Testing Two Independent Samples

Testing Two Independent Samples
What is Meant by Two Independent Samples?
 Oftentimes, we encounter questions like: are there differences in

the yield of two varieties, which of the two teaching methods is
more effective, etc.
 We analyze the difference between two sample statistics.
 We compare means or proportions of two samples from
specific sub-groups of the population.
 This is the question under consideration: Is the difference
between the samples large enough to allow us to conclude (with
a known probability of error) that the populations represented by
the samples are different?
 The H0 indicates that the populations are the same.

 Assuming that the H0 is true, there is no difference between
the parameters of the two populations.
 On the other hand, we reject the H0 and say there is a difference
between the populations.
 If the difference between the sample statistics is large enough.
 Or if the size of the estimated difference is unlikely.
 The use of two related samples may not be appropriate in some

cases. The use of two-independent samples may be
considered.
 In this design, the two samples may be obtained by either of two
methods:
(a) Drawn at random from two (2) populations.
(b) Assignment at random of two (2) treatments to the members
of some sample.
 Giventwo (2) populations, we get a sample in each population

and based on these two samples, we make inferences on the
populations where these samples were drawn.
N1 = N2
Are they
equal?
n1 n2
Derive a test-statistic to
determine if the 2
populations are equal
The Population of The Population of

Same or
all Men all Women
µ=? Different? µ=?
A sample is A sample is
selected from selected from
all men all women
Sample
_ of Men Sample
_ of Women
x1 = 6.2 Is the difference between x2 = 6.5
sample means statistically
significant?
Procedure for Testing the Hypothesis
 Make assumptions and meet the test requirements.

 Define the null hypothesis.
 Select the sampling distribution and establish the critical region.
 Compute the test statistic.
 Make a decision and interpret the results.
t-Test for Independent Samples
 The usual parametric technique for analyzing data from two

independent samples is to apply a t test to the means of the two
groups.
 The t test is used to test the difference between means when
population standard deviations are not known and when one or
both samples are less than 30.
 The observations are measured on at least interval scale.
 There are two options for the use of t test. One option is used
when the variances of the populations are not equal and the
other option is used when the variances are equal.
Formulas for the t Tests-- For Testing the Difference
Between Two Means-- Small Samples
Variances are assumed to be unequal:
( X1 - X2) - (µ1 - µ2)

t =
√
2
s 21 + s 2
n1 n2
where the degrees of freedom are equal to the smaller of

n1 – 1 or n2 – 1.
Formulas for the t Tests-- For Testing the Difference

Between Two Means-- Small Samples
Variances are assumed to be equal:
(X1 - X2) - (µ1 - µ2)

t=
√ √
(n1 - 1)s12 + (n2 - 1)s 22 1 + 1
n1 n2
n1 + n2 - 2
where the degrees of freedom are equal to the smaller of

n1 + n2 – 2.
Sample Problem 13
Do families that reside in the center-city have more children than

families that reside in the suburbs? A study obtained the following
results. Test your hypothesis at  = 0.05.
Statistics Suburbs Center-City

