TH 227 Statstics (Mon wed 7.00 to 9 05 p m) Homework: homework 417 (Due Wea\'esd

Question

TH 227 Statstics (Mon wed 7.00 to 9 05 p m) Homework: homework 417 (Due Wea'esday, December 13, at 7:30 Score: 083 of 5 pts AVN Score 4656%·25 EQuesbor of .ctre..es notes-ni 6 of 11 (11 compete) 9.4.9-T Listed below vo ages of actresses and actors " une mat they won un awardmecnegones or Best Actress nd Best Acta. Use the sanple dat, to test.awem. te-ense wward Use 0 10 signféicance leel Assume that the pared sample data is a simgle randoh samgle and that the dferences haes datobution hat is imely sma Actor's age 4 hypothesis test? 47 65 503 dentily the test ststistic '-(Round to two decm places as needed ) Enter your answer in the answer box and mertick Check Answer Clew Al

Explanation / Answer

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, = 0.1
from standard normal table, two tailed t /2 =2.132
since our test is two-tailed
reject Ho, if to < -2.132 OR if to > 2.132
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -19.4
We have d = -19.4
pooled variance = calculate value of Sd= S^2 = sqrt [ 2581-(-97^2/5 ] / 4 = 13.22
to = d/ (S/n) = -3.28
critical Value
the value of |t | with n-1 = 4 d.f is 2.132
we got |t o| = 3.28 & |t | =2.132
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.2811 ) = 0.0305
hence value of p0.1 > 0.0305,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -3.28
critical value: reject Ho, if to < -2.132 OR if to > 2.132
decision: Reject Ho
p-value: 0.0305

X Y X-Y (X-Y)^2 19 45 -26 676 25 47 -22 484 29 65 -36 1296 48 50 -2 4 27 38 -11 121 -97 2581

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