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Sagot :
The value of t obtained is 4.76
Given,
Mean of group 1, [tex]M_{1}[/tex] = 10
Mean of group 2, [tex]M_{2}[/tex] = 12
[tex]SS_{1}[/tex] = 5
[tex]SS_{2}[/tex] = 8
Number of individuals per group, n = 6
The degrees of freedom are calculated as:
[tex]df = (n_{1} -1)+(n_{2} -1)[/tex] = ( 8-1) + (8-1) = 7 + 7 = 14
The difference between sample means [tex]M_{d}[/tex] is:
[tex]M_{d}[/tex] = [tex]M_{1} -M_{2}[/tex] = 10 - 12 = 2
Here n is equal for both groups which is = 8
Now,
The standard deviation of sample 1 :
[tex]s_{1} =\sqrt{\frac{SS_{1} }{n_{1} -1} } = \sqrt{\frac{5}{8-1} } = \sqrt{\frac{5}{7} } = 0.85[/tex]
The standard deviation of sample 2 :
[tex]s_{2}= \sqrt{\frac{SS_{2} }{n_{2}-1 } } = \sqrt{\frac{8}{8-1} } = \sqrt{\frac{8}{7} } = 1.07[/tex]
Now, we have to calculate the standard error for the difference of the means.
MSE = [tex]\frac{(n_{1}-1)s^{2} _{1}+(n_{2}-1)s^{2} _{2} }{(n_{1}-1)+(n_{2}-1) }[/tex]
= [tex]\frac{(8-1)(0.85^{2})+(8-1)(1.07^{2}) }{(8-1)+(8-1)}[/tex]
= [tex]\frac{10.1318}{14}[/tex]
MSE = 0.7237
Then, the standard error can be calculated as:
[tex]s_{M_{d} } = \sqrt{\frac{2MSE}{n} } = \sqrt{\frac{2 * 0.7237}{8} } = 0.42[/tex]
Now we can calculate t :
[tex]t=\frac{M_{d} }{s_{M_{d} } } = \frac{2}{0.42} = 4.76[/tex]
Learn more about standard deviation here : https://brainly.com/question/16403666
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