Two Sample Tests

Nutan S. Mishra

Department of Mathematics and Statistics

University of South Alabama

Points to ponder

• µ1=µ2 can be rewritten as µ1-µ2 =0

• µ1< µ2 can be rewritten as µ1-µ2 < 0

• µ1 > µ2 can be rewritten as µ1 - µ2 > 0

• Equivalent values indecimal percentage.01 1%

.05 5%

.10 10%

.025 2.5%

Description of the problem• There are two populations 1 and 2

• Parameters of 1 are (µ1,σ1) and that of 2 are (µ2,σ2)

• We want to compare the average performances of these two populations.

• That is we want to test if µ1=µ2

• In other words-- to test if µ1-µ2 =0

Description of the problem• Recall that at this point the true values of µ1 and µ2

are unknown.• But we are not interested in knowing the absolute

values of µ1 and µ2

• We are interested in estimating and testing about the difference between the two namely µ1-µ2

• In order to conduct this test of hypothesis we collect two samples of size n1 and n2.

• Samples collected may be dependent or independent.

Dependent samplesDependent samples : selection of one sample affects the

selection of the other.

Example: collecting data on length of sleep hours of insomnia patients before and after certain medication.

First sample consists of values (3.2, 4.5, 4.5, 4.7) and the second samples consists of values (4.8, 5.3, 4.2, 4.7)

Note that both samples are of same size n1 = n2

Patient’s name

Before medication

(sample I)

After medication

(sample 2)

P1 3.2 4.8

P2 4.5 5.3

P3 4.5 4.2

P4 4.7 4.7

Independent samples Selection of one sample is independent of the other.

Example: collecting data on emission of harmful gases from the vehicles in the state of California and in the state of Alabama

Sample S1 consists of emission data on n1 vehicles from the state of California

Sample S2 consists emission data on n2 vehicles from the state of Alabama.

n1 may or may not be equal to n2

We may collect a sample of 100 vehicles from one state and a sample of 85 vehicles from the other.

Testing of hypothesisWe will discuss them as different cases as follows

Two samples

Independent Dependent

Large samples

Small samples

We mostly discuss the small samples

Formulae may be applied to large sample cases as well

Testing hypothesis about difference of means

There are two populations 1 and 2

• Parameters of 1 are (µ1,σ1) and that of 2 are (µ2,σ2)

• Sample statistics are (n1, , s1) and (n2, , s2)

• To test H0: µ1=µ2 equivalently H0: µ1-µ2 =0

• The point estimator for µ1-µ2 is

• The test statistic would be

• When σ12 and σ2

2 are unknown we replace them by their point estimators

namely s12 and s2

2 and the test statistic becomes

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Exercise 10.11X= hotel room rate per day in united states

1: all such room rates in 2000 (parameters (µ1,σ1) )

2 : all such room rates in 2001 ( parameters (µ2,σ2))

To test H0: µ1=µ2 Vs H1: µ1 > µ2 equivalently we want

To test H0: µ1 - µ2 =0 Vs H1: µ1 - µ2 > 0

To test these hypotheses, two samples are collected from the respective populations with the following statistics:

Sample S1 (from first population) = ( n1= 1000, = 85.69, s1 = 18.50)

Sample S2(from second population= ( n2 = 1100, = 84.58 s2 = 18.00)

The statistic would be

It’s a right tailed hypothesis thus the p-value corresponding to z = 1.39 (from normal table ) is .0823

Since p-value > .01 ( which is given LOS) we decide not to reject H0

And conclude that at 1% LOS, the data supports the null hypothesis, that is room rates in 2000 and 2001 were not significantly different.

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Two small independent samplesThere are two populations 1 and 2

• Parameters of 1 are (µ1,σ1) and that of 2 are (µ2,σ2)• Sample statistics are (n1, , s1) and (n2, , s2)• To test H0: µ1=µ2 equivalently H0: µ1-µ2 =0• The point estimator for µ1-µ2 is For small n1 and (or) n2 , the test statistic is developed under following

assumptions– Both populations are approximately normal– The population variances for both populations are unknown and

EQUAL σ1 = σ2 ( say both of them equal to σ which is unknown)If these conditions are fulfilled then we can estimate the common

unknown parameter σ by pooling the information (that is data coming from both samples) and call that estimate as pooled estimate of σ

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Two small independent samplesNow we define the test statistic under the two above conditions

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Exercise 10.28(b)1 : male customers at a supermarket

2: female customers at the same super market.

XF= amount spent by a female customer.

XM= amount spent by a male customer

µM : average amount spent by all the male customers ( value is unknown)

µF : average amount spent by all the female customers (value is unknown)

σM: population standard deviation of male population spending ( value is unknown)

σF: population standard deviation of female population spending ( value is unknown)

Assumptions: 1. XM and XF have normal distribution .

2. σM = σF (both are unknown) let us say σM = σF =σ

To test H0: µM =µF Vs H1: µM < µF in other words

To test H0: µM -µF =0 Vs H1: µM-µF < 0

Exercise 10.28(b) (continued)

Two samples are collected, one from each population.

Information from samples:

nM = 25 ( a sample of 25 males)

= $80 average spending by 25 males

sM = $17.50 sample standard deviation of spending of 25 males

nF = 20 ( a sample of 20 females)

= $96 average spending by 20 females

sF = $14.40 sample standard deviation of spending of 20 females

Now since both the samples are small, under the two assumptions above, the test statistic would be.

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Exercise 10.28(b) (continued)

Value of t-test statistic would be

To compute sp

Plugging in the value of Sp in the above formula we get the t-value as

at 5% LOS the value of t at 43 d.f. is (from table) = 1.645

Since the alternative is left sided, our critical region lies in the left tail and the critical value is – 1.645. Comparing is the test statistic with the critical value, we see that test statistics falls in the rejection region

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Exercise 10.28(b) (continued)

Thus we make a decision to reject the H0 at 5% Level of significance

Conclusion: We conclude that at 5% LOS, the sample data does not support the null hypothesis, this implies that the average spending by male customers is significantly lower than average spending by female customers.

Paired t-test Designed to compare the averages of two populations when two samples

drawn are dependent ( or matched samples or paired samples).Here the population consists of same members (or objects or items)We collect two observations on each member of the population– before the

treatment and after the treatment. Thus we generate two samples.

X1: observation before the treatment

X2: observation after the treatment

d= difference in the measurement = x2-x1

d is a new random variable with µd = average of all the differences σd = standard deviation of all d-values in the population

= mean of the d-values from the two samples each of size n

sd = sample standard deviation of all d-values obtained from the two samples

To test the hypothesis H0 : µd = 0 vs H1: µd < or > or ≠ 0Under the assumption that population of all d-values is approximately normal, the

test statistics is

d

Paired t-test (continued)

The test statistics

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Exercise 10.55(b)X1 : self confidence score before attending the course (scale 1-15)X2: self confidence score after attending the course (scale 1-15)d= x2-x1 = change in the confidence score after attending the courseH0: µd = 0 vs H1: µd > 0 = 1.42857, sd = 1.988 test statistic

Now since alternative hypothesis is right sided, the critical region lies on the right side of the t-curve.The critical value of t at 1% LOS is = 3.143Comparing the critical value with test statistic, we observe that test statistic falls in non

rejection region thus we make a decision not to reject null at 1% LOS since at 1% LOS sample data supports the null , we conclude that the course is not

much effective as far as improvement in the confidence scores is concerned.

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