Plz, Help me with this question.

lutik1710 [3]

-0.272727272727272727272727272727

4 0

3 years ago

Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal

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 $z>1.96 \cup z$

7 0

4 years ago

How do I find the answer?

11Alexandr11 [23.1K]

First simplify the number part, 14/2 = 7/1
then the x part, $\frac{x}{x^ \frac{3}{4} } = x^ \frac{1}{4} 
$
y doesn't need to be simplified
then simplify the z part, z^-6 = z^6

put it all together...
$7x^ \frac{1}{4}y^ \frac{1}{3} z^6$

4 0

3 years ago

James and his family spend a morning picking peaches. They fill several bags. This line plot shows the weight of each bag.

Alja [10]

Answer:

Step-by-step explanation:

um i no but i can not tell u

8 0

3 years ago

Solve pls brainliest

Sophie [7]

First is 3 second is 2.

3 0

3 years ago