Business Analysis Paper: Hypothesis Testing and Statistical Analysis

This paper discusses hypothesis testing and statistical analysis for various scenarios including delivery of groceries, performance of cars, mutual funds data, and seasonal deaths. The paper includes calculations of test statistics, critical values, and degrees of freedom, as well as interpretation of results.

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Question one
Data of delivery of groceries by Local Store and Supermarket
No.
Local Store
(X)
Supermarket
( Y)
1 17.3 21.8
2 12.2 15
3 16.1 18.5
4 17.2 15.4
5 18 20.6
6 18.6 19.3
7 14.6 16.8
8 22.3 19.3
9 14.4 16.3
10 21.3 23.8
11 16.5
n((n1∧n2 respectively) 11 10
Mean(( X ∧Y respectively) 17.14 18.68
Variance( sx
2 ∧s y
2 respectively) 8.67 8.22
a. Stating the null and alternate hypotheses to determine if the local store delivers
groceries faster than the supermarket chain.
The hypotheses will be
H0 : μ0=μ1
H1=μ0 ≠ μ1

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b. Calculation of test statistic to test your hypotheses and report your result
Due to an unequal sample size of the two variables, t-statistics (unpaired test), equal
variance, will be conducted.
t= ( X −Y )
√ ( sx
2
n1
+ sy
2
n2 )
w h ere , sx
2∧s y
2 are t h e variance of X∧Y respectively , X ∧Y are t h e means of X
Y are t h e means of X∧Y respectively , n1∧n2 are t h e sample ¿ X a nd Y
Note the degree of freedom in this case will be given by df =n1 +n2−2=11+10−2=19
From the table above
t= ( 17.14−18.68 )
√ ( 8.67
11 + 8.22
10 ) = −1.54
√ 1.6096
¿− 1.54
1.2687 =−1.2167
Therefore t-computed = -1.2167

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c. Specifying and justifying an appropriate probability of committing a Type I error
(α).
To make the decision critical value of t at 95% significance level from t-table is needed.
This will set the probability of incorrectly rejecting the null hypothesis at 5%, which is
the chance of committing type I error. This is reasonable level at which the end results of
the statistical test are reasonable.
For the above case, the critical value of t at 95% significance level will be,
t0.05 (19 df ) ( twotailed )=¿ 1.729
d. Decision and clear explanation of the result.
Since, t-computed |1.2167| < critical value 1.729, null hypothesis H0 is accepted. This
shows that the equivalence of population mean. This suggest that there’s no difference
in the delivery of groceries between the local store and the supermarket chain.
Thus the local store does not deliver groceries faster than supermarket chain. In
conclusion, the two businesses deliver groceries at the same pace.

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Question Two
Regular(I/100km)
(X)
Premium (I/100km)
(Y)
D= X−Y D2
1 18 15 3 9
2 14 13 1 1
3 13.5 12 1.5 2.25
4 13 12 1 1
5 12.5 11.5 1 1
6 13 11.5 1.5 2.25
7 10.5 11 -0.5 0.25
8 11.5 11 0.5 0.25
9 10.5 10 0.5 0.25
10 10 9 1 1
11 12.5 11.5 1 1
n 11 11
mea
n 12.63636 11.59091
Sum of D and D^2 respectively 11.5 19.25
a. Stating the hypotheses to determine if there is a significant difference in the
performance of cars using Premium versus regular petrol
The hypotheses are
H0 : μ0=μ1
H1=μ0 ≠ μ1
b. Computation of test statistic to test the hypotheses
Here the t statistic (paired test) will be applied

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t= ∑ D
√ n ∑ D2− ( ∑ D )
2
n−1
w h ere t h e D=difference between X∧Y , n=sample ¿ 11
In this case, the degree of freedom will be given by n−1=11−1=10
From the table above
t= 11.5
√ 11 ( 19.25 )−11.52
11−1
= 11.5
√ 211.75−132.25
10
¿ 11.5
√7.95 = 11.5
2.8196
t=4.0786
Hence the tcomputed is 4.0786
c. Specification and justification of an appropriate probability for committing
Type I error (α)
The probability of committing type I error will be 5%, which is probability of
incorrectly rejecting null hypothesis. This will be attained at the significance level of
95%, which is the probability of rejecting or accepting the null hypothesis accurately.
To make the decision, in this case, the critical value of need to determine from
t-tables .

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t0.05 ( 10 df ) ( twotailed ) =1.812
d. Reporting of the decision and clearly explain your result.
The tcomputed=4.0786 ¿ tα=1.812 , therefore, null hypothesis,H0 t h e : μ0=μ1 will be
rejected. This suggest that there is significant difference in performance of cars
using Premium and Regular petrol. In this case, alternative hythe pothesis
H1=μ0 ≠ μ1 is accepted. In conclusion the performance, of the two model of cars
is not the same, the t-test conducted reveals this.

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Question Three
Mutual funds Data:
No. Short term
( Group 1)
Long-term
( Group 2)
World Bond
( Group 3)
X j d= ( X j−X j )d^2 X d= ( X j−X j )d^2 X d= ( X j−X j ) d^2
1 2.34 0.6225 0.3875 3.98 1.218 1.4835 1.43 -0.0917 0.008403
2 1.47 -0.2475 0.0613 4.25 1.488 2.2141 0.52 -1.0017 1.003336
3 1.59 -0.1275 0.0163 2.78 0.018 0.0003 1.41 -0.1117 0.012469
4 1.47 -0.2475 0.0613 1.51 -1.252 1.5675 0.98 -0.5417 0.293403
5 1.29 -1.472 2.1668 4.48 2.9583 8.751736
6 0.31 -1.2117 1.468136
n 4 5 6
Mean( X j) 1.7175 2.762 1.5217
Sum of Squares 0.5263 7.4323 11.5375
a. Stating the null and alternate hypotheses to determine if there is a significant
difference between the different mutual funds.
The hypotheses will be:
H0 : μ0=μ1 =μ2
H0 : μ0 ≠ μ1∨μ0 ≠ μ2∨μ1=μ2
The concern will be whether there are differences among the means of three funds

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b. Calculation of test statistic to test your hypotheses and report your result
For the case above , one-way ANOVA is required, where F-statistic will be determined,
because the sample has more than two groups, has three. This involves comparison of
the between the groups and within the groups.
F= Between Group Variability
Wit h∈Group Variability
Within-group Sum of Square ( SSW )
SSW =∑
j=1
p
∑
i=1
n
( Xi , j −X j ) 2
w h ere , Xi , j are t h e observations∈groups∧X j average of group J
Therefore, SSW =0.5263+7.4323+11.5375=19.496
Here Sum of Squares (SSb) will be computed
SSb =n∑
j=1
p
( X j−X )2
w h ere X j is average group J ∧X is Grand average
Grand Average ( X )=1.7175+ 2.762+1.5217
3 =2.0004
Therefore,
SSb =4 ( 1.7175−2.0004 )2 +5 ( 2.762−2.0004 )2+ 6 ( 1.5217−2.0004 )2
¿ 4.5954

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From the above values, F-statistic is computed as follows
F=
SSb
Df b
SSW
Df W
= Mean Squarebetween
Mean Squarewit h ∈¿ , ¿
Df b=degree of freedom between groups=k −1, w h ereis k t h e number of groups=3
Df b=3−1=2
Df W =degree of freedom between groups=N−k ,
w h ere N is t h e total number of observations among t h e groups=4+5+6=15
Df W =15−3=12
Mean Squarebetween = SSb
Df b
= 4.5954
2 =2.2977
Mean Squarewit h ∈¿= 19.496
12 =1.6247¿
Therefore,
Fcomputed= Mean Squarebetween
Mean Squarewithin
=2.2977
1.6247
¿ 1.4143
c. The probability of committing type I error

Business Analysis paper
To make the decision critical value of F at 95% significance level from F-table is needed.
This will set the probability of incorrectly rejecting the null hypothesis at 5%, which is
the chance of committing type I error. This is reasonable level at which the end results of
the statistical test are reasonable.
For the above case, the critical value of F at 95% significance level will be,
Fα ( 3 ,12 df )=3.89
d. Decision
Since FComputed=1.4143 ¿ Fα =3.89 , the null hypothesis H0 : μ0=μ1 =μ2 will be
accepted. This shows that the equivalence of population mean. This suggest that there’s
no there is a significant difference between the different mutual funds Thus, the returns
among the three funds are likely to be equal, because there’s no difference among
there population means. For this reason, the investors can choose any of the three funds
as they have same level of viability.
Question Four

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a.
Season Spring Summer Autumn Winter
Proportion of
deaths(Observed)
78 71 87 86
i. Stating the hypotheses to test if there is a difference in population
The hypotheses will be:
H0: There’s no difference between in population
H1: There’s difference between in population
ii. Computation of the test statistic to measure the discrepancies between
the observed and the expected result
Here Chi-square statistic will be adopted,
χ2=∑ ( Observed−Expected )2
Expected
No. Observed
Results
(O)
Expected
Results (E)
O−E ( O−E )2 ( O−E ) 2
E
1 78 81 -3 9 0.1111
2 71 81 -10 100 1.2346
3 87 81 6 36 0.4444
4 86 81 5 25 0.3086
Sum(Chi-Square Values) 2.0988

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To make the decision the critical value of Chi-Square need to be determined at
the significance probability level of 5% at the degree of freedom (df ) of n−1 ,
w h ere nis t h e number of observations made=4 , df =4−1=3
χ2
α =0.05 ,df =3=7.81
The critical value of Chi-Square is 7.81
iii. Interpretation of the results
Computed Chi-Square, 2.0988 ¿ χ2
0.05 , 7.81, therefore, the null hypothesis is
accepted. This suggests that expected value of population of 81 among the
four seasons satisfies our prediction that there’s no difference in the
population among the four population. This also signifies that any
significance probability level less than 5% cannot allow us to reject the null
hypothesis.

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b. Closing Prices of Bitcoin for the first ten hours on 13Th October 2017
i. Using a three-point simple moving average to estimate the closing price for
13/10/2017 11:00.
i.
The closing price of Bitcoin is $ 5, 663.24
No
.
Date &Time Closing Price ($) Moving Average
Ft= ∑ Last three Term
3
1 13/10/2017 01:00 5512.59
2 13/10/2017 02:00 5713.08
3 13/10/2017 03:00 5729.3
4 13/10/2017 04:00 5739.95
5651.6567
5 13/10/2017 05:00 5525.25
5727.4433
6 13/10/2017 06:00 5545.39
5664.8333
7 13/10/2017 07:00 5625.61
5603.53
8 13/10/2017 08:00 5648.32
5565.4167
9 13/10/2017 09:00 5715.02
5606.44
10 13/10/2017 10:00 5626.39
5662.9833
11 13/10/2017 11:00
5663.2433

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ii. Using a three-point weighted moving average to estimate the closing price for
13/10/2017 11:00 (use weightings of 4.0 for the most recent year, 3.0 for
the next year, and 2.0 for the last year).
From excel
The closing price of
Bitcoin is $ 5,665.68
iii. Using
Exponential
Smoothing
to estimate the
closing price
for
13/10/2017
11:00.
Use a
smoothing
constant (α)
of 0.4, assume a
forecasted
No
.
Date
&Time
Closin
g Price
($)
weigh
t
Weighted Average
Ft=
W 1 At−1 +W 2 At−2 +W 3 At −3
∑ of weights (w)
1 13/10/201
7 01:00
5512.5
9
4
2 13/10/201
7 02:00
5713.0
8
3
3 13/10/201
7 03:00
5729.3 2
4 13/10/201
7 04:00
5739.9
5 5627.577778
5 13/10/201
7 05:00
5525.2
5 5724.457778
6 13/10/201
7 06:00
5545.3
9 5687.505556
7 13/10/201
7 07:00
5625.6
1 5625.147778
8 13/10/201
7 08:00
5648.3
2 5554.265556
9 13/10/201
7 09:00
5715.0
2 5595.003333
10 13/10/201
7 10:00
5626.3
9 5653.048889
11 13/10/201
7 11:00 5665.68
Sum of weights 9

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value of $5,500.00 for
13/10/2017 01:00.
The next Forecasted value is given by
Ft+ 1=α At + ( 1−α ) Ft , where At is the actual value , Ft is the forecasted value
Date &Time Actual Closing Price
(At)
Forecasted closing price
Ft+ 1=α At + ( 1−α ) Ft
¿0.4 At + ( 0.6 ) Ft
13/10/2017 01:00 5512.59 $5,500
13/10/2017 02:00 5713.08 $5,505.04
13/10/2017 03:00 5729.3 $5,588.25
13/10/2017 04:00 5739.95 $5,644.67
13/10/2017 05:00 5525.25 $5,682.78
13/10/2017 06:00 5545.39 $5,619.77
13/10/2017 07:00 5625.61 $5,590.02
13/10/2017 08:00 5648.32 $5,604.25
13/10/2017 09:00 5715.02 $5,621.88
13/10/2017 10:00 5626.39 $5,659.14
13/10/2017 11:00 $5,646.04
α 0.4
1-α 0.6

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The closing price of Bitcoin is $ 5, 646.04
iv. Interpretation of the results
The three forecasted closing value of Bitcoin, predicted using the 3 methods are $ 5,
663.24, $ 5, 665.68, and $ 5, 646.04 respectively. This suggests that the expected
value of Bitcoin at 11.00 lie within the range of
$ 5 ,646.04 ≤ X ≤ $ 5 ,665.68
From the same result, it's revealed that the price of Bitcoin will increase from the
price value (at 10.00), this is supported by the fact that the three forecasted values
are above the value at 10.00 ($5626.39).

