Answer:
Yes, he would take the same decision.
Step-by-step explanation:
Consequently, because the decision is taken on the test based on the use of alpha equals 0.025, the p-value of the test must have been greater than the given amount of importance that is 0.025 since the test is not applicable to us. So, p > 0.025.
If we know that p > 0.025, that would not mean p > 0.1 as well, because we do not know with the details given he had to make the same decision for 0.1 degree of meaning.
As for the 0.01 significance point, we 're sure p > 0.01 is greater than 0.025, so the test does not matter.
Answer:
57.5
Step-by-step explanation:
The expected frequency of West Campus and Failed :
Let :
Failed = F
East Campus = C
West Campus = W
Passed = P
Frequency of FnW :
[(FnE) + (FnW) * (PnW) + (FnW)] / total samples
[(52 + 63) * (63 + 37)] / 200
[(115 * 100)] / 200
11500 / 200
= 57.5
n(Failed n East campus)
Answer:
it rise over run so put 2y on graph count down 4 and move right 1
Step-by-step explanation:
Answer: Its 516. Hope this helps!
Step-by-step explanation:
