Question

For an independent t-test, if the mean of group 1=10, mean of group 2=12, ss1 =5, ss2=8, there are 6 individuals per group. what is the value of t obtained?

Answer 1

The value of t obtained is 4.76

Given,

Mean of group 1, $M_{1}$ = 10

Mean of group 2, $M_{2}$ = 12

$SS_{1}$ = 5

$SS_{2}$ = 8

Number of individuals per group, n = 6

The degrees of freedom are calculated as:

$df = (n_{1} -1)+(n_{2} -1)$ = ( 8-1) + (8-1) = 7 + 7 = 14

The difference between sample means $M_{d}$ is:

$M_{d}$ = $M_{1} -M_{2}$ = 10 - 12 = 2

Here n is equal for both groups which is = 8

Now,

The standard deviation of sample 1 :

$s_{1} =\sqrt{\frac{SS_{1} }{n_{1} -1} } = \sqrt{\frac{5}{8-1} } = \sqrt{\frac{5}{7} } = 0.85$

The standard deviation of sample 2 :

$s_{2}= \sqrt{\frac{SS_{2} }{n_{2}-1 } } = \sqrt{\frac{8}{8-1} } = \sqrt{\frac{8}{7} } = 1.07$

Now, we have to calculate the standard error for the difference of the means.

MSE = $\frac{(n_{1}-1)s^{2} _{1}+(n_{2}-1)s^{2} _{2} }{(n_{1}-1)+(n_{2}-1) }$

       = $\frac{(8-1)(0.85^{2})+(8-1)(1.07^{2}) }{(8-1)+(8-1)}$

       = $\frac{10.1318}{14}$

  MSE  = 0.7237

Then, the standard error can be calculated as:

$s_{M_{d} } = \sqrt{\frac{2MSE}{n} } = \sqrt{\frac{2 * 0.7237}{8} } = 0.42$

Now we can calculate t :

$t=\frac{M_{d} }{s_{M_{d} } } = \frac{2}{0.42} = 4.76$

Learn more about standard deviation here : brainly.com/question/16403666

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