STOCKHOLM UNIVERSITY
Department of Statistics
Fall 2022
Cover page: Hand-in Assignment 3, Basic Statistics for Economists
3. Econometrics
Assignments’ teacher:
Group (1-10):
Assignments’ Group
(1-15):
Data:
Part 3A: Data 7 (from Assignment 1)
Part 3B: Data 1, Beer production Australia
Note! Always save your own version of the report
Group Members:
Name:
Date of birth:
E-mail
Result after first deadline:
Pass
Fail
Comments:
Results after the second deadline:
Pass
Comments:
Fail
Part A Regression analysis
1.
a) The highest absolute correlation with the dependent variable (the number of items sold) is the
independent variable called “Price difference”, which is 0,884926419. This is shown in a scatter plot
diagram below.
In the scatter plot, the price difference is represented by the x-axis and the number of items sold is
represented by the y-axis. The number of items sold increases when the price difference itself
increases and this result is obtained when the competitor's price is subtracted from the retailer's price.
One can also see that the correlation coefficient is close to 1, which can be an indication that it is a
strong positive linear relationship.
b) The lowest absolute correlation with the dependent variable (the number of items sold) is the
independent variable called “Retailer’s price”, which is -0,079295618. This is shown in a scatter plot
diagram below.
Just as before, in the scatter plot, the retailer’s price is represented by the x-axis and the number of
items sold is represented by the y-axis. In this diagram, one can see that the correlation coefficient is
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close to “0”, which indicates no correlation. This means there is no linear relationship between the
two variables. In other words, the variables do not depend on each other and therefore, the linear
relationship in this scatter plot is almost non-existent.
2. Note that:
The chosen independent variable: Price difference
The chosen dependent variable: Quantity - number of items sold
a) In the simple linear regression model above, the coefficient of determination (R2) = 0,783094767.
This could be an explanation of the variation from the price difference, which in turn means that
approximately 78,31% of the variance in the dependent variable can be explained by the independent
variable. A larger value of R 2 could indicate a better regression model, and 78,3% could be considered
a high value (Newbold et al. 2020, p.439).
It is possible to calculate the R2 value without estimating the regression model. It can be calculated
with the help of the formulas below:
(Formula sheet, p.3)
b) The linear equation used in the regression model:
.
The quantity numbers of items sold will have the intercept coefficient Bo when the Price difference =
0. This in turn means that the quantity - number of items sold will have an average value of 1435,43
when the price difference is 0. Additionally, the regression coefficient B↿ is an indication of the slope
of the regression line. So when the coefficient increases with 1 unit, the quantity - number of items
sold will have an average increase of 69,63 (Newbold et al. 2020, p.421-423).
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c) The testing of the regression coefficient is done through a formal hypothesis test. Note that this test
is a two-sided-test which means that the absolute value can be used in the testing and it also gives a
possibility for the slope (slope = B↿) to vary from 0 in different ways (positive vs negative or both).
Also, the significance level for testing is at a 5% level (alpha = 0,05), divided in two as it is a
two-sided test. That gives a significance level testing of 5%/2 = 2,5% (alpha = 0,025).
Observations: n = 30
Hypotheses: The null hypothesis formulated as:
H૦ : B↿= 0
H↿ : B↿ ≠ 1
(Newbold et al, 2020, p.351-354).
Test Statistic:
b1:69,62950823,
β1*:0,
Sb1= 6,925357,
K:1 (regression),
t(n-K-1): (b1- β1*)/Sb1
Critical Value
T (n - 2,α/2) = t (28, 0,025) = 2,048 (from Table 3 in the formula sheet with alpha 0,025 and 28
degrees of freedom).
T (30-2, α/2 = t (28,0,025) = 2,048
Decision Rule
b1 = 69,62950823, β1* = 0, Sb1= 6,925357, T(n-K-1) = 30-1-1 = 28
If T obs; t(n - K -1; 0,025) is > T critical value; (T (n - 2,α/2) then a rejection of the null hypothesis
can be made.
T obs (28) = 69,62950823 / 6,925357 = 10,0542843 ≈ 10,05 (28, according to table 3 in the formula
sheet)
Conclusion: The null hypothesis can be rejected since the observed value (= 10,05) is higher than the
critical value (= 2,048). One explanation of this is that the “Price Difference” is correlated with the
number of “ quantity - number of items sold” at the level of significance of 5%.
