Business Statistics: Analysis of Occupation, Age Group and Income

This article provides a detailed analysis of occupation, age group and income using descriptive statistics, frequency charts, and confidence intervals. The article covers topics such as proportion of people based on age group, maximum frequency of occupation, inter quartile range, and point estimate of mean for income.

BUSINESS STATISTICS
BB108
Student Name
[Pick the date]

Task 1
Frequency column chart to represent the proportion of people based on the age group.
Relative frequency pie chart to represent the proportion of people based on the age group.
Occupation frequency column
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Transport Frequency table
(a) Number of LGA used public transport = 22
(b) Maximum frequency of occupation = 4 (Education)
(c) Proportion of people with age group of 30-34 years = 0.18
Task 2
(a) Sorted data of occupation 4 (Education) V15
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(i) 70th percentile
(ii) First and third quartile
 First quartile Q1
P = 25th percentile
First quartile Q1
 Third quartile Q3
P = 75th percentile
Third quartile Q3 = 17
(b) The 70th percentile implies 70% of the sample values of the given variable would have a value
equal to or less than 17 (Hair et. al., 2015).
(c) Inter quartile range
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Inter quartile range = Third quartile – First quartile = 17 -0 = 17
This refers that 50% of the sample values would lie within a range of 17 administrators.
Task 3
(a) Descriptive statistics of variable occupation 4
(b) Upper and lower inner fence limits
And
(c) On the basis of the above limits, it is apparent that the given variable would not contain any
outlier as the maximum value is 25 which is lower than the upper limit of 42.5. Also, the lowest
value of the variable as apparent from the descriptive statistics is 0 which is greater than -25.5.
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In such a scenario, the appropriate measure of central tendency would be mean while that of
dispersion would be standard deviation. This is because the distorting effects of outliers would
be absent here (Eriksson & Kovalainen, 2015).
(d) The mode value is 0.00. The mean is 8.5 and marginally greater than median amount of 6.5
which is also indicative of presence of positive skew. Owing to the presence of skew, the given
distribution cannot be considered as normal. Also, with regards to dispersion, it is apparent that
it is quite high considering the value of standard deviation almost equals mean (Flick, 2015).
Task 4
(a) Descriptive statistics of variable occupation 4
The given variable is not normally distributed because of the following reasons (Hillier, 2016).
 Presence of skew which ought to be zero
 Non-convergence of the central tendency measures
 The values lying between +/- 1 standard deviation of the mean exceed 68%.
(b) The z value for 1.5 is 0.4322 (Fromm standard normal table).
Therefore,
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It can be seen that 86.64% of the population would lie between the ranges -1.5 to +1.5. Therefore, it can
be concluded that 43 of the people from occupation 4 would fall in this range.
(c) The upper and lower limit based on mean and standard deviation is calculated below:
Lower limit = Mean – (1.5 * Standard deviation)
Lower limit = 8.50 – (1.5 * 8.51) = -4.257
Similarly,
Upper limit = Mean + (1.5 * Standard deviation)
Upper limit = 8.50 + (1.5 * 8.51) = 21.257
The lower and upper bound comes out to be -4.257 and 21.257. It can be said that nearly 43 people
from occupational 4 comes out in this range. Therefore, this is evidence of the correctness of the result
of part (b). But this result does not confirm with the conclusion drawn in part (a).
Task 5
(a) Descriptive statistics of variable occupation 4
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(i) Point estimate of the mean for occupation 4 is comes out to be 8.50.
(ii) The 90% confidence interval for occupation 4 is calculated below:
Lower limit of 90% confidence interval = Mean – Confidence level
Upper limit 90% confidence interval = Mean + Confidence level
Hence, the 90% confidence interval for occupation 4 is [6.48 10.52].
(iii) We can conclude with 90% confidence that population mean of occupation 4 would lies between
6.48 and 10.52 (Hair et. al., 2015).
(b) If the population mean is 59, then the interval estimate would not be considered as appropriate as
this lies outside the estimated interval for population mean.
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Task 6
(a) Descriptive statistics for variable V9 (Income 1 i.e. $650-$800 weekly income)
(j) Point estimate of the mean for variable V9 (Income 1 i.e. $650-$800 weekly income 57.38.
(ii) The 99% confidence interval for Income 1is calculated below:
Lower limit of 99% confidence interval = Mean – Confidence level
Upper limit 99% confidence interval = Mean + Confidence level
Hence, the 99% confidence interval for Income 1 $650-$ 800 income earners are [36.08 78.68].
(b) 95% confidence interval for $650-$ 800 income earners based on empirical rule.
Confidence interval = (Sample Statistic) + {(Critical z or t) * (Standard Error of the Sample Statistic)}
The population standard deviation is known and this z value will be used.
Hence,
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The z value for 95% confidence interval = 1.96
Through empirical rule
For 95% confidence interval
μ ± 2σ = 57.38 ± (2*7.95)
Hence,
(c) The 99% confidence interval is more precise and hence the interval is wider in comparison to the
95% confidence interval (Flick, 2015).
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References
Eriksson, P. & Kovalainen, A. (2015) Quantitative methods in business research 3rd ed. London: Sage
Publications.
Flick, U. (2015) Introducing research methodology: A beginner's guide to doing a research project. 4th
ed. New York: Sage Publications.
Hair, J. F., Wolfinbarger, M., Money, A. H., Samouel, P., & Page, M. J. (2015) Essentials of business
research methods. 2nd ed. New York: Routledge.
Hillier, F. (2016) Introduction to Operations Research 6th ed. New York: McGraw Hill Publications.
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