Two Samples t Test Lecture 43

Two Samples t Test

Lecture 43

Introduction

The students' two samples  t test is a statistical method used to determine that the two population means are identical or significantly different from each other. it is widely used when the sample sizes are small and the standard deviations are unknown but identical. 

When to use the Students’ Two Samples t-Test

The students’ two samples t-test is used when:

i. The sample sizes are small.

ii. Both samples are random and independent.

iii. The standard deviation is unknown but assumed to be identical.

iv. The samples selected from normal are approximately normal.

Confidence Interval between μ1 - μ2 when σ1 = σ2 but Identical

Assume that two independent random small samples of size n1 and n2 drawn from two normal populations with means of μ1 and μ2, and unknown but identical standard deviations σ1, σ2, yield unbiased estimates of μ1 and μ2, respectively, denoted by X¯1 and X¯2. The standard deviation is the same but unknown, and both samples are small. The sampling distribution of  (X¯1 - X¯2) approaches the t distribution with v1 and v2 degrees of freedom.

Where:

sp is the pooled estimate of σ1 and σ2.

 

Now to construct a 100 (1 - α) confidence interval for (μ1 - μ2). Choose two values (-tα/2, tα/2) and make the following probability statement.

Example 10.6: The two samples 25 male and 29 females are selected from male and females’ population with identical standard deviations of sleeping hours. The sample means are 6.9 and 6.6 hours of sleep, with standard deviations of 1.8 and 1.5, respectively. Construct 98 % confidence interval between the average sleep’s hour for male and female.

Solution:

1- α = 0.98

 α = 0.02

t  α/2 (v) = t0.01 (52) = 2.678

Example 10.7: From an area planted in one variety of guayule, a random sample of 27 plants is made from an area planted with a single variety of guayule. 15 were off types, and 12 were aberrant. These plants' percentages of rubber were:

Off types	Aberrant
6.21	4.28
5.70	5.71
6.04	6.48
6.55	5.42
6.02	4.37
4.45	6.20
6.75	5.06
5.88	6.40
6.82	7.00
6.09	4.91
5.59	5.51
6.06	5.36
7.45
6.74
7.60

Construct a 98% confidence interval between the population means of the rubber percentages. Assume that the rubber percentages are approximately normal with identical variances.

Solution: Let X1 & X2 represent off types and aberrant, respectively.

1-α = 0.98

α = 0.02

tα/2 (n1 + n2 - 2) = t 0.01(25) = 2.787

X1	X1^2	X2	X2^2
6.21	38.5641	4.28	18.3184
5.70	32.49	5.71	32.6041
6.04	36.4816	6.48	41.9904
6.55	42.9025	5.42	29.3764
6.02	36.2404	4.37	19.9809
4.45	19.8025	6.20	38.44
6.75	45.5625	5.06	25.6036
5.88	34.5744	6.40	40.96
6.82	46.5124	7.00	49.00
6.09	37.0881	4.91	20.1601
5.59	31.2481	5.51	30.3601
6.06	36.7236	5.36	28.7296
7.45	55.5025
6.74	45.4276
7.60	57.76
93.98	596.8803	66.70	375.5236

The sample mean of the rubber percentage of off-type plants:

The sample mean of the rubber percentage of aberrant plants:

The sample variance of the rubber percentage of off-type plants:

The sample variance of the rubber percentage of aberrant plants:

The pooled standard deviation can be obtained as:

98 % confidence interval for the difference between off-types and aberrant population means

The difference between population means of off-types and aberrants is from 4% to 13.6 %.

Hypothesis Testing between μ1 - μ2 when σ1 = σ2 but Identical

Assume that two independent random small samples of size n1 and n2 drawn from two normal populations with means of μ1 and μ2, and unknown but identical standard deviations σ1, σ2, yield unbiased estimates of μ1 and μ2, respectively, denoted by X¯1 and X¯2. The standard deviation is the same but unknown, and both samples are small. With v1 and v2 degrees of freedom, the sampling distribution of (X¯1-X¯2) is close to the t distribution.

