Question

4. A HR specialist is especially interested in two traits of workers: whether they are diligent (or lazy) and whether they are creative (or not). We assume that the two traits are independent, and that the probabilities that a worker is diligent and lazy are each equal toWe assume that the probability that a worker is creative is equal to p, where pE (0,1) is unknown. A group of workers were asked to take a test; the results are summarized in the table below: Type of worker Diligent Lazy and not creative and creative not creative N3=20 Diligent Lazy and and creative Number of workers Assumed probabilities N2 30 1p) N4= 40 (1-p) N1=10 p Find the formula for the MLE of p (depending on Ni, N2, N3, N4) .and the value of the estimator for the above sample: Using the chi-squared test (and the estimate from the previous point), verify whe- ther the assumed probability scheme is valid for a significance level 0.05. The value of the appropriate test statistic is equal to nul hypothesis the statistic has a distribution with degrees of freedom and a critical value equal to TO REJECT (underline the appropriate) the null hypothesis .., under the so we REJECT /HAVE NO GROUNDS

Answer 1

Anuwer ca) HR Specialist intested in tun taitr r udke fntepenclent two tais me a croative P,-P), pe Co,1) net As per the, test , the Can assunme eoch of. te Tesults ane ux populations ar Pocus ue mdepenent automatically datnbuud So, ft con be. treatid These ae 4 gadups and 4. the paobabiletes for Cacpj of P(NNN3NyP) Ny (n, N2 N 3 Ny Ni+NA NIN2N3NU Litehoad r () ap NITN2 N2INY (-1) L tama ukehood etimata otP NitN2 NotNy NtN3)- P C+Nu) NtN3(NtNg tN3+Ny) NrtN3 NIEN2T N3tNy NitN3 NLE NitN2NgtNy

as per the ven data N vas i NLE 10+90 10430+90+40 30 - 3 100 0.3 b) un the chi-squae hove to verify asumed paobabllity test we is valid tor ifgnificonce level o 0-0 observed requency expectei frequeny -nxpi the x-tet statistic K t-n mp; mpi approximatty fellaa a x(-) datbution we liste 1o, tgz30, +z= n: loo m () 35 tatistics Compare tet -n 30 -l00 35 IS aS EMr NI

104-619 -I00 -the Compudıd value of tut Statistics 4.619 ID x o0 or, F-i- 3 The citical value equal DS, X-6tat47619 7:82 mull Teason d evfdarxe td eject Ercugh hypothes ausumed probobilt. inal we have not to Here,the value o appauirmāle test- stoilistic is <qual bypothess Stalistic tet ha distributio ofth 4-1-3 degrees factron to 82. equal NOlue coitical and to REJECT the Noll HYAOT HESIS goounds Hence Iwe have no