One Tailed Vs Two Tailed T Test

Let us see about the topic is one tailed vs two tailed t test. Now we can first see about the two tailed, when the two tails of the sampling distribution of the normal curve are used, the relevant test is known as two – tailed test. The second one is the alternative hypothesis H1: µ1≠ µ2 is taken in two – tailed test for H0: µ12 . when there only one tail of the sampling distribution of the normal curve is used, the test is described as one tail test.


Two – Tailed Vs One-tailed T Tests:

 

One- Tailed Test:

 one tailed Vs two tailed t test

Two Tailed Test:

 one tailed Vs two tailed t test


Example Problem for One Tailed Vs Two Tailed T Test:

 

Example 1:- using one tailed vs two tailed t test

The average hourly wage of a sample of 150 workers in plant A was Rs. 2.56 with a S.D. of Rs. 1.08. the average wage of a sample of 200 workers in plant B was Rs.2.87 with a S.D of Rs. 1.28. we can applicant safely assume that the hourly wage paid by plant B are higher than those paid by plant A?

Solution:

Let x1 and x2 denotes the hourly wages paid to workers in plant A and plant B respectively

We set up

H0: µ12

H1: µ1< µ2 (one – tailed)

L.O.S: α = 0.05

Test Statistic

Z  = `(barx^1-barx^2)/(sqrt((s^1/(2n^1))+(s^2/(2n^2))))`

`(2.56-2.87)/(sqrt((1.08)^2/150 + (1.28)^2/200))`

|Z|= 2.453

Critical Value:

The table value of Z at 5% level in case of one-tailed test is Z = 1.645

Conclusion:

 Since |Z| > 1.645, H0 is rejected at 5% level

The hourly wage paid by plant B is higher than those paid by plant A.

 

I am planning to write more post on Paired Samples T Test, binomial probability distribution, Keep checking my blog.


Example 2:- using one tailed vs two tailed t test


Given that on the average 4% of insured men of age 65 die within a year and that 60 of a particular group of 1000 such men died within a year. We can this group be regarded as a representative sample.

Solution:

Null hypothesis  H0: p = 0.04

H1: p ≠ 0.04

Test Statistic:

p’ = 60/ 1000

 = 0.06

p=0.04

q = 0.96

z = `(0.06 - 0.04)/(sqrt((0.04 * 0.96)/(1000)))`

   =  `0.2/ (sqrt((0.004*0.96)))`

=3.2

Conclusion:

Since |Z| > 3, H0 is rejected.

The group chosen is not a representative sample.