Mean 2.37 2.78
Standard Deviation 0.63 0.95
Sample Size 42 37
Sample Problem 13
Step 1: Assumptions and test requirements.
 Independentrandom sampling.
 The samples must be independent of each other.
 Level
of measurement is interval-ratio.
 Number of children can be considered as interval-ratio.
 Populationvariances are equal.
 As long as the two samples are approximately the same, we can
make this assumption.
Sample Problem 13
Step 1: Assumptions and test requirements.
 Sampling distribution in normal size.
 Because we have to small samples (n<30), we have to add the
previous assumption in order to meet this assumption.
Step 2: Define the null hypothesis.
 Null hypothesis, H0: µ1 = µ2
 Alternative hypothesis, H1: µ1 < µ2
Sample Problem 13
Step 3: Sampling distribution and critical region.
 Sampling distribution.
 Student’s t distribution
 Significance level:  = 0.05 (one-tailed)
 Degrees of freedom
 n1 + n2 - 2 = 42 + 37 - 2 = 77
 Critical t
 t (critical) = -1.671
Sample Problem 13
Step 4: Compute the test-statistic.
Method
μ₁: mean of Sample 1
µ₂: mean of Sample 2
Difference: μ₁ - µ₂
Equal variances are not assumed for this analysis.
Descriptive Statistics
Sample N Mean StDev SE Mean
Sample 1 42 2.370 0.630 0.097
Sample 2 37 2.780 0.950 0.16
Sample Problem 13
Step 4: Compute the test-statistic.
Estimation for Difference
95% Upper Bound
Difference for Difference
-0.410 -0.103
Test
Null hypothesis H₀: μ₁ - µ₂ = 0
Alternative hypothesis H₁: μ₁ - µ₂ < 0
T-Value DF P-Value
-2.23 61 0.015
Sample Problem 13
Step 5: Decision and interpretation of results.
t(obtained) = -2.23
 This is beyond t (critical) = -1.671
 The obtained test-statistic falls in the critical region, so we reject
the H0; p=0.015 <  = 0.05.
 The difference between the number of children in center-city
families and the suburban families is statistically significant.
 The difference between the sample means is so large that we
can conclude (at  = 0.05) that a difference exists between the
populations represented by the samples.
Steps in using Minitab:

(a) Open the Minitab worksheet
and select Stat > Basic
Statistics > 2-Sample t.
(b) Enter the data in the Two-
Sample t for the Mean
dialogue box based on the
problem.
(c) Click Options.

(d) After clicking Options, enter
95.0 in the Confidence
level; 0.0 in Hypothetical
difference; and select the
Alternative hypothesis:
Difference > hypothesized
difference.
(e) Click OK.

(f) Click OK.
(g) The t-test for independent

samples yields the following
results as shown in the right
side:
t-value = -2.23
p-value = 0.015
Hence, Ho is rejected.
Sample Problem 14
The data below shows the distances of the home runs hit in record-
setting seasons by Mark McGwire and Barry Bonds. Assume that
we have simple random samples from large populations and use  =
0.05 significance level to test the claim that the distances come from
populations with different means.
Statistics McGwire Bonds
Mean 418.5 403.7
Standard Deviation 45.5 30.6
Sample Size 70 73
Sample Problem 14
Method
μ₁: mean of Sample 1
µ₂: mean of Sample 2
Sample 1 70 418.5 45.5 5.4
Sample 2 73 403.7 30.6 3.6
Sample Problem 14
Test
Alternative hypothesis H₁: μ₁ - µ₂ ≠ 0
T-Value DF P-Value
2.27 120 0.025
 Decision: Because p=0.025 <  = 0.05, reject the H0.

 Interpretation: There is sufficient evidence to support the claim
that there is a difference between the mean home run distances
of Mark McGwire and Barry Bonds.
Sample Problem 15
A researcher wishes to test the claim that on the average more
juveniles than adults are classified as missing persons. Records for
the last five years are shown below. At  = 0.05, is there enough
evidence to support the claim?
Juveniles 65,513 65,934 64,213 61,954 59,167

Adults 31,364 34,478 36,937 35,946 38,209
Sample Problem 15
Method
μ₁: mean of Juveniles
µ₂: mean of Adults
Juveniles 5 63356 2808 1256
Adults 5 35387 2631 1177
Sample Problem 15
Test
Alternative hypothesis H₁: μ₁ - µ₂ > 0
T-Value DF P-Value
16.25 7 0.000
Sample Problem 16
From the sample data below, use  = 0.05 significance level to test
the claim that the mean amount of tar in filtered king-size cigarettes
is less than the mean of tar in non-filtered king-size cigarettes. All
measurements are in milligrams and the data are from the Federal
Trade Commission.
16 15 16 14 16 1 16 18 10 14 12
Filtered
11 14 13 13 13 16 16 8 16 11
Non-filtered 23 23 24 26 25 26 21 24
Direct and Broker-Purchased Mutual Funds
Millions of investors buy mutual funds, choosing from thousands of
possibilities. Some funds can be purchased directly from banks or
other financial institutions while others must be purchased through
brokers, who charge a fee for this service. This raises the question:
Can investors do better by buying mutual funds directly than by
purchasing mutual funds through brokers? To help answer this
question, a group of researchers randomly sampled the annual
returns from the mutual funds than can acquired directly and those
that are bought from brokers and recorded the net annual returns.
Can we conclude at  = 0.05 that directly-purchased mutual funds
outperform mutual funds bough through brokers?
Direct-Purchased Mutual Funds Broker-Purchased Mutual Funds

