Case Study 14-1
A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women
experiment and randomly divided into two groups. One of the groups is subjected to high stress for two mont
stress-free environment. The professor measures the menstrual cycle (in days) of each woman during the sec
obtained.
High stress 20 23 18 19 22
Relatively stress free 26 31 25 26 30
Find Mean and Standard deviation of above –
Set Up Hypothesis
There Is No Significance between them - Under The Null Hypothesis Ho: u1 = u2
There Is Significance between them - Under The Alternate Hypothesis H1: u1 != u2
Test Statistic
X(Mean)=20.4
Standard Deviation(s.d1)=2.0736
Number(n1)=5
Y(Mean)=27.6
Standard Deviation(s.d2)=2.7019
Number(n2)=5
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
t cal=20.4-27.6/Sqrt((4.29982/5)+(7.30026/5))
t cal=-4.73
| t cal | =4.73
Critical Value
The Value of |t tab| with (n1+n2-2) i.e. 8 d.f is +2.306,-2.306
We got |t cal| = 4.727 & | t tab | = 3.2498
Make Decision
Hence Value of | t cal | > | t tab| and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -4.727 ) = 0.002
Image Format:
A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women
experiment and randomly divided into two groups. One of the groups is subjected to high stress for two mont
stress-free environment. The professor measures the menstrual cycle (in days) of each woman during the sec
obtained.
High stress 20 23 18 19 22
Relatively stress free 26 31 25 26 30
Find Mean and Standard deviation of above –
Set Up Hypothesis
There Is No Significance between them - Under The Null Hypothesis Ho: u1 = u2
There Is Significance between them - Under The Alternate Hypothesis H1: u1 != u2
Test Statistic
X(Mean)=20.4
Standard Deviation(s.d1)=2.0736
Number(n1)=5
Y(Mean)=27.6
Standard Deviation(s.d2)=2.7019
Number(n2)=5
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
t cal=20.4-27.6/Sqrt((4.29982/5)+(7.30026/5))
t cal=-4.73
| t cal | =4.73
Critical Value
The Value of |t tab| with (n1+n2-2) i.e. 8 d.f is +2.306,-2.306
We got |t cal| = 4.727 & | t tab | = 3.2498
Make Decision
Hence Value of | t cal | > | t tab| and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -4.727 ) = 0.002
Image Format:
