The degree of freedom for t statistic is 11.
According to the given question.
For a repeated-measure study, comparing two treatments with 12 scores in each treatment .
So, we can say that sample size, n = 12.
We know that, when you have a sample and estimate the mean, we have
n – 1 degrees of freedom, where n is the sample size.
Therefore,
The degree of freedom for the given sample test will be
d.f = n -1
⇒ d.f = 12 - 1
⇒ d.f = 11
Hence, the degree of freedom for t statistic is 11.
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21 is .368
22 is .71
23 is .66
24 is .1, 3/9, 2/3, 4/5, and 1.0
25 is .4, .42, .5, .55, .6
Answer:
value of x is 59.5
Step-by-step explanation:
Solution.
Given, Right angled triangle
where reference angle given 48
hypotenuse(h)=80
perpendicular(p)=x
Now ,
I) sin48=p/h
or,sin48×h=p
or0.978×80=x
.°. 59.5=x
When rates expressed as a quantaty of 1, such as 2 feet per second or 5 miles per hour