9.33 4.68 4.23 14.69 10.29 3.24 3.71 16.40 4.36 9.43
6.94 3.09 10.28 -2.97 4.39 -6.76 13.15 6.39 -11.07 8.31
16.17 7.26 7.10 10.37 -2.06 12.80 11.05 -1.90 9.24 -3.99
16.97 2.05 -3.09 -0.63 7.66 11.10 -3.12 9.49 -2.67 -4.44
5.94 13.07 5.60 -0.15 10.83 2.73 8.94 6.70 8.97 8.63
12.61 0.59 5.27 0.27 14.48 -0.13 2.74 0.19 1.87 7.06
3.33 13.57 8.09 4.59 4.80 18.22 4.07 12.39 -1.53 1.57
16.13 0.35 15.05 6.38 13.12 -0.80 5.60 6.54 5.23 -8.44
11.20 2.69 13.21 -0.24 -6.54 -5.75 -0.85 10.92 6.87 -5.72
1.14 18.45 1.72 10.32 -1.06 2.59 -0.28 -2.15 -1.69 6.95
x2-Test for Independence
 The chi-square independence test can be used to test the

independence of two variables.
For example, a new post-operative procedure is administered
to a number of patients in a large hospital. One can ask the
question: “Do the doctors feel differently about this procedure
from the nurses, or do they feel basically the same way?” To
answer this question, a researcher selects a sample of nurses
and doctors and tabulates the data as shown below:
 Since the main question is whether there is a difference in

opinion, the null hypothesis is stated as follows:
H0: The opinion about the procedure is independent of the
profession.
H1: The opinion about the procedure is dependent on the
profession.
 To test the null hypothesis using the chi-square independence

test, one must compute the expected frequencies, assuming
that the null hypothesis is true. These frequencies are
computed by using the observed frequencies in the table.
 When data are arranged in table form for the chi-square
independence test, the table is called contingency table. It is
made up of R rows and C columns.
 The contingency table is designated as an R x C (row times
columns) table. In the foregoing example, R = 2 and C = 3;
hence, this is a 2 x 3 contingency table.
 Each block in the cell is called a cell and is designated by its

row and column position. The cells are shown below.
 The degrees of freedom for any contingency table are (rows -

1) x (columns -1) or (R - 1) x (C - 1).
 Using the previous table containing the data, computed the

expected frequencies for each block (or cell).
 Find the sum of each row and column and the grand total.
 Foreach cell, multiply the corresponding row sum by the

column sum and divide by the grand total to get the expected
value.
Row Sum x Column Sum
Expected value =
Grand Total
For example, for C1,2, the expected value denoted by E1,2 is:
(200) x (200)
E1,2 = = 100
400
Compute the expected value for each cell using the example
above.
 The expected values can now be placed in the corresponding

cells along with the observed values as shown below.
 The rationale for the computation of the expected frequencies

for a contingency table uses proportions. For C1,1 a total of
150 out of 400 people prefer the new procedure.
 The formula for the test value for the independence test is the
same as the one used for the goodness-of-fit test as shown
below.
(O – E)2
X2 = E
(100 – 75)2 + (80 – 100)2 (20 – 25)2 (50 – 75)2

= + +
75 100 25 75
(120 – 100)2 (30 – 25)2
+ +
100 25
= 26.67
 Decision. Since 26.67 > 5.991, the decision is to reject the

null hypothesis.
Chi-Square Test for Association: Worksheet rows, Worksheet columns
Rows: Worksheet rows Columns: Worksheet columns
New Old No
Procedure Procedure Preference All
1 100 80 20 200
75 100 25
2 50 120 30 200
75 100 25
All 150 200 50 400
Chi-Square Test
Chi-Square DF P-Value
Pearson 26.667 2 0.000
Likelihood Ratio 27.058 2 0.000
Sample Problem 17
The data below summarizes results from tests of the accuracy of
polygraphs. Use α = 0.05 significance level to test the claim whether
the subject lies is independent of the polygraph indication. What do
the results suggest about the effectiveness of polygraphs?
Category Polygraph Polygraph
Indicated Truth Indicated Lie
Subject actually told the truth 65 15
Subject actually told a lie 3 17
Sample Problem 18
A researcher wishes to determine whether there is a relationship
between the gender of an individual and the amount of alcohol
consumed. A sample of 68 people is selected and the following data
are obtained. At α = 0.05, can the researcher conclude that alcohol
consumption is related to gender?
Gender Alcohol Consumption Total
Low Moderate High
Male 10 9 8 27
Female 13 16 12 41
Total 23 25 20 68
Sample Problem 19
A study was conducted to determine whether there is a relationship
between jogging and blood pressure. A random sample of 210
subjects was selected and were classified as shown in the table
below. At α = 0.05, test the claim that jogging and blood pressure
are not related.
Jogging Blood Pressure

