hypothesis testing
TRANSCRIPT
Introduction
•A statistical hypothesis is an assumption about an
unknown population parameter.•It is a well defined procedure which helps us to decide
objectively whether to accept or reject the hypothesis
based on the information available from the sample.•In statistical analysis, we use the concept of probability to
specify a probability level at which a researchers
concludes that the observed difference between the sample
statistics and population parameter is not due to chance.
Hypothesis Testing Procedure
Step 1: - Set Null and Alternative Hypothesis
Step 2:-Determine the appropriate Statistical Test
Step 3: -Set the level of Significance
Step 4: -Set the Decision Rule
Step 6: - Analyze the data
Step 7: - Arrive at a statistical conclusion
Step 5: - Collect the Sample data
• The null hypothesis is denoted by Ho, is the
hypothesis which is tested for the possible rejection
under the assumption that it is true.
• Theoretically, Ho is set as no difference considered
true, until and unless it is proved wrong by the
collected sample data.
• The alternative hypothesis is denoted by H1 or Hα, is
a logical opposite of the Ho.
Step 1: - Set Null and Alternative Hypothesis
1. H0: = 0 versus
Ha: 0 (two-sided)
2. H0: 0 versus
Ha: < 0 (one-sided)
3. H0: 0 versus
Ha: > 0 (one-sided)
Often H0 for a one-sided test is written as H0: = 0. Remember a p-value is computed assuming H0 is true, and 0 is the value used for that computation.
• After setting he hypothesis, the researches has to
decide on an appropriate statistical test that will be
tested for the statistical analysis.
• The statistic used in the study (mean, proportion,
variance etc.) must also be considered when a
researchers decides on appropriate statistical test,
which can be applied for hypothesis testing in order
to obtain the best results.
Step 2: - Determine the appropriate Statistical Test
• The level of significance is denoted by α is the probability,
which is attached to a null hypothesis, which may be
rejected even when it is true.
• The level of significance also known as the size of the
rejection region or the size of the critical region.
• Level of significance must be determined before we draw
samples, so that the obtained result is free from the bias of a
decision maker.
• 0.01, 0.05, 0.010
Step 3: -set the level of significance
• Next step is to establish a critical region, which is
the area under the normal curve . These regions are
termed as acceptance region (when the Ho is
accepted) and the rejection region or critical region.
• If the computed value of the test statistic falls in the
acceptance region , the null hypo is accepted .
• Otherwise Ho is rejected.
Step 4: - Set the decision Rule
• In this stage data are collected and appropriate
sample statistics are computed.
• The first 4 steps should be completed before
collecting the data for the study.
Step 5: - Collect the sample data
• In this step the researcher has to compute the test
statistic. This involves selection of appropriate
probability distribution for a particular test.
• For Example- When the sample is small, then t-
distribution is used. If sample size is large then use
Z-test.
• Some commonly used testing procedures are F, t, Z,
chi square.
Step 6: - Analyze the data
• In this step the researcher draw a conclusion. A
statistical conclusion is a decision to accept or reject
a Ho. This depends whether the computed statistic
falls in the acceptance region or rejection region.
Step 7: - Arrive at a statistical conclusion
Critical Region
The critical region (or rejection region) is the set of all values of the test statistic that cause us to reject the null hypothesis.
Significance Level
The significance level (denoted by ) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. Common choices for are 0.05, 0.01, and 0.10.
Critical Value
A critical value is any value separating the
critical region (where we reject the H0) from
the values of the test statistic that does not
lead to rejection of the null hypothesis, the
sampling distribution that applies, and the
significance level . For example, the critical
value of z = 1.645 corresponds to a
significance level of = 0.05.
Two-tailed,Right-tailed,
Left-tailed Tests
The tails in a distribution are the extreme regions bounded by critical values.
Two-tailed TestH0: =
H1:
is divided equally between the two tails of the critical
region
Means less than or greater than
Right-tailed Test
H0: =
H1: > Points Right
Left-tailed Test
H0: =
H1: <
Points Left
Conclusions in Hypothesis Testing
We always test the null hypothesis.
1. Reject the H0
2. Fail to reject the H0
Accept versus Fail to Reject
Some texts use “accept the null hypothesis.”
We are not proving the null hypothesis.
The sample evidence is not strong enough to warrant rejection (such as not enough evidence to convict a suspect).
Reject H0 if the test statistic falls within the critical region.
