4. A HR specialist is especially interested in two traits of workers: whether they are diligent (or lazy) and wheth...
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 (1-p) N4= 40 (1-p) N1=10 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 nul hypothesis .., under the so we REJECT /HAVE NO GROUNDS
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 (1-p) N4= 40 (1-p) N1=10 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 nul hypothesis .., under the so we REJECT /HAVE NO GROUNDS