Paired t-Test Lecture 44

 

 Paired t-Test

Lecture 44

Introduction

The paired t-test is also called the dependent two-sample t-test and is used to determine the confidence interval for the difference between population means or test the hypothesis about the difference between population means. This is applicable when the observations of both samples are dependent or match-paired. This happens when the observations are paired either naturally or by design.

i.  Natural Pairing

Natural pairing occurs when two measurements are taken on a single object or individual at two different points in time. In natural pairing, the words before and after are used in the statement of hypothesis.

Examples:

a. The performance of students before and after attending the coaching classes.

b. The weight of recruits before and after the training program.

c. The temperature of the patient before and after using the medicine.

ii. By Design Pairing

By design pairing, the effects of non-interest are eliminated or brought under control.

Examples:

a. Suppose we compare the yield capabilities of two varieties of wheat, and both varieties are treated with the same conditions.

b. Let the performance of two groups of statements be compared, and both groups were taught in the same environments and with the same teachers.

When used to pared Test

i. Both samples' observations depend on one another.

ii. The observations are paired with matches.

iii. The difference between paired observations is normally distributed or approximately normally distributed.

The test statistic:

Where:

 d− is the mean of match-paired observations.

 sd is the standard deviation of the difference of paired observations.

Confidence Interval for μ1-μ2  of Dependent Samples

Let d be the difference between matched paired observations of samples selected from normally distributed populations with population means difference (μD) and unknown difference of standard deviations (SD). The population standard deviation of the difference is unknown, so it is replaced by the sample standard deviation "sd" of the difference of paired observations.

The sampling distribution of d-bar approaches the t distribution with (n-1) degrees of freedom.

Now to construct a confidence interval for the difference of population means. Choose two values from the t-table and make the following probability statement.

Example 10.11: The scores of 5 batsmen before and after receiving the batting tips are given below:

Batsman	Score Before	Score After
1	25	45
2	80	77
3	56	62
4	88	80
5	49	56

Construct a 98% confidence interval for the difference between before and after receiving the betting tips.

Solution:

1-α =0.98

α = 0.02

tα/2(n-1) = t0.01(4) = 3.747

The sample mean of the difference:

The sample standard deviation of the difference:

The confidence interval for the difference of population mean before and after receiving batting tips:

Hypothesis Testing about the Difference between Population Means of Dependent Samples

Let d be the difference between matched paired observations of samples selected from normally distributed populations with population means difference (μD) and unknown difference of standard deviations (SD). The population standard deviation of the difference is unknown, so it is replaced by the sample standard deviation "sd" of the difference of paired observations.

The sampling distribution of d-bar approaches the t distribution with (n-1) degrees of freedom.

Testing Procedure:

i. State null and alternative hypothesis

H0: μD = 0 vs H1: μD ≠ 0

H0: μD ≥ 0 vs H1: μD < 0

H0: μD ≤ 0 vs H1: μD > 0

ii. The significance level: α

iii. The test statistic: The samples are dependent

iv. Critical Region:

Reject H0 when |t|  ≥ tα/2(n-1)
When t < - tα(n-1), reject H0.

Dismiss H0 when t > tα(n-1)

v. Computation

vi. Remarks.

Example 10.12: The weight of 6 recruits before and after a tough physical exercise by the army is recorded as follows:

Recruit	Weight Before	Weight After
1	195	201
2	160	158
3	140	145
4	170	176
5	178	190
6	195	190

Use a 5% significance level to test the hypothesis that the weight of the recruit is significantly different before and after the physical exercise.

Solution:

i. State null and alternative hypothesis

H0: μD = 0 vs H1: μD ≠ 0

ii. The significance level: α = 0.05

iii. The test statistic: The samples are dependent

iv. Critical Region:

Reject H0 when |t|  ≥ t0.025(5) = 2.571

v. Computation:

Recruit	X1	X2	d = X1 - X2	d^2
1	195	201	-6	36
2	160	158	2	4
3	140	145	-5	25
4	170	176	-6	36
5	178	190	-12	144
6	195	190	5	25
			-22	270

The mean of the difference of sample observations:

The standard deviation of the difference of sample observations:

vi. Remarks: The computed t value falls in the acceptance region; the sample data does not provide sufficient evidence to reject the null hypothesis. Thus, it is concluded that the weights of recruits before and after are not significantly different.

Example 10.13: In a certain experiment to compare two types of sheep food, A and B, the following results of increase in weights were observed:

Sheep	Food A	Food B
1	49	52
2	53	55
3	51	52
4	52	53
5	47	50

Test the hypothesis at the 5% significance level: the effect of food A on weight gain is better than that of food B.

Solution:

i. State the null and alternative hypotheses.

H0: μA ≥ μB vs H1: μA < μB

OR

H0: μD ≥ 0 vs H1: μD < 0

ii. The significance level: α = 0.05

iii. The test statistic: The samples are dependent

iv. Critical Region:

Reject H0 when t < -t₀.₀₅(4) = - 2.132

v. Computation:

Sheep	X1	X2	d=X1-X2	d^2
1	49	52	-3	9
2	53	55	-2	4
3	51	52	-1	1
4	52	53	-1	1
5	47	50	-3	9
			-10	24

The mean of the difference between the sample data:

The standard deviation of the difference between the sample data:

vi. Remarks: The computed t value falls in the rejection region; the sample data does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that food A on weight gain is better than food B.

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