Determine the critical value(s) of the test statistic for each of the following small sample tests for the population mean where the assumption of normality is satisfied. round your answer to four decimal places.

step 1 of 3:

left-tailed test,α=0.01,n=19

step 2 of 3:

right-tailed test ,α=0.1,n=13

step 3 of 3:

two-tailed test, α=0.05,n=17

Step 1  t  = - 2.552

Step 2 t   =  1.356

Step 3  two values     t  =  - 2.12 on the left   and   t = 2.12 on the right

Step-by-step explanation:

Normal Distribution

size sample  < 30

Table to use  t-student table

a)  left-tailed test       α  =  0,01            n = 19      gl   = 19 -1      gl = 18

Directly from  t table      

t  = - 2.552

b) Right-tailed test     α  = 0,1     n  = 13     gl  = 13 - 1    gl =12

Directly from  t table

t  = 1.356

c)  Two-tailed test   α  = 0,05   but two tails means   we use α/2

    α/2  =  0,025    n = 17   gl  = n - 1   gl  = 17  -1     gl  = 16

From table:

t =  - 2.12  on the left tail  and  t = 2.12 on the right tail

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584