Question

A personnel director claims that the distribution of the reasons withers leave their job is different from the distribution: 41% limited advancement; 25% lack of recognition; 15% low salary/benefits; unhappy with management 10%; 9% bored. You randomly select 200 workers who recently left their jobs and record each worker’s reasons for doing so. The table below show the results. At the 0.01 level of significance, test the personnel director’s claim.

Survey Results
Response	Frequency
Limited advancement	78
Lack of recognition	52
Low salary/benefits	30
Unhappy with management	25
Bored	15

Expected Frequencies for each?

Answer 1

Null hypothesis:

The distribution of the reasons workers leave their job follows the distribution: 41% limited advancement; 25% lack of recognition; 15% low salary/benefits; unhappy with management 10%; 9% bored.

Alternate hypothesis:

The distribution of the reasons workers leave their job is different than the distribution.

Categories	Observed	Expected	(Fo-Fe)2/Fe
Limited advancement,	78	200*0.41=82	(78-82)2/82 = 0.195
Lack of recognition,	52	200*0.25=50	(52-50)2/50 = 0.08
Low salary/benefits,	30	200*0.15=30	(30-30)2/30 = 0
Unhappy with management,	25	200*0.10=20	(25-20)2/20 = 1.25
Bored	15	200*0.09=18	(15-18)2/18 = 0.5
Total =	200	200	2.025

Critical value for test.

The significance level is α=0.01,

Number of degrees of freedom is df = 5 - 1 = 4

The critical value is 13.277.

So, the rejection region for this test is .

Test statistic for test

The Chi-Squared statistic is computed as follows:

= 0.195+0.08+0+1.25+0.5

= 2.025

Conclusion:

Since we observe that , the null hypothesis cannot be rejected as there is not enough evidence to reject the null hypothesis.

Hence, the personnel director’s claim cannot be accepted that the the distribution of the reasons withers leave their job is different from the distribution: 41% limited advancement; 25% lack of recognition; 15% low salary/benefits; unhappy with management 10%; 9% bored.