Complete parts (a) through (c) below.
(a) determine the critical value(s) for a right-tailed test of a population mean at the alphaequals0.01 level of significance with 10 degrees of freedom.
(b) determine the critical value(s) for a left-tailed test of a population mean at the alphaequals0.10 level of significance based on a sample size of nequals15.
(c) determine the critical value(s) for a two-tailed test of a population mean at the alphaequals0.01 level of significance based on a sample size of nequals12.

a) t =  2.7638

b) t = - 2.6245

c) t = 3.1058      on the right side    and

   t  = -3.1058    on the left

Step-by-step explanation:

a)Determine critical value for a right-tail test  for α = 0.01 level of significance and 10 degrees of fredom

From t-student table we find:

gl  =  10   and α = 0.01       ⇒  t =  2.7638

b)Determine   critical value for a left-tail test  for α = 0.01 level of significance  and sample size n = 15

From t-student table we find:

gl  =  14   and α = 0.01           gl  =  n - 1      gl  = 15 - 1    gl = 14

t = - 2.6245

c) Determine critical value for a two tails-test  for α = 0.01 level of  significance the  α/2   =   0.005  and sample size  n = 12

Then

gl  =  11   and α = 0.005

t = 3.1058      on the right side of the curve and by symmetry

t = - 3.1058  

From t-student table we find:

e) none of the above because the measure of angle cba is 101°

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: