Question

Question one

A researcher in a large supermarket wishes to study sickness absences among its employees. The
organisation has branches in all the provinces, each branch keeps full records of sickness leave. A random sample of ten such branches produced the following data showing the number of days
of sickness per branch in the year 2017.
18 23 26 30 32 35 39 45 48 54
Required:
a) Using the above data

a). Calculate (manually and using the computer software such as EXCEL, SPSS etc), a
95% confidence interval for the mean amount of sickness days per branch.

b). Estimate the number of branches that should be included in a simple random sample so that a 95% confidence interval for the mean number of days sickness should not have a width greater than 4 days 5 marks
c)) After the sample was collected, it became apparent that the branches fell into three natural groups in terms of sales-small, medium and large. From the data on all of the branches in the provinces, the researcher found that of 210 randomly selected staff, 90 worked in small branches, 36 in medium sized branches, and the rest worked in large branches. In total, 96 of the selected staff had no days off for sickness, of which 52 worked in small branches, and 29 worked in large sized branches.
i) Form a table showing the information clearly. 5 marks
ii) Carry out an appropriate statistical test to investigate whether the size of branch influences the occurrence of sickness absence, interpret your results clearly.

Answer 1

a). Calculate (manually and using the computer software such as EXCEL, SPSS etc), a
95% confidence interval for the mean amount of sickness days per branch.

The required confidence interval for the population mean by using excel is given as below:

Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation	11.5181017
Sample Mean	35
Sample Size	10
Confidence Level	95%
Intermediate Calculations
Standard Error of the Mean	3.642343568
Degrees of Freedom	9
t Value	2.2622
Interval Half Width	8.2396
Confidence Interval
Interval Lower Limit	26.76
Interval Upper Limit	43.24

b). Estimate the number of branches that should be included in a simple random sample so that a 95% confidence interval for the mean number of days sickness should not have a width greater than 4 days

WE are given

Margin of error = 4

Estimate for standard deviation = σ = 11.5181017

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Sample size formula is given as below:

n = (Z*σ/E)^2

n = (1.96*11.5181017/4)^2

n = 31.85327

Required sample size = 32

c)) After the sample was collected, it became apparent that the branches fell into three natural groups in terms of sales-small, medium and large. From the data on all of the branches in the provinces, the researcher found that of 210 randomly selected staff, 90 worked in small branches, 36 in medium sized branches, and the rest worked in large branches. In total, 96 of the selected staff had no days off for sickness, of which 52 worked in small branches, and 29 worked in large sized branches.
i) Form a table showing the information clearly.

The required table is given as below:

	Sale
	Small	Medium	Large	Total
No days off	52	15	29	96
Off	38	21	55	114
Total	90	36	84	210

ii) Carry out an appropriate statistical test to investigate whether the size of branch influences the occurrence of sickness absence, interpret your results clearly.

Here, we have to use Chi square test for independence.

Null hypothesis: H₀: The size of branch not influences the occurrence of sickness absence.

Alternative hypothesis: H_a: The size of branch influences the occurrence of sickness absence.

We assume level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 1*2 = 2

α = 0.05

Critical value = 5.991465

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	Small	Medium	Large	Total
No off	52	15	29	96
off	38	21	55	114
Total	90	36	84	210

Expected Frequencies
	Column variable
Row variable	Small	Medium	Large	Total
No off	41.14286	16.45714	38.4	96
off	48.85714	19.54286	45.6	114
Total	90	36	84	210

Calculations
(O - E)
10.85714	-1.45714	-9.4
-10.8571	1.457143	9.4

(O - E)^2/E
2.865079	0.129018	2.301042
2.412698	0.108647	1.937719

Chi square = ∑[(O – E)^2/E] = 9.754203

P-value = 0.007619

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the size of branch influences the occurrence of sickness absence.