Descriptive Statistics for Business Data

This assignment focuses on applying descriptive statistical techniques to various business datasets. It covers determining data types and measurement levels for attributes like gender, GPA, working hours, and family size. Students calculate measures of central tendency (mean, median, mode) and standard deviation, identify outliers using the empirical rule, and construct 95% confidence intervals for sample means. Additionally, they analyze probability scenarios related to alternating failures and interpret the meaning of confidence intervals in a business context.

STATISTICS FOR BUSINESS
Assignment
STUDENT ID
[Pick the date]

Question 1
1. Gender
Data type: Categorical
Measurement level: Nominal
This is because gender type male and female is measured at nominal scale.
2. Approximate undergraduate college GPA
Data type: Continuous
Measurement level: Ratio
This is because undergraduate college GPA such as 1.0 to 4.0 would be a mathematical operation
and hence, ratio is the measurement level.
3. Number of hours per week
Data type: Continuous
Measurement level: Ratio
This is because number of hours per week would be a mathematical operation and hence, ratio is
the measurement level.
4. Ideal number of children for a married couple
Data type: Discrete
Measurement level: Ratio

This is because number of children would be a mathematical operation and hence, ratio is the
measurement level.
5. Describe parents
Data type: Categorical
Measurement level: Ordinal
This is because the given statistics shows the order 1 for the respective month which dominate up
to order 5 for father and hence, the ordinal is the measurement level.
Question 2
Given data
(a) Computation of mean, median and mode
These values have been computed with the help of excel functions i.e.
Mean = AVERAGE ()
Median = MEDIAN ()

Mode = MODE ()
(b) It can be seen that measures of central tendency i.e. mean, median and mode are not
equal and hence, the measure of central tendency is not agreed.
Mean ≠ Median≠ Mode
(c) Standard deviation
This has been computed with the help of excel functions STDEV ().
(d) Sort and standardized the data
x=724.67
s=114.28
For each of the given value, it is essential to find the standardized score as highlighted below:
zi= ( xi−x
s )

(e) Empirical rule suggests that unusual observation are those which are beyond x ± 2 s and
the outliers are those which are beyond x ± 3 s .
Unusual observations:
x ± 2 s=724.67 ± ( 2∗114.3 ) =496.1 , 953.3

The observation 1030 fall outsides the range (496.1953.3)and hence, it is termed as unusual
observation.
Outliers:
x ± 3 s=724.67 ± ( 3∗114.3 )=381.8 1067.6
There are no outliers present because based on empirical rule none of the value lies outside the
above highlighted range.
(f) Based on the three sigma empirical rule, it can be said that the data is from normal
population because nearly 99.7% of the data is fall within the range of three standard
deviation of the mean.
x ± 3 s=724.67 ± ( 3∗114.3 )=381.8 1067.6 (All values fall under this range)
Question 3
(a) Computation of mean, median and mode
For quiz 1
Quiz 1: 60, 60, 60, 60, 71, 73, 74, 75, 88, 99.
Mean= 60+60+60+60+71+73+74 +75+88+ 99
10 =720
10 =72
Median= ( 71+73 )
2 =72
Mode=60
For quiz 2

Quiz 2: 65, 65, 65, 65, 70, 74, 79, 79, 79, 79
Mean= 65+65+65+65+70+ 74+79+79+79+79
10 = 720
10 =72
Median= ( 70+74 )
2 =72
Mode=65∧79
For quiz 3
Quiz 3: 66, 67, 70, 71, 72, 72, 74, 74, 95, 99
Mean= 66+67+70+71+72+72+74+ 74+95+ 99
10 =760
10 =76
Median= ( 72+72 )
2 =72
Mode=72∧74
For quiz 4
Quiz 4: 10, 49, 70, 80, 85, 88, 90, 93, 97, 98
Mean=10+ 49+70+80+ 85+88+ 90+93+97+98
10 = 760
10 =76
Median= ( 85+88 )
2 =86.50
Mode=¿ There is no mode for the data of quiz 4.

(b) It is apparent that measures of central tendency i.e. mean, median and mode are not same
and hence, the measures of central tendency are not agreed.
(c) The value of mean, median has been computed for all the quizzes. However, quiz four
does not have mode.
(d) For quiz 1 – Data is skewed and direction is towards the left.
For quiz 2 – Data is roughly symmetric.
For quiz 3 – Data is skewed and direction is towards the right.
For quiz 4 – Data is skewed and direction is towards the left.
(e) For quiz 1 – The avg. performance is 72.
For quiz 2 - The avg. performance is 72.
For quiz 3 - The avg. performance is 76.
For quiz 4 - The avg. performance is 76.
The average performance of students in quiz 1 and quiz 2 are lower than the performance in quiz
3 and 4.
Question 4
Probability that in a given alternator will fail on a one hour flight = 0.02
(a) Probability that both will fail
P ( bothwill fail )=¿

P ( bothwill fail )=0.0004
(b) Probability that neither will fail
P ( neither will fail ) =¿
P ( neither will fail ) =(0.982)
P ( neither will fail ) =0.9604
(c) Probability that one or the other will fail
P ( one∨the other will fail ) =1− { P ( neither will fail ) }
¿ 1−(0.9604)
¿ 0.0396
Question 5
Given data

(a) 95% confidence interval
Mean = 3,279
Standard deviation = 97.05
Sample size = 18
As per central limit theorem, the sample size is lower than 30 and hence, t value would be used.
Degree of freedom = Sample size -1 = 18-1 = 17
The value of t stat for 95% confidence interval and 17 degree of freedom = 2.1098
Now,
95 % confidence interval=Mean ±t stat ¿
Upper limit of 95% confidence interval = 3279+2.1098 ( 97.05
√18 )=3326.98

Lower limit of 95% confidence interval = 3279−2.1098 ( 97.05
√18 )=3230.45
95% confidence interval [3230.45 3326.98]
(b) Sample size =?
In order to obtained an error of ± 20 steps with the 95% confidence.
Now,
Margin of error E=t stat ¿
20=2.1098 ¿
√ Sample ¿ ¿ 204.75
20 ¿
Sample ¿ ¿
Sample ¿ 104.81 105
Therefore, the requisite sample size is 105.
(c) Line chart of the data set

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
2,900
3,000
3,100
3,200
3,300
3,400
3,500
Line Chart
Time
Distance recorded
The above highlighted line chart indicates that number of steps decreases with respect to time.