Answer:
The degrees of freedom is 11.
The proportion in a t-distribution less than -1.4 is 0.095.
Step-by-step explanation:
The complete question is:
Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4 if the samples have sizes 1 = 12 and n 2 = 12 . Enter the exact answer for the degrees of freedom and round your answer for the area to three decimal places. degrees of freedom = Enter your answer; degrees of freedom proportion = Enter your answer; proportion
Solution:
The information provided is:
Compute the degrees of freedom as follows:
Thus, the degrees of freedom is 11.
Compute the proportion in a t-distribution less than -1.4 as follows:
*Use a <em>t</em>-table.
Thus, the proportion in a t-distribution less than -1.4 is 0.095.
300/250 can be divide by 50 which makes it 6/5
You add them and then it would be 30.6898033492
Answer:
idk lol but is this markel?
Step-by-step explanation:
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
