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OverLord2011 [107]
3 years ago
7

Which one of the following is not a integer A-5 B.-3 C-1/2 D-10

Mathematics
2 answers:
Trava [24]3 years ago
7 0
C ; because that is a fraction and the rest are whole numbers.
Karolina [17]3 years ago
4 0
-1/2 is not a integer
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Usimov [2.4K]

Answer:

a) We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

b) We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

Step-by-step explanation:

Part a

We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

Part b

We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

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Answer:

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