0.9 sample proportion in the favor of flexible
work timings
= 50/500= 0.1 sample proportion not in the favor of
flexible work timings
Chi- square test
x
2
Chi Square test
• Symbolized by Greek
• pronounced “Ki square”
• A Test of STATISTICAL
x
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SIGNIFICANCE for TABLE data
What do tests of statistical
significance tell us?
• Are OBSERVED RESULTS
• SIGNIFICANTLY
DIFFERENT than would be expected
• BY CHANCE
• Criteria
ά < .05
Testing Hypothesis
Chi- square test
• Evaluates whether observed frequencies
for a qualitative variable (or variables)
are adequately described by hypothesized
or expected frequencies.
– Qualitative (or categorical) data is a set of
observations where any single observation is
a word or code that represents a class or
category.
Testing Hypothesis
Chi- square test
•X2 - test for
– Test of deviation from expected frequencies:
Test whether the observed frequencies
deviate from expected frequencies (e.g. using
a dice, there is an a priori chance of 16.67%
for each number)
– Test of association: Finding relationship
between two or more independent variables
(e.g. test relation between gender and the use
of high or low accents?)
Chi- square distribution
 ( fo  fe )
   
fe

2
2



Testing Hypothesis
Chi- square test
• In the test of significance of mean we are comparing
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the mean of one sample with the hypothesized
population mean. In the test of significance of
difference between two means, we are comparing the
means of two samples.
In chi-square test, we can check the equality of more
than two population parameters (like proportions,
means).
If we classify a population into several categories with
respect to two attributes (such age & job performance)
we can then use a chi-square test to determine whether
the two attributes are independent of each other.
Testing Hypothesis
Chi- square test
•Using the Chi- square test:
• Chi square test enables to test
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for the equality of
several proportions.
Chi square is test of statistics used to test a
hypothesis that provides a set of theoretical
frequencies with which observed frequencies are
computed.
Its really just a comparison between expected
frequencies and observed frequencies among the
cells in a crosstabulation table.
Testing Hypothesis
Chi- square test
Conditions for a the application of Chi- square test (x2):
• All raw data for X2 must be frequencies / actual
numbers (not percentages & proportions)
• The expected frequency of cell should be more
than 5.
• The chi square test work only when the sample
size is large enough (n > 50).
• The observation drawn need to be random and
independent.
• A constraint must be linear.
Chi- square test
• Properties of Chi- square test:
• As t-distribution there is different chi- square
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distribution for each different number of degree
of freedom.
For very small number of degree of freedom, the
x2 distribution is severely skewed to the right. As
the number of degree of freedom increases, the
curve rapidly becomes more symmetrical until
the number reaches large values, at which point
the distribution can approximated by the normal.
The chi-square distribution is a probability
distribution.
The chi-square distribution involves squared
observations and hence it is always positive .
Chi-Square
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 ( fo  fe )2
   
