Question

Interpret the differences in conducting two-tailed , upper tailed and lower tailed t-tests. Please provides an example of how at least two of these type of t-tests have been used in quantitative and qualitative research.

Answer 1

Paired and unpaired t-tests are just some of the statistical tests
that can be used to test quantitative data.
One of the most common statistical tests for
qualitative data is the chi-square test (both the goodness of fit test and test of independence)

example for quantitative data research

Assumed data,

t test for paired data
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -6
We have d = -6
pooled variance = calculate value of Sd= √S^2 = sqrt [ 804-(-60^2/10 ] / 9 = 7.02
to = d/ (S/√n) = -2.7
critical Value
the value of |t α| with n-1 = 9 d.f is 2.262
we got |t o| = 2.7 & |t α| =2.262
make Decision
hence Value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.7014 ) = 0.0243
hence value of p0.05 > 0.0243,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -2.7
critical value: reject Ho, if to < -2.262 OR if to > 2.262
decision: Reject Ho
p-value: 0.0243
we have enough evidence to support the claim

Assumed data,
t test for independent samples
mean(x)=4.3757
standard deviation , s.d1=1.6423
number(n1)=14
y(mean)=7.1227
standard deviation, s.d2 =2.0761
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.771
since our test is left-tailed
reject Ho, if to < -1.771
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.3757-7.1227/sqrt((2.69715/14)+(4.31019/15))
to =-3.965
| to | =3.965
critical value
the value of |t α| with min (n1-1, n2-1) i.e 13 d.f is 1.771
we got |to| = 3.96495 & | t α | = 1.771
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -3.965 ) = 0.00081
hence value of p0.05 > 0.00081,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -3.965
critical value: -1.771
decision: reject Ho
p-value: 0.00081

qualitative approach example:
Assumed data,

Given table data is as below MATRIX col1 col2 TOTALS row 1 346 3628 3974 row 2 495 3563 4058 TOTALS 841 7191 N = 8032 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N ------------------------------------------------------------------ expected frequencies calculated by applying E - table matrix formula E-TABLE col1 col2 row 1 416.102 3557.9 row 2 424.898 3633.1 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 346 416.102 -70.102 4914.3 11.81 3628 3557.9 70.102 4914.3 1.381 495 424.898 70.102 4914.3 11.566 3563 3633.1 -70.102 4914.3 1.353 chisqr^2 o = 26.11 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, alpha = 0.05 from standard normal table, chi square value at right tailed, chisqr^2 alpha/2 =3.841 since our test is right tailed,reject Ho when chisqr^2 o > 3.841 we use test statistic chisqr^2 o = Σ(Oi-Ei)^2/Ei from the table , chisqr^2 o = 26.11 critical value the value of |chisqr^2 alpha| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.841 we got | chisqr^2| =26.11 & | chisqr^2 alpha | =3.841 make decision hence value of | chisqr^2 o | > | chisqr^2 alpha| and here we reject Ho chisqr^2 p_value =0 ANSWERS --------------- null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent test statistic: 26.11 critical value: 3.841 p-value:0 decision: reject Ho

we have enough evidence to support the claim