Estimate the value of √97.5÷1.96

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

13. What is the highest number of degrees that an angle can have? 180 360 90 270

kykrilka [37]

Answer:

360

Step-by-step explanation: A full angle is 360. 90*4=360.

8 0

3 years ago

Please hurry it’s missing and I need it fast

sattari [20]

2. 83 R 2
3. 82 R 1111 etc

3 0

2 years ago

Read 2 more answers

Six times a number is greater than 20 more than that number. What are the possible values of that number? n less-than 4 n greate

MrRissso [65]

Answer: answer is b

Step-by-step explanation: because ik

7 0

3 years ago

Read 2 more answers

Please help with this

ioda

Answer:

Some have multiple choices. See below for additional comments.

a) iii
b) i
c) v
d) ii
e) iv

Step-by-step explanation:

i -- slope-intercept form. Shows the slope and the y-intercept

ii -- slope-intercept form. Shows the slope and the x-intercept

iii -- intercept form. Shows both the x- and y-intercepts

iv -- standard form.

v -- point-slope form. Shows the slope and one point on the line.

vi -- another point-slope form. Shows the slope and one point on the line

__

With the above forms in mind, ...

a) both intercepts are best shown in form iii

b) the slope is shown in forms i, ii, v, vi. Form i is the most uncluttered.

c) points are shown in forms v and vi. Form v may be least confusing.

d) the sign can be readily determined in form ii.

e) testing the given point is easily done using forms ii and iv. Form iv may be the best for this.

_____

Additional comments

Since each of these forms of the equation has utility in different situations, it can be worthwhile to learn the different forms. For example, I like the standard form for making perpendicular lines, though any of the "slope-" forms can be useful for that, too. The useful form to start with will be the one that best relates to the information given. (For two points given, there is a "2-point form" not shown here.)

6 0

3 years ago