Asked by Danny
In two independent samples from populations that are normally distributed, x¯1 =35.0, s1 = 5.8, n1 =12 and x¯2 = 42.5, s2 = 9.3, n2 = 14. Using the 0.05 level of significance, test H0: µ1 = µ2 versus H1: µ1 ≠ µ2
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MathMate
Here's how I would proceed.
First, we have a case of two independent unpaired samples from normally distributed populations, where
x̄<sub>1</sub>=35
s<sub>1</sub> = 5.8
n<sub>1</sub>=12
x̄<sub>2</sub>=42.5
s<sub>2</sub> = 9.3
n<sub>2</sub>=14
H: μ<sub>1</sub> = μ<sub>2</sub>.
We calculate the t-statistic
t = (x̄<sub>2</sub> - x̄<sub>1</sub>)/sqrt(s<sub>1</sub>²/n<sub>1</sub>+s<sub>2</sub>²/n<sub>2</sub>)
=(42.5-35)/sqrt(5.8²/12+9.3²/14)
=7.5/2.99686
=2.503
The number of degree of freedoms is given by:
dof
=((s<sub>1</sub>²/n<sub>1</sub>) + (s<sub>2</sub>²/n<sub>2</sub>))²/((s<sub>1</sub>²/n<sub>1</sub>)²/n<sub>1</sub> + (s<sub>2</sub>²/n<sub>2</sub>)²/n<sub>2</sub>)
=(5.8²/12+9.3²/14)²/((5.8²/12)²/11 + (9.3²/14)²/13)
=80.662/3.65
= 22.1
The two-tail t-statistic of 2.503 and a degree of freedom of 22 has a probability of 0.020, less than the significance level of 5%, so the hypothesis is rejected.
A similar calculation applies to the hypothesis of μ<sub>1</sub> ≠ μ<sub>2</sub>.
Please check my calculations.
There is a very good article on the student's t-test, and an interactive t-statistic table.
http://en.wikipedia.org/wiki/Student's_t-test
http://www.tutor-homework.com/statistics_tables/statistics_tables.html#t
First, we have a case of two independent unpaired samples from normally distributed populations, where
x̄<sub>1</sub>=35
s<sub>1</sub> = 5.8
n<sub>1</sub>=12
x̄<sub>2</sub>=42.5
s<sub>2</sub> = 9.3
n<sub>2</sub>=14
H: μ<sub>1</sub> = μ<sub>2</sub>.
We calculate the t-statistic
t = (x̄<sub>2</sub> - x̄<sub>1</sub>)/sqrt(s<sub>1</sub>²/n<sub>1</sub>+s<sub>2</sub>²/n<sub>2</sub>)
=(42.5-35)/sqrt(5.8²/12+9.3²/14)
=7.5/2.99686
=2.503
The number of degree of freedoms is given by:
dof
=((s<sub>1</sub>²/n<sub>1</sub>) + (s<sub>2</sub>²/n<sub>2</sub>))²/((s<sub>1</sub>²/n<sub>1</sub>)²/n<sub>1</sub> + (s<sub>2</sub>²/n<sub>2</sub>)²/n<sub>2</sub>)
=(5.8²/12+9.3²/14)²/((5.8²/12)²/11 + (9.3²/14)²/13)
=80.662/3.65
= 22.1
The two-tail t-statistic of 2.503 and a degree of freedom of 22 has a probability of 0.020, less than the significance level of 5%, so the hypothesis is rejected.
A similar calculation applies to the hypothesis of μ<sub>1</sub> ≠ μ<sub>2</sub>.
Please check my calculations.
There is a very good article on the student's t-test, and an interactive t-statistic table.
http://en.wikipedia.org/wiki/Student's_t-test
http://www.tutor-homework.com/statistics_tables/statistics_tables.html#t
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