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Question Five
a. Data of monthly average prices (in US dollars) of a barrel of Brent Crude oil and
an ounce of gold:
Column1
Gold(X) Crude
Oil(Y)
X^2 Y^2 XY
1 1267 50 1605289 2500 63350
2 1238 46 1532644 2116 56948
3 1157 54 1338649 2916 62478
4 1192 55 1420864 3025 65560
5 1234 55 1522756 3025 67870
6 1231 52 1515361 2704 64012
7 1267 53 1605289 2809 67151
8 1246 51 1552516 2601 63546
9 1260 47 1587600 2209 59220
10 1237 49 1530169 2401 60613
11 1283 51 1646089 2601 65433
12 1314 55 1726596 3025 72270
n 12 12
sum ∑ X=14926∑ Y =618 ∑ X2=18583822∑ Y 2=31932
∑ X Y =768451
Mean( X ∧Y ) 1243.8 51.5
Standard Deviation 40.8608 3.0896

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(s¿¿ x∧s y)¿
Using a Simple Regression Model to compute the following for the above data:
i. The correlation coefficient (r)
It is given by the formula,
r xy= n ∑ XY −∑ X ∑ Y
√ [ n ∑ X 2− (∑ X )2
] [ n∑ Y 2− (∑ Y )2
]
r xy= 12 ( 768451 )− ( 14926 ) ( 618 )
√ [ 12 ( 18583822 )− (14926 )2 ] [ 12 ( 31932 )−6182 ]
¿ 9221412−9224268
√ ( 223005864−222785476 ) ( 383184−381924 )
¿ −2856
√ ( 220388 ) ( 1260 ) = −2856
√ 277688880 = −2856
16663.99951
¿−0.17138742692
Therefore, r ≈−0.1714( 4 d . p)
ii. The slope of a line of best fit for this data
From the linear regression function, y=a+bx ,
w h ere a is y −intercept w h ile b ist h e sklope of t h e regression line

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The slope of the line of best fit will be given by
b=r s y
sx
,
w h ℜ , sx∧sy are standard deviations of x∧ y respectively ,
r is correlationis coefficient
From the table above
X =1243.833 ,Y =51.5 , sx=40.8608∧sy =3.0896
Therefore, b=r s y
sx
¿−0.1714 ( 3.0896
40.8608 )=−0.1714∗¿0.075612
¿−0.01296
Hence b ≈−0.013
iii. The Y-intercept,a
a=Y −b X
¿ 51.5 —−0.013 ( 1243.833 ) =51.5+16.12
¿ 67.62
Thus y-intercept is 67.62

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iv. The value of Y (crude oil)) given a value of X = $52.50
From the ii and iii above the linear regression will be given by,
y=67.62−0.013 x
Therefore, the value of Y (crude oil) atX =$ 52.5, will be,
Y =67.62−0.013 ( 52.5 )=67.62−0.6825
¿ 66.9375
Hence the value of crude oil at X =$ 52.5 is $66.94
b. The simple linear regression equation for these data is y = 0.0021x + 22.069 and the
correlation
coefficient is r = 0.80
Description of the relation between the two variables

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The two variables are linearly related since the correlation of coefficient (0.8) is greater
than 0. At the same time, since the correlation coefficient is closer to 1 than to 0, two
variables are said to be highly associated. Moreover, since the correlation coefficient
(0.8) and the gradient of the linear regression (0.0021) are positive, the two variables are
said to have a positive linear relationship. Finally, the positive gradient (0.0021)
indicates that the line of two variables has a positive slope, thus left-right upward slope
on the chart.
In conclusion the two variables, GDP per Capita and Social media usage have a high
positive linear relationship due to their postive correlation coefficient (0.80), which is
close to one, and positive slope (0.0021). This suggests that two variables are directly
related, thus a change in one, leads to change in the other as demonstrated by the left-
right upward slope of the linear chart in the question.

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