A study of potential age discrimination considers promotions among middle manage
ID: 3293384 • Letter: A
Question
A study of potential age discrimination considers promotions among middle managers in a large company. The data are as follows: The data set is called PROB5.sav. a. Find the expected numbers in each cell for the promoted row under the hypothesis of independence of age and promoted for the row Promoted. Neatly write the expected value to two decimals inside the cells in the table above b. Justify the indicated degrees of freedom. C. What is the Chi Square value for the cell where Not Promoted and 40- 49 come together (The 2, 3) cell? d. Reject the null hypothesis x^2_call > x^2_10/1 = ____ (write in the degrees of freedom value) if > (write in the degrees of freedom value) e. Is there significant evidence of a relation between age and promotion? alpha = 0.10. Show the SPSS output that supports your answer. f. What is the p-value and the conclusion? P-value =Explanation / Answer
a) applying chi square test of independence.
expected values are given in expected block,.
b) degree of freedom =(rows-1)*(columns-1)=(2-1)*(4-1)=3
c)chi sqaure value =0.753
d)reject the null hypothesis for X2 >6.251
e)as test stat 13.025 is greater then critical value ; therefore there is significant evidence of a relation between age and promotion.
f) p value =0.0046
Observed O under 30 30-39 40-49 50 and over Total promoted 9 29 32 10 80 not promoted 41 41 48 40 170 Total 50 70 80 50 250 Expected E=rowtotal*column total/grand total under 30 30-39 40-49 50 and over Total promoted 16.000 22.400 25.600 16.000 80 not promoted 34.000 47.600 54.400 34.000 170 Total 50 70 80 50 250 chi square =(O-E)^2/E under 30 30-39 40-49 50 and over Total promoted 3.063 1.9446 1.6000 2.2500 8.857 not promoted 1.441 0.915 0.753 1.059 4.168 Total 4.504 2.860 2.353 3.309 13.025Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.