Question

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

3 years ago

Find the area of 7cm 20cm 14cm 11cm

FinnZ [79.3K]

Multiply all 4 an the answer is gonna be the area

3 0

3 years ago

Could anyone help me with these?

lora16 [44]

Part A:

Derrick is incorrect.
A yard is equivalent to 3 feet.
Thus, 20 yards will be equivalent to 60 feet.

The width is 20 yard and 2 feet. It's the same as 60 feet and 2 feet, or 62 feet. Thus, Derrick is incorrect.

Part B:

Let's calculate the area of the parking lot.

We know the width is 62 feet from part A.
The length is 30 yards, or 90 feet.

Multiply the length and width to get the area in square feet.

$90 \times 62=5580$

The parking lot has an area of 5580 square foot.
It costs 2 dollars per square foot to repave the parking lot. Multiply the area by 2 dollars.

$5580 \times 2=11160$

It will cost $11,160 to repave the whole parking lot.

4 0

3 years ago

Read 2 more answers

your trying to save up for a trip to hawaii. Your trying to budget for a hotel, and you found a really nice one for 1,200. And t

stepladder [879]

Answer:

4800 good luck

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Gift tax falls under which tax category?