fe

2



Cannot be negative because all discrepancies are squared.
Will be zero only in the unusual event that each observed
frequency exactly equals the corresponding expected
frequency.
Other things being equal, the larger the discrepancy between
the expected frequencies and their corresponding observed
frequencies, the larger the observed value of chi-square.
It is not the size of the discrepancy alone that accounts for a
contribution to the value of chi-square, but the size of the
discrepancy relative to the magnitude of the expected
frequency.
The value of chi-square depends on the number of
discrepancies involved in its calculation.
Chi- square as a test of Goodness of fit
• Chi- square test developed by Karl Pearson in 1990.
• Chi- square as a test of Goodness of fit, which is
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used to test whether or not the observed
frequency results support a particular hypothesis.
The test can be used to identify whether the
deviation between the observed and estimated
values can be because of a chance.
In some situations researchers would like to see
how well the observed frequency pattern will fit
in to the expected frequency pattern. In such
cases the chi square test is used to test whether the
fit between the observed distribution and the
expected distribution is good.
Testing Hypothesis
Steps in Chi- square test
An Example:1. Chi- square test
• Ex: Suppose that 60 children were asked as to
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which ice -cream flavour they liked out of the
three flavours of vanilla, strawberry & chocolate.
The answer are recorded as follows. (ά = 0.05)
Flavours
Numbers
Vanilla
17
Strawberry
24
Chocolate
19
• x2 =
∑ (Fo – Fe)
Fe
An Example:2. Chi- square test
• The following table depicts the expected
sales (Fe) and actual sales (Fo) of television
sates for company. Test whether there is a
substantial difference between the
observed and expected values. Using Chisquare test. (ά = 0.05)
(Fo)
57
69
51
83
44
48
35
37
(Fe)
59
76
55
75
39
53
30
48
Analysis of Variance
ANOVA
Analysis of Variance
ANOVA
• Till now we delt with the research
problems where tow means are involved.
However if the problem requires the
comparison of the means of more than
two populations using z-test, t-test
become more complex and tedious. To
avoid this tedious process we can use
Analysis of Variance
ANOVA developed by R. A. Fisher it is
also called the F-test.
Analysis of Variance
ANOVA
•Objective of ANOVA:
• The objective of the ANOVA test is to test
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whether there is any significant difference
between the means is of various samples
ANOVA test uses the variability between the
sample means as the basis for analysis.
ANOVA measures the variability in data point
within samples and it also measures the
variance between the sample means.
• These two variations are compared using
the F-test.
Analysis of Variance
ANOVA
• In its simplest form, it is used to compare
means for three or more categories.
– Example:
+Life Happiness scale and Marital Status
(married, never married, divorced)
• Relies on the F-distribution
– Just like the t-distribution and chi-square
distribution, there are several sampling
distributions for each possible value of df.
What is ANOVA?
• If we have a categorical variable with 3+
categories and a metric/scale variable, we
could just run 3 t-tests.
– The problem is that the 3 tests would not be
independent of each other (i.e., all of the
information is known).
A better approach: compare the
variability between groups (treatment
variance + error) to the variability
within the groups (error)
Analysis of Variance
ANOVA
•The methodology of ANOVA is based on the
following assumptions:
1.Each sample size ‘n’ is drawn randomly &
each sample is independent of the other
samples.
2. The population are normally distributed.
3. The population from which the samples are
drawn have equal variances.
Two Sources of Variability ANOVA
• In ANOVA, an estimate of variability between groups
is compared with variability within groups.
– Between-group variation is the variation among the means
of the different treatment conditions due to chance (random
sampling error) and treatment effects, if any exist.
– Within-group variation is the variation due to chance
(random sampling error) among individuals given the same
treatment.
ANOVA
Total Variation Among Scores
Within-Groups Variation
Variation due to chance.
Between-Groups Variation
Variation due to chance
and treatment effect (if any existis).
The F Ratio ANOVA
Between GroupVaria bility
F
Within GroupVaria bility
ANOVA (F)
Total Variation Among Scores
Within-Groups Variation
Variation due to chance.
Between-Groups Variation
Variation due to chance
and treatment effect (if any existis).
The F Ratio ANOVA
ANOVA (F)
Total Variation Among Scores
Within-Groups Variation
Variation due to chance.
Between-Groups Variation
Variation due to chance
and treatment effect (if any existis).
Mean Squares Within
Mean Squares Between
MSb etween
F
MSwith in
“mean squares between”
“mean squares within”
The F Ratio ANOVA
MSbetween
F
MSwithin
SSwi thin
MSwi thin 
dfwi thin
SSbetween
MSbetween 
df between
“sum of squares total”
SStotal  SSbetween  SSwithin
“degrees of freedom total”
df total  df between  dfwithin
F-distribution
• F-test is always a one-tailed test.
