Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .95, df 5 10 b. Central area 5 .95, df 5 20 c. Central area 5 .99, df 5 20 d. Central area 5 .99, df 5 50 e. Upper-tail area 5 .01, df 5 25 f. Lower-tail area 5 .025, df 5 5

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got

b)  "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got

e) "=T.INV(1-0.01,25)"

And we got

f) "=T.INV(0.025,5)"

And we got

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got

c. Central area =.99, df = 20

 For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got

d. Central area =.99, df = 50

  For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means

We can use the following excel code:

"=T.INV(0.025,5)"

And we got

366

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.