d)
b0: 1435,42731 (from regression model)
b1: 69,62950823 (from regression model)
b1X: 69,62950823*3 = 1644,3158
T(n - 2,α/2) = t(28,0.025) = 2,048
n : 30
K:1
xbar: =()Mean price difference in Excel = 1,24
s2x: =()SVariance price difference in Excel = 3,09
X = Competitor’s price – Retailer's price = 31 SEK – 28 SEK = 3 SEK
= 120180,682/30-1-1= 4292,1672
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= 4292,1672*((1+1/30)+ ((3-1,24)^2)/((30-1)*3,09))))
= (1435,42731+69,62950823*3) ± 2,048√4292,1672*((1+1/30)+
((3-1,24)^2)/((30-1)*3,09)))) = 1644,3158 ±138,6544285
X1 = 1644,3158 + 138,6544285 = 1782,9703
X2 = 1644,3158 – 138,6544285 = 1505,6614
Conclusion: The interval of the values is between 1782,9703 and 1505,6614 which indicates a result
of a 95% prediction level given that x = 3.
e)
Calculation:
b0: 1435,42731 (from regression model)
b1: 69,62950823 (from regression model)
b1X: 69,62950823*3 = 1644,3158
T(n - 2,α/2) = t(28,0.025) = 2,048
n : 30
K:1
xbar: =()Mean price difference in excel = 1,24
s2x: =()SVariance price difference in excel = 3,09
X = Competitor’s price – Retailer's price = 31 SEK – 28 SEK = 3 SEK
= 120180,682/(30-1-1)= 4292,1672
= 4292,1672 ( (1/30)+ ((3-1,24)^2)/((30-1)*3,09) = 291,07269
= (1435,42731+69,62950823*3) ± 2,048√4292,1672*((1/30)+
((3-1,24)^2)/((30-1)*3,09)))) = 1644,3158 ± 34,9627872
- X1 = 1644,3158 + 34,9627872 = 1679,2786
- X2 = 1644,3158 - 34,9627872 = 1609,3530
Conclusion: The interval of the values between 1679,2786 and 1609,3530 indicates a result of a 95%
confidence level given that x = 3.
3 a) The three variables in the data set that are linear combinations of each other are: Competitor’s
price, retailer’s price and price difference. The price difference is computed through: Competitor’s
price minus the retailer’s price.
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b) The three variables above should not be used as independent variables in the same model because
of a multicollinearity that could arise, which occurs when variables like these are correlated.
Therefore, a model containing variables that are closely correlated to one another would be
misleading as it could weaken the significance of the statistics and give rise to incorrect calculations.
There is also a chance that the standard errors could be of a higher level, due to the multicollinearity
that could arise when using datasets who are linear combinations of each other (Newbold et al. 2020,
p. 578-580)
4. For this question, the same dataset as previously is used. The dependent variable is the number of
items sold and the three independent variables are retailers prices, price difference and advertising
costs.
Fig 4.1
a) i) The coefficient of determination (R2) is approximately 0,8477 which in other words means that
84.77% of the dependent variable can be explained by the chosen independent variables. In addition,
the high coefficient of determination value shows that the model is a good fit for the data.
ii) The coefficient of determination of the multiple regression model (R2 = 0.8477) is higher than that
for the linear regression model (R2 = 0.7831). Compared to question 2b, there was an R2 of 78.31%
which was lower. The higher R2 in the second model could be explained by the fact that there are more
independent variables and thus more data to explain the dependent variable, and without analyzing the
data, this should be expected. Nevertheless, more variables are not necessarily better. There could be a
situation in which more variables means a lower coefficient of determination where the independent
variable does not contribute to forecasting the dependent variable.
iii) An alternative to the coefficient of determination, is the adjusted R Squared (
) which in this
case is a similar measure but adjusted for adding more variables and more data to the dataset. It aims
to eliminate false correlations between the variables so that it will take into account the true
correlation. It is therefore fitting in this case for multiple regression where there are multiple
variables.
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b) The coefficient for the retailer's price is 46.13. This indicates that every 1 unit increase in
retailer’s price will lead to 46.13 units increase in the sold items if both advertising costs and
price difference are held constant.
The coefficient for advertising cost is -0.042. This means that every 1 unit increase in
advertising costs will lead to a 0.04 units decrease in the sold items if both retailer’s price and
price difference are held constant.
The coefficient for price difference is 91.87. This indicates that every 1 unit increase in the
price difference will lead to 91.87 units increase in the sold items if both advertising costs
and retailer’s price are held constant.
The intercept value is equal to 420.68 and this means that the predicted value of items sold is
420.68 if all independent variables within this model are equal to zero. This means that when
advertising cost as well as retailer’s price and price difference equal zero.
c) Yes, the 95% confidence interval for the advertising costs coefficient contains the value
zero as the values are -0.099 and 0.016. This means that were the experiment to be run again,
there would be a high chance of finding no correlation in the data. Which, in other words,
implies that the actual coefficient value can be zero, meaning that the predictor has no
relationship with the dependent variable.
d) Fig 4.2
Number of Items sold:
(28*46.12989498)+(3*91.87085393)+(2500*-0.041838779)+420.6837777=1883.33645143
=1883.34
Given the assumed values for the independent variables, the output for the dependent variable is
approximately 1883. The number of items sold will therefore be 1883.
