Mean, Standard Deviation & Confidence Intervals

This assignment delves into the calculation of mean, standard deviation, and confidence intervals. It explains their significance in finance for assessing risk and return profiles of assets. The document includes a dataset with weekly meat consumption data for six individuals, showcasing how to calculate summary statistics. Additionally, it provides examples illustrating the application of these concepts using historical stock returns and discusses the coefficient of variation as a measure of risk-adjusted return.

Computing Assignment
Bus105
Student id
[Pick the date]

Section 1
Data
a) Scatter plot highlighted below:
The variables income and annual contribution is showing linear positive relationship with
strength of moderate to strong.
b) Regression line from scatter plot

y=0.1197+1829.1
y represents: annual contribution
x represent: Income
Annual contribution needs to be determined for income x = $200,000
y = (0.1197*200000) +1829.1 = $25769.1
Therefore, the value annual contribution for given income ($200,000) is $25769.1.
c) The z score needs to be determined.
Given average of the estimates (μ) = $27,000
Standard deviation (σ ) = $2,100
x ( above ) =$ 25769.1
Formula for z score : z={ x−μ
(σ) }
z score=25769.1−27000
2100
Vaue of z score=−0.58614
Hence, z score=−0.58614 −0.59
d) Value of P (Z < Z score) needs to be determined.
Value would be computed by using excel function “NORMSDIST () ”
P (Z < Z score) = P (Z<−0.58614 ¿

¿ NORMSDIST ( −0.58614 )
¿ 0.27889
e) Value of estimated rank would be shown below:
Value of estimated rank = {P (Z < Z score)}*10000
¿ P ( Z ←0.58614 )∗10000
¿ ( 0.27889 ) ∗10000
¿ 2788.9 2789
Section 2
a) The first pivot table indicates high risk investment (riskier) and loss and profit relation.
The respective sample size for riskier type n1=81
Proportion for riskier investment that would create a loss P1= 12
81 =0.1481
The second pivot table indicates low risk investment (safer) and loss and profit relation.

The respective sample size for safer type n2 =19
Proportion for safer investment that would create a loss P2= 3
19 =0.1578
b) Graphical representation for the calculated proportion
r (riskier); 0.15
s (safer); 0.16
Pie Chart
c) From the computations conducted above, it seems that the probability of loss and the
underlying investment risk are correlated in an inverse manner. The sample data clearly hints
at lower proportion of risky investments making loss in comparison with corresponding
proportion for safe investments.
d) (i) The estimated value P1−P2
P1−P2=0.1481−0.1578
P1−P2=−0.010

(ii) The value of average of the estimates ( μ ) = 0.1
Standard deviation (σ ) = 0.0743
x ( above ) =−0.010
z= { x −μ
( σ ) }
z= { −0.010−0.1
0.0743 }
The z score=−1.477
(iii) P( Z<Z score)=?
Value would be computed by using excel function “NORMSDIST () ”
P (Z < Z score) = P (Z< −1.477 ¿
¿ NORMSDIST (−1.477 )
¿ 0.0698
(iv) Value of estimated rank would be shown below:
Value of estimated rank = {P (Z < Z score)}*10000
¿ P ( Z ←1.477 )∗4000
¿ ( 0.0698 )∗4000
¿ 279.311

e) Given significance level = 5%
(i) Hypotheses
Null hypothesis H0 :(P¿¿ 1−P2)=0¿
Alternative hypothesis H1 :(P1−P2 )≠ 0
(ii) The p value with respect to the proportion values and the sample size is computed as
0.963.

(iii) It is apparent from the above output than p value is 0.963 which is higher than level of
significance and hence, insufficient statistical evidence is present to reject null
hypothesis.
(iv) Hence, it may be concluded that no relationship is observed between the investment
making loss and their underlying risk.
Section 3
a) The first pivot table represents low risk investment (which is shown by value N in pivot
table) is highlighted below:
The value of sample ¿ n1 ¿=77
Avg .return of low risk investment (x ¿¿ 1)=0.035 ¿
Standard deviation ( s1 ) =0.003
The second pivot table represents high risk investment (which is shown by value Y in pivot
table) is highlighted below:
The value of sample ¿ n2 ¿=23
Avg .return of low risk investment (x ¿¿ 2)=0.084 ¿