An Example: ANOVA
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A study compared the felt intensity of unrequited love among
three groups: individuals who were currently experiencing
unrequited love, individuals who had previously experienced
unrequited love and described their experiences
retrospectively, and individuals who had never experienced
unrequited love but described how they thought they would
feel if they were to experience it. Determine the significance
of the difference among groups, using the .05 level of
significance.
Imagined
Retrospective
Current
7
12
8
6
8
10
5
9
12
6
11
10
An Example: ANOVA
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A psychologist interested in artistic preference randomly assigns
a group of 15 subjects to one of three conditions in which they
view a series of unfamiliar abstract paintings. The 5 participants
in the “famous” condition are led to believe that these are each
famous paintings. The 5 participants in the “critically
acclaimed” condition are led to believe that these are paintings
that are not famous but are highly thought of by a group of
professional art critics. The 5 in the control condition are given
no special information about the paintings. Does what people are
told about paintings make a difference in how well they are
liked? Use the .01 level ofFamous
significance. Critically Acclaimed No Information
10
5
4
7
1
6
5
3
9
10
7
3
8
4
3
Testing Hypothesis
Test of Significance of Mean for Small Sample
•
Examples:5. In order to receive the accident incurrence
rates for automobiles, an insurance company want to
assess the damage caused by accidents at the speed of
120Km/ hour. A sample of 16 new cars was selected at
random and the company crashed each one at the speed of
120 Km/hour. The cars so damages were repaired and it
was found that the average repair amount was Rs 2500
with SD of Rs 950. Estimate true average damage in terms
of rupee of all cars due to crash at 120 Km/ hour. Assume
 = 0.05
Testing Hypothesis
• Examples:
• Suppose we are interested in a population 20
industrial units of the same size all of which are
experiencing excessive labour turnover problems.
The past record shows that the mean of the
distribution of annual turnover is 320 employees with
SD of 75 employees. A sample of 5 of these
industrial units is taken at random which gives a
mean of annual turnover as 300employees. Is the
sample mean consistent with the population?
Testing Hypothesis
•
Ex: A psychologist is working with people who have had a
particular type of major surgery. The psychologist
proposes that people will recover from the operation more
quickly if friends and family are in the room with them for
the first 48 hours after the operation (based on several
other studies on social support), but acknowledges that the
presence of friends and family may also slow recovery
time, due to the added activity and possible stress
associated with visitors. It is known that time to recover
from this kind of surgery is normally distributed with a
mean of 12 days and a standard deviation of 5 days. The
procedure of having friends and family in the room for the
period after the surgery is done with 9 randomly selected
patients. The patients recover in an average of 8 days.
Using the .01 level of significance, what should the
researcher conclude?
Testing Hypothesis
• State the research hypothesis: Do patients who have
friends and family with them following surgery recover more or
less quickly than people who do not?
• State the statistical hypothesis:
• Set decision rule: Z crit  2.58
• Calculate the test statistic: Z 
H o :   12
H A :   12
8  12
 2.40
5
9
• Decide if results are significant: Retain H0, -2.40 > -2.58
• Interpret results as it relates to the statistical
hypothesis: Patients who have friends and family with
them following surgery do not recover significantly
faster, or slower, than patients who do not have social
support
Hypothesis Testing
Example 8.6. A horticulturist knows from
experience that the honeybees visiting her
orchard weigh .87 gram on the average.
Feeling that this year’s honeybees look bigger,
she decides to weigh a random sample of n =
50 of the bees all together and she gets an
average weight of .91 grams per bee with s =
.15 gram.
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Summary
A hypothesis is a statement that is considered to be true, till it is proved false. The testing of
hypothesis is a process of testing the significance of a parameter of the population on the
basis of a sample.
A hypothesis is tested by determining the null and alternative hypothesis.
In hypothesis testing, the significance level is the criterion used for rejecting the null
hypothesis.
There are two types of errors in testing of hypothesis - Type I and Type II errors. Rejecting a
null hypothesis when it is true is called as type I error, accepting a null hypothesis when it is
false is called as type II error.
Two tailed tests of a hypothesis will reject the null hypothesis if the sample mean is
significantly higher than or lower than the hypothesized population mean.
The test of significance is done to determine whether the means of two samples drawn from
two different sources differ significantly or not.
In testing of significance where sample distribution does not follow normal distribution chisquare test is used