Fail to reject H0 if the test statistic does not fall within the critical region.
Decision Criterion
The Two Types of Errors and Their Probabilities
When the null hypothesis is true, the probability of a type 1 error, the level of significance, and the -level are all equivalent.
When the null hypothesis is not true, a type 1 error cannot be made.
Type I ErrorA Type I error is the mistake of
rejecting the null hypothesis when it is true.
The symbol (alpha) is used to represent the probability of a type I error.
Type II ErrorA Type II error is the mistake of failing
to reject the null hypothesis when it is false.
The symbol (beta) is used to represent the probability of a type II error.
Type I and Type II Errors
Controlling Type I and Type II Errors
For any fixed , an increase in the sample size
n will cause a decrease in
For any fixed sample size n , a decrease in will cause an increase in . Conversely, an increase in will cause a decrease in .
To decrease both and , increase the sample size.
DefinitionPower of a Hypothesis Test
The power of a hypothesis test is the probability (1 - ) of rejecting a false null hypothesis, which is computed by using a particular significance level and a particular value of the population parameter that is an alternative to the value assumed true in the null hypothesis.
Trade-Off in Probability for Two Errors
There is an inverse relationship between the probabilities of the two types of errors.Increase probability of a type 1 error =>
decrease in probability of a type 2 error
Type 2 Errors and Power
Three factors that affect probability of a type 2 error1. Sample size; larger n reduces the probability of a type 2
error without affecting the probability of a type 1 error.
2. Level of significance; larger reduces probability of a type 2 error by increasing the probability of a type 1 error.
3. Actual value of the population parameter; (not in researcher’s control. Farther truth falls from null value (in Ha direction), the lower the probability of a type 2 error.
When the alternative hypothesis is true, the probability of making the correct decision is called the power of a test.
Z-Test (For large Samples)
• Hypothesis testing for large samples is based on the
assumption that the population, from which the
sample is drawn, is normal. As a result the sampling
distribution of the mean is also normally distributed.
• Even when the population is not normally
distributed the sampling distribution of mean for a
large sample size is normally distributed.
Formula for single population mean
n
xz
Where, µ =population mean,
σ = population standard Deviation,
n = sample size and
x bar is sample mean.
Example• A marketing research firm conducted a survey 10 years ago
and found that the avg. household income of a particular
region is Rs. 10000. Mr. X, Who recently joined the firm as
vice president has expressed doubts about the accuracy of
the data. For verifying the data, the firm has decided to take
a random sample of 200 households that yields a sample
mean (income) of Rs. 11000. Assume that the population
S.D. of the household income is Rs. 1200. Verify Mr. X’s
doubts using the 7 step hypothesis testing. Let α=0.05.
SolutionStep-1: Set null and alternative hypothesis
Here researcher is trying to verify whether there is any
change in the avg. household income within 10
years. The Ho is set as no difference, i.e. the avg.
household income has not changed.
Ho: µ=10000, H1: µ#10000
Step-2: Determine Statistical Test
Sample size is > than 30, and the sample mean is used
as statistic. So Z formula is used
n
xz
SolutionStep-3: Set the level of significance (α)
It is known as size of the rejection region or size of
the critical region. Here it is 0.05
Step-4: Set the decision rule
H1 shows that we have two-tailed test (means
household income can be less than or more than
10000 Rs.) and the level of significance is 0.05. The
acceptance region covers 95% of the area and the
rejection region covers the remaining 5% of the
area at the two ends of the distribution. (Zα/2=1.96)
0
0.025 0.025
1-.05=0.95
+1.96-1.96
0.47500.4750
SolutionStep-5: Collect the sample data
Sample size is 200=n, and Sample Statistic (Sample
Mean) = 11000.
Step-6: Analyze the data.
n=200, µ=10000, σ =1200, x bar or mean = 11000
79.11
200
1200
1000011000
z
SolutionStep-7:Arrive at a Statistical Conclusion
Calculated value of Z is 11.79
And tabulated value of Z is 1.96
As Calculated > Tabulated , We reject the null Hypo
Means Mr. X’s doubts about this average
household income was right.
Example / Case Study
• A soft drink company produces 2 liters bottles of
one of its popular drinks. The quality control
department is responsible for verifying that each
bottle contains exactly 2 liters of soft drink. The
result of a random check of 40 bottles undertaken by
the quality control officer are given in table.
Use 0.01 level of significance to test whether each
bottle contains exactly 2 liters of soft drink
Bottle Sr. No.
