The price of oranges went from $.50 per Ib to $1.80 per lb in four years. Find

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

How many square inches in a foot

Cloud [144]

Answer:

There are 144 square inches in a square foot.

Step-by-step explanation:

A square foot is a square that is 1 foot by 1 foot. Since there are 12 inches in 1 foot, and to get the area you need to multiple 2 adjacent sides together, you do 12 times 12 which is 144 square inches. Many people have the misconception that a square foot has 12 square inches, but this is wrong since a square foot has dimensions of 12 inches by 12 inches, but when you multiply them together, you can see that the answer is actually 144.

5 0

2 years ago

Hat is the position of 7 in the number 876,543? A. The ten-thousands place B. The hundreds place C. The tens place D. The thousa

Vesna [10]

As you can see in this diagram, the answer is A, the ten-thousands place.

7 0

2 years ago

the cost of a school banquet is $95 plus $15 for each person attending. write an equation that gives total cost as a function of

AleksandrR [38]

Im sorry if I misunderstood the problem but heres my take.

Answer:

1155 for the cost of the people and 1250 for the total including the $95

Step-by-step explanation:

Multiply it ( 77 x 15 )

then add the answer to the $95 ( 95 + N )

= $N

8 0

2 years ago

3 An open cone has a vertical angle measuring 40°

belka [17]

Answer:

13.41 cm

Step-by-step explanation:

The radius, height, and HALF of vertical angle makes a triangle.

Where

Radius will be the base of the triangle

Height will be the "height" of the triangle

Top angle will be HALF of vertical angle (40/2 = 20)

From the 20 degree angle, the side "opposite" would be the radius, which is 30. The side "adjacent" would be the height, let it be h.

<em>Which trig ratio relates opposite and adjacent? Yes, that is TAN. Thus we can write:</em>

Tan(20) = opposite/adjacent

Tan(20) = 30/h

h = 30/Tan(20)

h = 13.41

So

The height of the cone would be around 13.41 cm

3 0

2 years ago