Question

Question:

The survey is conducted to calculate average number of sick days.The employees are categorised into nurses, doctors and administrators. Stratified random sampling, is used with each group forming a separate strata in order to conduct survey. The no. of employees in the strata are given in table below:

Table-1:

Number of employees in each category

	Doctors	Nurses	Administrators
No. of employees	120	80	10

Number of Sick days for 9 doctors, 7 nurses and 4 administrators

Table-2:

Group	No of Sick Days									Mean	S²
Drs	8	7	10	12	19	6	13	12	9	10.67	15.5
Nurses	4	8	3	12	7	10	15			8.43	18.29
Adm	5	12	10	11						9.5	9.67

a.  Calculate the sample size from each strata if proportional allocation is done for selecting total 25 employees from table-1

b.  Using the data from table-2 calculate the mean number of sick days for that year.

c.  Place a 95% confidence bound on the mean.

d. Calculate the total number of sick days for the year.

e.  Place a bound on the total number of sick days and interpret this bound.

Answer 1

a)

Population size, N =N_i =120+80+10 =210

Required sample size, n =25

From doctors =(n/N)*120 =(25/210)*120 =14

From nurses =(25/210)*80 =10

From administrators =(25/210)*10 =1

(So, 14+10+1 =25)

b)

For Drs, No. of sick days =(8+7+.....+9) =96

For Nurses, No. of sick days =(4+8+.....15) =59

For Adm, No. of sick days =(5+.... +11) =38

Total number of sick days for the year =96+59+38 =193

Mean number of sick days for the year, =193/(9+7+4) =193/20 =9.65

c)

For 95% confidence level, for a two-tailed case, the critical value of Z is: Z-critical =1.96

Standard deviation, S for the data in table 2: 8,7,..........,10,11 is as follows:

S = where n =20 and =9.65

S =3.8563

Standard Error, SE =S/ =3.8563/ =0.8623

Margin of Error, MoE =Z-critical*SE =1.96*0.8623 =1.69

95% confidence bound on the population mean number of sick days for the year is:

= =9.651.69 =(7.96, 11.34)

d)

Total number of sick days for the year, =N* =210*9.65 =2026.5

e)

Let the confidence level =95%. So, Z-critical =1.96

95% confidence bound on the total number of sick days is:

T =[N(Z-critical)S/] =[N(Z-critical)SE] =2026.5[210*1.96*0.8623] =2026.5354.9 =(1671.6, 2381.4)

We can round it to (1671, 2382).

Interpretation: "If we conduct this survey many number of times and find the 95% confidence bounds, we would expect that about 95% of the bounds obtained contain the true total number of sick days for the year and about 5% of the bounds do not contain the true total number of sick days".

Thus, we are 95% confident that the above bound (1671, 2382) contains the true total number of sick days for the year.