Quantity in liters
1 1.97
2 1.98
3 1.99
4 2.01
5 2.02
6 2.03
7 2.01
8 1.97
9 1.96
10 2.04
11 2.00
12 2.01
13 2.02
14 1.99
15 2.00
16 1.97
Bottle Sr. No.
Quantity in liters
17 1.98
18 2.03
19 1.98
20 1.99
21 2.01
22 2.05
23 2.03
24 2.04
25 2.01
26 1.97
27 1.98
28 1.99
29 1.98
30 2.03
31 2.01
32 1.99
Bottle Sr. No.
Quantity in liters
33 1.97
34 1.96
35 2.02
36 2.03
37 2.04
38 1.98
39 1.99
40 2.01
When population S.D. is unknown. The population
S.D. (σ) is replaced by sample S.D. (s).
Then
The above formula of Z is based on the assumption
that the sample is drawn from an infinite population.
If population is finite then Z formula is
n
sx
z
1.
NnN
n
xz
Confidence Level (1-α)%
α One-tailed test Two-tailed test
90% 0.10 1.28 1.645
95% 0.05 1.645 1.96
99% 0.01 2.33 2.575
Hypo Testing for a Population Proportion
In business research information is generally expressed in
terms of proportions. For ex. Market share of a company,
quality defects, consumer preferences etc.
This kind of data is highly dynamic in nature. Business
researchers sometimes want to test the hypothesis about
such proportions.
Formula
npq
ppz
Where, n= sample size, p = population proportion, q=1-p and p bar is sample proportion
Example
The production manager of a company that manufactures
electric heaters believes that atleast 10% of the heaters are
defective. For testing his belief, he takes a random sample
of 100 heaters and finds that 12 heaters are defective. He
takes the level of significance at 5% for testing the
hypothesis. ( table value of Z at 5% is 1.96)
Sol.
Ho : p=0.10
H1 : p#0.10 (two-tailed test)
Example / Case Study
p= population proportions = 10/100=0.10
q= 1-p = 1- 0.10 = 0.90
p bar = sample proportion = 12/100 = 0.12
n = 100
67.0
1000)(0.10)(0.9
0.1012.0
npq
ppz
Ho is accepted as Z calculated is < then Z tabulated (0.67<1.96)
Hypo Testing for the difference between two means usingFormula
When Population S.D. are not given
2.Population ofMean sample2. ofMean x
1.Population ofMean sample1. ofMean x
2.Population of S.D. σ 2. sample of size then
1.Population of S.D. σ 1. sample of size then
nσ
nσ
μμxxz
22
11
22
11
2
2
2
1
2
1
2121
sample2. of S.D. s
sample1. of S.D. s
ns
ns
μμxxz
2
1
2
2
2
1
2
1
2121
The means of two single large sample of 1000 and 2000
members are 67.5 inches and 68.0 inches respectively. Can
the samples be regarded as drawn from the same population
of S.D. 2.5 inches.(Z value at 5% is 1.96)
Sol.
Ho : (samples are from the same population
H1 :2121 or 0
21#
Example / Case Study
1.55.1-Z
-5.1
2000)5.2(
1000)5.(2
00.865.67
nσ
nσ
μμxxz
0.86x 67.5. x
2.5 σ .0002n 2.5 σ .0001n
22
2
2
2
1
2
1
2121
21
2211
As Z calculated is > Z Tabulated .
So Ho is rejected. W conclude that samples are not drawn from the same population.
Small Sample testing
When we have sample size < 30. to test the hypothesis about
population parameter we use t-test.
The t-statistic is a standardized score for measuring the difference between the sample mean and the null hypothesis value of the population mean:
ns
xt 0
error standard
valuenullmean sample
This t-statistic has (approx) a t-distribution with df = n - 1.
Royal tyres has launched a new brand of tyres for
tractors and claims that under normal circumstances
the average life of the tyres is 40000 km. A retailer
wants to test this claim and has taken a random
sample of 8 tyres. The results obtained are presented
in the table. He tests the life of the tyres under
normal conditions Test the hypothesis at 5% . (t
value with 7 d.f. at 5% is 2.365).
Example / Case Study
• He tests the life of the tyres under normal conditions and found a mean life of 39750 km. and S.D. (s) 2618.61.
Tyres Km
1 35000
2 38000
3 42000
4 41000
5 39000
6 41500
7 43000
8 38500
Sol.