Standard deviation(s2 )=0.097
b) Graphical representation of returns data through box and whisker plot
c) There is a huge difference in risk which is essnetially represented by the stndard deviaiton.
The standard deviaiton seems to be very high for risky investment which is not compensated
with proportionately higher returns. Hence, from an investor perspective, lowe risk
investments would be superior.
d) (i) Estimate for μ1−μ2
μ1−μ2=x1 −x2
μ1−μ2=0.035−0.084=−0.0489
(ii) The value of average of the estimates ( μ ) = −0.0256
Standard deviation (σ ) = 0.0173

x=x1−x2 ( above ) =−0.0489
z= { x −μ
( σ ) }
z= { −0.0489−(−0.0256)
0.0173 }
The z score=−1.348
(iii) P(Z< Z score)=?
Value would be computed by using excel function “NORMSDIST () ”
P (Z < Z score) = P (Z<−1.348 ¿
¿ NORMSDIST ( −1.348 )
¿ 0.0887
(iv) Value of estimated rank would be shown below:
Value of estimated rank = {P (Z < Z score)}*2000
¿ P ( Z ←1.348 )∗2000
¿ ( 0.0887 )∗2000
¿ 177.55
e) (i) Given significance level = 5%
Hypotheses

Null hypothesis H0 :(μ¿¿ 1−μ2)=0 ¿
Alternative hypothesis H1 :(μ1−μ2)≠ 0
(ii) The p value with respect to the proportion values and the sample size is computed as 0.
(iii) It is apparent from the above output than p value is 0, which is lower than level of
significance and hence, sufficient statistical evidence is present to reject null hypothesis and to
accept alternative hypothesis.
(iv) It may be concluded that returns are dependent on the underlying risk associated with the
investment and hence the average returns on both risky and safe investments cannot be assumed
to be same.
Section 4
a) Pivot table to represent the total number of people who said no or yes for the proposed
change.
b) The value of sample size for people who said yes (n) = 114

Proportion for people who said yes ^p= 114
198 =0.5757
c) (i) the z score needs to be determined.
Given average of the estimates (μ) = 0.6
Standard deviation (σ ) = 0.0357
x ( above ) =0.5757
Formula for z score : z={ x−μ
(σ) }
z score= 0.5757−0.6
0.0357
Vaue of z score=−0.679
Hence, z score=−0.679
(ii) Value of P (Z < Z score) needs to be determined.
Value would be computed by using excel function “NORMSDIST () ”
P (Z < Z score) = P (Z<−0.679 ¿
¿ NORMSDIST (−0.679 )
¿ 0.2485
(iii) Value of estimated rank would be shown below:
Value of estimated rank = {P (Z < Z score)}*10000
¿ P ( Z ←0.679 )∗1000

¿ ( 0.2485 )∗1000
¿ 248.55
d) 95% confidence interval
Formula used
For lower limit
For upper limit
The value of t stat for 95% confidence interval =1.96
Lower limit ¿ 0.5757− ( 1.96 )∗
√ 0.5757 ( 1−0.5757 )
198 =0.5068
Upper limit ¿ 0.5757+ (1.96 )∗
√ 0.5757 (1−0.5757 )
198 =0.6445
95% confidence interval = [0.5068 0.6445]
Section 5
a) The dataset is self-made and intends to capture the weekly meat consumption of six
individuals and mentions their respective genders.

b) The summary variables include gender and consumption of meat on a weekly basis.
c) The relevant dataset and summary pivot table is listed below.
Data set
Section 6
In finance, especially related to portfolio theory mean and standard deviation computation has
high degree of relevance and utility. Based on the computation of mean and standard deviation,
the risk return profile of a particular asset can be obtained which then can be compared with
other asset classes to make an optimum choice. Essentially the mean is indicative of the returns
while risk associated with a security is captured using the standard deviation. Typically the
security or asset which offers the least coefficient of variation is chosen as it tends to offer a
given return for the least amount of risk.
This can be demonstrated with the aid of the following example where the risk return
characteristics of three securities have been provided in the tabular format as highlighted below.
Stock Average Return (%) Standard Deviation (%) Coefficient of Variation

X 15 20 1.33
Y 20 60 3
Z 30 50 1.66
From the choices available at hand, the most superior investment seems to be X even enough it
offers the least returns from the available choices. This is on account of X having the lowest
coefficient of variation.
Also, in certain cases, it may be required that standard deviation and mean need to be computed.
One of this is based on the historical data associated with the stock. Since probabilistic data
usage is comparatively less common, hence an example is being taken to illustrate the
computation of mean and standard deviation with the use of historical data.
Consider that returns for the stock over the last three years be 10%, 20%, 30% respectively.
Therefore, average return on the stock = (10+20+30)/3 = 20%
Also derivation of standard deviation can be carried with the aid of the table shown below.
X (X- Mean) (X-mean)^2
0.1 -0.1 0.01
0.2 0 0
0.3 0.1 0.01
TOTAL 0.02
Standard deviation = √(0.02/(3-1)) = 10% pa
Reference: https://www.youtube.com/watch?v=UwO4JvB9OpE