5) A company is most interested in earnings thus if one wishes to analyze a company from a
business analytical perspective this is what should be the central focus. The figure which is to
be examined in the model is items sold since there is a relationship between the number of
items sold and earnings. However, one ought to analyze earnings, as it does not matter how
many items are sold when it does not impact the business's earnings. Thus sales and earnings
would be of most interest. Earnings are the product of the number of sold items and the price
which the items are sold for adjusted for the acquisition cost and overheads. So the core of
the analysis should be the relationship between these items. The model as of now is focusing
on sold items as the core. The optimal analysis would plot earnings against sold items, selling
price and related costs. In this case, it is assumed that profitability is the ultimate goal. If one
breaks down decision-making in strategic parts, there might be subgoals that contribute
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towards profitability, but in themselves do not express profit. The point made is that one
should analyze the variables that contribute to the end goal which is higher earnings.
This can also be expressed as follows : if one considers the dependent variable the end goal,
then one should identify independent variables which contribute to a high R2. In other words,
one wants to use independent variables that contribute well to plotting the dependent one. In a
business situation, this would also have the added benefit of identifying divers for earnings if
one uses that as the dependent variable.
It should also be taken into account that one might not know all the variables which are
important in our analysis. If one were to omit an important value, the relationship between
the others would not correspond well to reality and the outputs of our model would be
misleading (Newbold et al. 2020, p. 575).
If earnings were to be forecasted without using the price of sold items, the relationships, if
one would analyze it, would only take into account sold items and related costs. This could
forecast earnings to some extent if one assumes an increase in sold items by itself leads to
higher earnings and in the same way lower cost means higher earnings. The fact is, however,
that if the price decreased this would change the situation and not guarantee the assumptions
just mentioned. This demonstrates that the model loses accuracy if one omits an important
independent variable and the impact on accuracy depends on the significance of the variable.
To conclude, it can be said that when one variable were to be analyzed there would only be
one relationship. Profit is dependent on the amount of goods sold, price and cost associated,
and if you sell more, higher profit can be assumed. If the equation for earnings is sold items
by price, then a higher price would mean higher earnings and it is also important to take into
consideration that lower costs would also result in higher earnings. That, however, would
assume that the other variables are unmoving. As posed in the question, management has
asked for the usefulness of the model to be discussed. When discussing the model with
management it ought to be argued that while one can see strong advantages with the given
model based on the “number of items sold” it would be appropriate to investigate further into
other variables that lead to higher profitability, too. There could still be bias if not
investigated properly, however, including more variables would allow for a more conclusive
picture to be formed.
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Part B
1. The data set chosen is quarterly beer production in Australia and it is measured in mega
liters. In the diagram above, time is on the x-axis where each quarter of each year is shown
and the total beer production is on the y-axis. The data set and time-series diagram have in
total 48 points and it spans over twelve years from January 1957 until December 1968. Each
point on the diagram corresponds to the quarterly beer production in Australia.
2. A positive upward trend and seasonal variations are present in the time-series graph. On
average, the beer production is increasing over time during the relevant time period. In the
data, there are seasonal variations present which are due to fluctuations in the beer
production. There is a pattern of recurring highs and lows each year. In every fourth quarter,
the manufacturing of beer increases rapidly which might be due to the fact that it is summer
in Australia. In the first quarter, the production goes down slightly and in the second quarter,
the production continues to decrease. There are no outliers present in the diagram. No
cyclical components or other major irregularities can be observed in the data. The model is
additive since the seasonal variations are around the same magnitude during the relevant time
span.
3. The time series data is reasonable since beer consumption varies depending on the season.
Beer manufacturers might forecast and plan how much beer will be consumed and change the
production of it accordingly. The beer companies know that the demand will increase during
summer and they prepare for this by increasing production. In the data, it is shown that the
beer production increases rapidly every fourth quarter which means there is a strong seasonal
component to it. This is recurring again and again in the data during the investigated 12 years.
Beer consumption depends on the season and it varies depending on what time of the year it
is. In the time series, the production has therefore increased and decreased over time which
can be very much expected.
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4. The Excel file “Beer production Australia” contains quarterly data and it is therefore
suitable to use the Excel sheet “Seasonally adjusted quarterly”. The blue line in the graph
above represents the original data while the orange line is the trendline and it is seasonally
adjusted. The trendline is trending upwards which means that the total beer production has
increased during the past 12 years investigated. In the graph, it is clearly shown that the
difference between the original values and the adjusted values are relatively small. The
seasonally adjusted series varies much less than the original series depending on which
quarter it is. The seasonal component is not as clear in the smoothed data compared to the
original data and the purpose of doing a seasonal adjustment of the data is to gain greater
understanding of the overall trend and make it more useful. For example, the trendline can be
used to forecast future production of beer which enables better planning of the business
(Newbold et al. 2020, p. 693-701).
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References
Newbold et al. (2020). Statistics for Business and Economics (9th edition). United Kingdom:
Pearson Education.
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