Which of the following is the quotient of the rational expression shown below

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

My little brother is in 6th grade we are a bit confused how to show the work with a picture for these problems.

Nezavi [6.7K]

You have to draw the numbers then you would have to divide them draw it

6 0

3 years ago

Read 2 more answers

Simplify 3(2f+6)+7 thanks

Naya [18.7K]

{Expand} 3(2f+6): 6f+18

6f+18+7

18+7=25

= 6f+25

4 0

3 years ago

Plz help Arithmetic Sequence

IgorC [24]

Answer:

it is a+b-6x

Step-by-step explanation:

2020edgenuityedge

3 0

3 years ago

Running out of time! Please help and show work! Thank you! Will mark brainliest!

VikaD [51]

(-3j²k³)²(2j²k)³ = When a power is raised to a power the exponents have to be multiplied.

= (-3²j⁽²*²⁾k⁽³*²⁾)(2³j⁽²*³⁾k³) = We can take out the constants

= (9)(8)(j⁴k⁶)(j⁶k³) = We can group the same variables

= 72(j⁴j⁶)(k⁶k³) = When multiplying two powers that have the same base, you have to add the exponents.

= 72 j⁽⁴⁺⁶⁾k⁽⁶⁺³⁾ = 72j¹⁰k⁹

Answer = 72j¹⁰k⁹

Hope this helps!

$\textit{\textbf{Spymore}}$

5 0

3 years ago