Status Low Moderate High
Joggers 34 57 21
Non-Joggers 15 63 20
Sample Problem 19
Chi-Square Test
Chi-Square DF P-Value
Pearson 6.789 2 0.034
Likelihood Ratio 6.955 2 0.031
Cramer’s Coefficient
 C is a value which estimates the degree of relationship between
two variables (when the variables are in at least nominal) when
measurement under these two variables are summarized in an r
x c contingency (frequency) table. The formula for getting the
c-coefficient is:
where:
√
c= x2 n = total number of subjects/ units
n(L) in samples
L = minimum of r & c (smaller of
the 2 values, r & c)
Ex: if r = 3; c = 4; then L = 3
Cramer’s Coefficient for Sample Problem 19
√ √ √
c= x2 = 6.789 = 6.789
n(L) 210(2) 105
=
√ 0.064657
= 0.254278
Wilcoxon Rank-Sum Test for Two Independent
Samples (or Mann-Whitney U Test)
 The Wilcoxon rank-sum test is a non-parametric test that uses
ranks of sample data from two independent populations.
 It is used to test the null hypothesis that the two independent
samples come from populations with the same distribution. That
is, the two populations are identical. The alternative
hypothesis is the claim that the two population distributions are
different in some way.
 Two samples are independent if the sample values selected from
one population are not related or somehow matched or paired
with the sample values from the other population.
 It is used when the variables are in the ordinal measurement.
 The key idea underlying the Wilcoxon rank-sum test is this: If two
samples are drawn from identical populations and the individual
values are all ranked as one combined collection of values, then
the high and low ranks should fall evenly between the two
samples.
 If the low ranks are found predominantly in one sample and the
high ranks are found predominantly in the other sample, we
suspect that the two populations are not identical.
 Suppose we have samples from two populations, X & Y. The null
hypothesis is that X and Y have the same distribution (or X & Y do
not differ). The alternative hypothesis H1 against we test H0 is that
X is stochastically larger than Y -- a directional hypothesis.
 H1 is accepted if the probability that a score from X is larger than a
score from Y is greater than one-half. That is, if X is one
observation from population X and Y is an observation from
population Y, then H1 is that:
P[X > Y] > ½
 If the evidence supports H1, this implies that the “bulk” of the
elements of population X are larger than the bulk of the elements
of population Y. Hence, the null hypothesis is H0: P[X > Y] > ½.
 It could also be that our hypothesis might instead be that Y is
stochastically larger than X.
In that case, the alternative hypothesis H1 would be that P[X > Y]
> ½.
 If the evidence supports H1, this implies that the “bulk” of the
elements of population X are larger than the bulk of the elements
of population Y. Hence, the null hypothesis is H0: P[X > Y] < ½.
Confirmation of this hypothesis would imply that the bulk of Y is
larger than the bulk of X.
 Fora two-tailed test, i.e., for a prediction of differences which does
not state the direction of the differences, H1 would be that P[X > Y]
≠ ½.
 Anotherway of stating the alternative hypothesis H1 is that the
median of X is greater than the median of Y, that is, H1: H0: 0X >
0Y.
Step 1: State the hypothesis.
H0: The populations do not differ.
Suppose we let X = response or scores of a population where a
smaller sample is drawn while Y = response or scores of the
population where a larger sample is drawn.
H0 can also be stated as:
H0 : The chance that X is larger than Y is the same as the
chance that X is smaller than Y.
Ha: Most of the Xs are larger than the Ys
(X is stochastically larger than Y)
or
P[X > Y] > ½.
Ha: Most of the Xs are smaller than the Ys
(X is stochastically smaller than Y)
or
P[X > Y] < ½.
Ha: P[X > Y] ≠ ½
(either most of the Xs are larger tha larger than Y or
most of the Xs are smaller than Y)
Step 2: Find the critical value. At α = __, use Wilcoxon Rank test.
Step 3: Compute the test value.
Use the test statistic:
Wx = sum of the ranks of the smaller samples
Wx = sum of the ranks of the smaller samples
 Let m = sample size of smaller sample (Xs)
n = sample size of larger sample (Ys)
(m + n) = total sample size
 Rank the responses or scores for both samples (Xs & Ys)
together in one order; rank 1 is assigned to the lowest
response and rank (m+n) to the highest response.
Example: m = 4 Xs and n = 6 Ys. Rank all observations 1
to 10.
Let Wx = sum of the ranks of the smaller sample or sum of
the ranks of the Xs.
Step 4: Decision rule: H0: P[X > Y] = ½
Reject H0 if Wx is large or P[Wx ≥ Observed Wx] ≤ α
Let Wx = sum of the ranks of the smaller sample or sum of
the ranks of the Xs.
Step 4: Decision rule: H0: P[X > Y] = ½
Reject H0 if Wx is large or P[Wx ≥ Observed Wx] ≤ α
Sample Problem 20
Suppose an extension worker would like to evaluate the level of
adoption of a certain technology of farmers and would like to
compare the level of adoption of farmers from two barangays. With
the following results, make a generalization at α = 0.05.
Barangay Rating
Barangay 1 10 50 45 30 40 60
Barangay 2 20 75 70 55 65
Sample Problem 20
H0: The level of adoption of farmers in the two barangays are
the same.
Ha: There is a difference between level of adoption of
farmers in the two barangays.
Step 2: Find the critical value. At α = __, use a two-tailed Wilcoxon
Rank Sum test.
Step 3: Compute the test value. Rank all ratings 1 to 11.
Sample Problem 20
m = 5 (smaller group; barangay 2)
n = 6 (larger group, barangay 1)
x = response of farmers in barangay 2 or smaller sample
Wx = sum of ranks in smaller sample (barangay 2)