If it is desired to test the null hypothesis as:

The test statistics become:

Testing Procedure:

i. State null and alternative hypotheses as:

H0: μ1-μ2 = Δ0 . H1: μ1-μ2 ≠ Δ0 

H0: μ1-μ2 ≥ Δ0. H1: μ1-μ2 < Δ0 

H0: μ1-μ2 ≤ Δ0. H1: μ1-μ2 > Δ0 

ii. The significance level; α

iii. The test statistic: The samples are selected randomly, independently, and have identical standard deviations. 

vi. Critical Region:

Reject H₀ , when |t| ≥ tα/2 (n₂ - 2)

Reject H₀ , when t ≥ tα (n1+n2-2)

Reject H0, when t ≤ -tα (n1+n2-2)

v. Computation:

vi. Remarks.

Example 10.8: Two random independent samples, each of size 11 and 6, are chosen from classes using methods A and B, respectively, in order to test how well they perform. Below are the results of a class test that the students took:

Method A	Method B
51	38
42	49
49	45
55	29
46	31
63	35
56
58
47
39
47

Test the null hypotheses: the performance of two teaching methods is identical at the 5% significance level.

Solution: Let X1 and X2 represent methods A and B, respectively. 

i. State null and alternative hypotheses as:

H0 : μ1-μ2 = 0 Vs. H1 : μ1-μ2 ≠ 0 

ii. The significance level; α = 0.05

iii. The test statistic: The samples are selected randomly, independently, and have identical standard deviations. 

vi. Critical Region:

Reject H0, when | t | ≥ t0.025 (14) = 2.145

v. Computation:

X1	X1^2	X2	X2^2
51	2601	38	1444
42	1764	49	2401
49	2401	45	2025
55	3025	29	841
46	2116	31	961
63	3969	35	1225
56	3136
58	3364
47	2209
39	1521
47	2209
553	28315	227	8897

The sample mean of the students marks taught by method A:

The sample mean of the students marks taught by method B:

The sample variance of the students marks taught by method A:

The sample variance of the students marks taught by method B:

The pooled standard deviation of methods A and B is given below:

The t statistic can be computed as:

vi. Remarks: The computed t value, which is 3.313, falls in the rejection region. The sample data does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that the populations of two teaching methods are significantly different.

Example 10.9: Test the hypothesis μ1-μ2 > 3  using the data from two normally distributed populations with equal standard deviations but unknown values provided below:

Sample	Size	Sample Mean	Sample Standard Deviation
1	15	12.74	0.412
2	12	8.75	1.331

Solution:

i. State null and alternative hypotheses as:

H0 : μ1-μ2 ≤ 3 Vs. H1 : μ1-μ2 > 3 

ii. The significance level; α = 0.05

iii. The test statistic: The samples are selected randomly, independently, and have identical standard deviations.

iv. Critical Region:

Reject H0, when t ≥ t0.05 (25) = 1.708

v. Computation:

The pooled estimate of standard deviations of samples 1 and 2.

vi. Remarks: The computed t value, which is 2.734, falls in the rejection region. The sample data does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that the difference between population means is more than 3.

Example 10.10: The yield capabilities of two varieties of potato are tested by the agriculture department. Two acres are planted at each research station to reduce the impact on the environment, and the yields, measured in maunds per acre, are recorded as follows:

Varity 1	Variety 2
28	45
23	25
35	31
41	38
44	32
29	33
37	25
31	30
38	33

Assume independence and test the hypothesis that the yield capabilities of variety 1 are less than variety 2 at the 5% significance level.

Solution:

i. State null and alternative hypotheses as:

H0 : μ1 ≥ μ2  Vs. H1 : μ1<μ2  

ii. The significance level; α = 0.05

iii. The test statistic: The samples are selected randomly, independently, and have identical standard deviations.

vi. Critical Region:

Reject H0, when t ≤ -t0.05 (16) = -1.746

v. Computation:

The sample mean of variety 1:

The sample mean of variety 2:

The sample variance of variety 1:

The sample variance of variety 2:

The pooled standard deviation of varieties 1 & 2:

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 yield capabilities of variety 1 are not less than those of variety 2 of potato.

• Read More: t Test for Dependent Samples