Ho : µ = 40000
H1 : µ # 40000X bar = 39750,
S = 2618.61
N = 8
27.0
82618.61
4000039750
ns
μxt
As t calculated is < t tabulated
We accept Ho
Hypo Testing for difference between two population means (small samples)
2121
2
2
21
2
1
2121
n1
n1
2nn1)(ns1)(ns
μμxxt
With degrees of freedom n1+n2 -1
Example
XYZ constructions is a leading company in the
construction sector in India. It wants to construct
flats in Raipur & Dehradun, the capitals of the
newly formed states of Chattisgarh and Uttarakhad,
respectively. The company wants to estimate the
amount that customers are willing to spend on
purchasing a flat in the two states. It randomly
selected 25 potential customers from Raipur and 27
from Dehradun and posed the question, “how much
areyou willing to spend on a flat?” The data
collected from the two cities is shown in tables.
The company assumes that the intention to purchase of
the customers is normally distributed with equal
variance in the two cities taken for the study. On the
basis of the samples taken for the study, estimate the
difference in population means taking 95% as the
confidence level.
Proposed Exp. On flats by customers from Raipur (in thousand Rs.)
125 165
130 170
126 130
127 145
150 130
135 140
140 150
160 160
120 140
150 145
155 165
145
140
165
135
130
Proposed Exp. On flats by customers from Dehradun (in thousand Rs.)
185 135
165 185
160 180
170 190
180 145
190 160
170 170
150 180
155 145
160
145
150
155
160
145
140
Statistical Inference about the difference between the means of two related population (matched samples)
Chi-Square Test for Categorical data
Categorical Data: It is defined as the counting of frequencies from one or more variables.
Chi square test is the category of non-parametric test. i.e. here we are not sure about the population distribution (whether it is normal or not). The statistical tests that do not require prior knowledge about the population are termed as non-parametric tests.
Product
Age
Mobile banking
Internet Banking
Personal Banking
Row Total
17-27 125 175 145 445
28-35 155 180 197 532
36-44 167 210 150 527
45-57 146 156 142 444
58-70 133 156 176 465
Column Total 726 877 810 2413
Preference of type of banking across different age groups
sFrequencie Expected f
sFrequencie Observed f
,f
ffχ
is StatisticTest χ
1
0
e
e02
2
Conditions for applying chi square
• The sample should consist of at least 50
observations and should be drawn randomly from
the population.
• The individual observation in a sample should be
independent to each other.
• Data should not be presented in the % or ratio form,
rather they should be expressed in original units.
• Sum of the frequencies must be 5 or more.
Goodness of fit Test2χ
This test compares the theoretical frequencies with
the observed frequencies to determine the difference
between theoretical and observed frequencies.
Q.- Five coins are tossed 3200 times and the
following results are obtained. Test the
hypothesis that coins are biased at 5%. (chi
square value at 5% with 5 d.f. is 11.070
No. of Heads 0 1 2 3 4 5
Frequency 80 570 1100 900 500 50
Ho : coin is unbiased
Let p = the probability of getting a head = ½
From binomial distribution, expected frequencies are:
F(x) = N.P(x), wherexnxqp
(x)!x)!-(n
n! P(x)
100 F(5) 500, F(4)
1000 F(3) 1000, F(2) ,500)1(.3200)1(
100)0(
(1/2)(1/2)(5)!0)!-(5
5!3200. 3200.P(0) head 0 ofFrequecny
qp(x)!x)!-(n
n! P(x)
1/2 p-1 q 1/2,p 5,n 3200,N
xnx
xnx
PF
similarly
F
No. of Heads
Observed Frequency (fo)
Expected Frequency (fe)
(fo-fe) (fo-fe)
0
1
2
3
4
5
80
570
1100
900
500
50
100
500
1000
1000
500
100
400
4900
10000
10000
0
2500
4
9.8
10
10
0
25
58.8
Example• A marketing research firm conducted a survey 4 years ago
and found that the avg. household income of a particular
region is Rs. 10050. Mr. X, Who recently joined the firm as
vice president has expressed doubts about the accuracy of
the data. For verifying the data, the firm has decided to take
a random sample of 200 households that yields a sample
mean (income) of Rs. 11000. Assume that the population
S.D. of the household income is Rs. 1500. Verify Mr. X’s
doubts using the 7 step hypothesis testing. Let α=0.05.