2 + 11 + 10 + 7 + 9 = 39
Sample Problem 20
Wx = sum of ranks in smaller sample (barangay 2)
= 2 + 11 + 10 + 7 + 9 = 39
(i) P[Wx ≥ 39] = 0.0628
(ii) P[Wx ≤ 39] = 0.9589
Step 4. Make a conclusion. Since 0.0628 > 0.025, we fail to reject
H0.
Step 5. Conclusion. There is no difference in the level of
adoption of farmers in the two barangays.
Sample Problem 20
Method
η₁: median of Brgy 1
η₂: median of Brgy 2
Difference: η₁ - η₂
Sample N Median
Brgy 1 6 42.5
Brgy 2 5 65.0
Estimation for Difference
CI for Achieved
Difference Difference Confidence
-20 (-45, 20) 96.42%
Sample Problem 20
Test
Null hypothesis H₀: η₁ - η₂ = 0
Alternative hypothesis H₁: η₁ - η₂ ≠ 0
W-Value P-Value
27.00 0.121
Sample Problem 21
The data below shows the Flesch Reading Ease scores for randomly
selected pages from each of two books: Harry Potter and the
Sorcerer’s Stone by J. K. Rowling and War and Peace by Leo
Tolstoy. Use the two sets of independent sample data with α = 0.05
significance level to test the claim that reading scores for pages from
the two books have the same distribution.
Rowling 85.3 84.3 79.5 82.5 80.2 84.6 79.2 70.9 78.6 86.2 74.0 83.7 71.4
Tolstoy 69.4 64.2 71.4 71.6 68.5 51.9 72.2 74.4 52.8 58.4 65.4 71.6
Sample Problem 22
Two independent samples of army and marine recruits are selected
and the time in minutes it takes each recruit to complete an obstacle
course is recorded as shown in the table below. At α = 0.05, is there
a difference in the times it takes the recruits to complete the course?
Army 15 18 16 17 13 22 24 17 19 21 26 6
Marines 14 9 16 19 10 12 11 8 15 18 25
Class Activity 8: Application of appropriate
statistical analysis to test the hypothesis
Many students have had the unpleasant experience of panicking on
a test because the first question was exceptionally difficult. The
arrangement of test items was studied for its effect on anxiety. The
following scores are measures of “debilitating test anxiety” which
most of us call panic or blanking out. Based on the data gathered, is
there a sufficient evidence to support the claim that the two
populations of scores have the same mean? Is there a sufficient
evidence to support the claim that the arrangement of the test items
has an effect on the score? Perform the test at  = 0.05 level of
significance.
Class Activity 8: Application of appropriate
statistical analysis to test the hypothesis
Questions Arranged from Easy Questions Arranged from Difficult
to Difficult to Easy
24.64 39.29 16.32 32.83 28.02 33.62 34.02 26.63 30.26
33.31 20.60 21.13 26.69 28.90 35.91 26.68 29.49 35.32
26.43 24.23 7.10 32.86 21.06 27.24 32.34 29.34 33.53
28.89 28.71 31.73 30.02 21.96 27.62 42.91 30.20 32.54
25.49 38.81 27.85 30.29 30.72
(a) Formulate the statements of the problem.

(b) Draw the conceptual framework based on the problem.
(c) Test the hypothesis indicating the following: assumptions and test
requirements, the H0 and H1, critical region, test-statistic
computation, and conclusion and interpretation.
Thank You!

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What is Meant by Two Independent Samples?

 Oftentimes, we encounter questions like: are there differences in

 The H0 indicates that the populations are the same.

 The use of two related samples may not be appropriate in some

 Giventwo (2) populations, we get a sample in each population

The Population of The Population of

 Make assumptions and meet the test requirements.

 The usual parametric technique for analyzing data from two

( X1 - X2) - (µ1 - µ2)

where the degrees of freedom are equal to the smaller of

Formulas for the t Tests-- For Testing the Difference

(X1 - X2) - (µ1 - µ2)

where the degrees of freedom are equal to the smaller of

Do families that reside in the center-city have more children than

Statistics Suburbs Center-City

Steps in using Minitab:

Steps in using Minitab:

Steps in using Minitab:

(g) The t-test for independent

 Decision: Because p=0.025 <  = 0.05, reject the H0.

Juveniles 65,513 65,934 64,213 61,954 59,167

Direct-Purchased Mutual Funds Broker-Purchased Mutual Funds

 The chi-square independence test can be used to test the

 Since the main question is whether there is a difference in

 To test the null hypothesis using the chi-square independence

 Each block in the cell is called a cell and is designated by its

 The degrees of freedom for any contingency table are (rows -

 Using the previous table containing the data, computed the

 Foreach cell, multiply the corresponding row sum by the

 The expected values can now be placed in the corresponding

 The rationale for the computation of the expected frequencies

(100 – 75)2 + (80 – 100)2 (20 – 25)2 (50 – 75)2

 Decision. Since 26.67 > 5.991, the decision is to reject the

Rows: Worksheet rows Columns: Worksheet columns

Jogging Blood Pressure

Cramer’s Coefficient for Sample Problem 19

Wx = sum of ranks in smaller sample (barangay 2)

(a) Formulate the statements of the problem.

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