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irinina [24]
3 years ago
10

Seriously need help!!????

Mathematics
1 answer:
Serga [27]3 years ago
3 0
50 would be a good choice of answer
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Read 2 more answers
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
eimsori [14]

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

#SPJ4

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