Time Series Analysis and Forecasting

This assignment focuses on time series analysis and forecasting. It includes tasks like calculating moving averages, building a regression model to predict sales based on various factors, evaluating the model's performance using mean square error and mean absolute deviation, and interpreting the results. The assignment also involves analyzing stock price prediction models and understanding the impact of variables like company stocks sold and New York Stock Exchange volume.

Running Head: HI6007 STATISTICS ASSIGNMENT
HI6007 Statistics Assignment
Name of the Student
Name of the University
Author Note

1HI6007 STATISTICS ASSIGNMENT
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................2
Answer 3..........................................................................................................................................3
Answer 4..........................................................................................................................................4
Answer 5..........................................................................................................................................4
Answer 6..........................................................................................................................................5
Answer 7..........................................................................................................................................5
Answer 8..........................................................................................................................................7
Answer 9..........................................................................................................................................7
Answer 10........................................................................................................................................8

2HI6007 STATISTICS ASSIGNMENT
Answer 1
(a)
Source of Variation Sum of
Square
s
Degrees of
Freedom
Mean Square F
Between treatments SSB =
90
(k – 1) = 3 MSB = (SSB / k
– 1) = 30
(MSB / MSE) =
5
Within treatments (Errors) SSE =
120
(N – k) = 20 MSE = (SSE / N
– k) = 6
Total TSS =
210
(N – 1) = 23
The critical value of F at α = 0.01 with 3 and 20 degrees of freedom is 4.94. The
calculated value of F is higher than the significance value. Thus, there is significant
difference between the means.
(b) The number of groups is k. Therefore, the number of groups in this problem is 4
(c) The total number of observations is N. Therefore, the number of observations in this
problem is 24.
Answer 2
The projection equation can be estimated with the help of regression analysis. The results of the
regression analysis as obtained from the excel is given in the following table:
Coefficients Standard Error t Stat P-value
Intercept 136 13.763 9.881 0.000
Year (t) 39.18 2.218 17.664 0.000

3HI6007 STATISTICS ASSIGNMENT
The linear trend expression to project the sales is given by:
T = 136 + 39.18 * t
The projected sales for t = 11 is 136 + (39.18 * 11) = 567
Answer 3
To test whether the price and the number of flash drives sold are related, the following
null and alternate hypothesis can be stated:
H0: The price and the number of flash drives sold are not related
H1: The price and the number of flash drives sold are related
Table 3.1: ANOVA
df SS MS F Significance F
Regression 1 37.157 37.157 29.624 0.003
Residual 5 6.271 1.254
Total 6 43.429
The test has been done using ANOVA. From the analysis, it can be seen that the
significance F value is 0.003 which is less than the level of significance (0.01). Thus, null
hypothesis is rejected. The price and the number of flash drives sold are related.
Table 3.2: Table of Regression Coefficients
Coefficients Standard Error t Stat P-value
Intercept 29.786 4.704 6.332 0.001
Price (x) -0.729 0.134 -5.443 0.003
This can also be tested with the help of regression analysis. The results are given in table
3.2. From the analysis, it can be seen that the p-value of the analysis performing the t-test is
0.003, which is less than the level of significance (0.01). Thus null hypothesis is rejected. There
is significant relationship between the price and the number of flash drives sold.

4HI6007 STATISTICS ASSIGNMENT
Answer 4
Source of Variation Sum of
Square
s
Degrees of
Freedom
Mean Square F
Between treatments (800 *
4) =
3200
(5 – 1) = 4 800.00 (800 / 822.22) =
0.973
Within treatments
(Errors)
(10600
– 3200)
= 7400
(13 – 4) = 9 (7400 / 9) =
822.22
Total 10600 (14 – 1) = 13
Answer 5
To test whether there is significant difference in the average sales of the three stores,
Analysis of Variance (ANOVA) test has been conducted. The following null and alternate
hypothesis can be framed for the test:
Null Hypothesis (H0): There is no significant difference between the average sales of the three
stores.
Alternate Hypothesis (H1): There is significant difference between the average sales of the three
stores.
Table 5.1: ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 324 2 162 40.5 0.000 4.256
Within Groups 36 9 4

5HI6007 STATISTICS ASSIGNMENT
Total 360 11
From the p-value, it can be seen that the value is less than level of significance (0.05).
Thus, the null hypothesis is rejected. There is significant difference in the average sales of the
three stores.
Answer 6
(a) The null and the alternate hypothesis for this analysis is given below:
Null Hypothesis (H0): There is no significant difference between the average sales of the
three boxes.
Alternate Hypothesis (H1): There is significant difference between the average sales of
the three boxes.
(b) The required ANOVA table is given in table 6.1.
Table 6.1: ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 24467.2 2 12233.6 53.711 0.000 3.885
Within Groups 2733.2 12 227.767
Total 27200.4 14
(c) The value of the F statistic is higher than the critical value of F. Moreover, the p-
value is less than the level of significance (0.05). Thus, the null hypothesis is rejected.
There are significant differences in the average sales for each of the boxes.
Answer 7
Brand A Brand B Brand C
Average Mileage 37 38 33
Sample Variance 3 4 2
Total Mileage (37 * 10) = 370 (38 * 10) = 380 (33 * 10) = 330

6HI6007 STATISTICS ASSIGNMENT
Number of tyres for each brand = 10.
Total number of tyres = (10 * 3) = 30.
Total mileage = (370 + 380 + 330) = 1080
Average mileage = (1080 / 30) = 36
Thus, sum of squares between groups (SSB) = Σ [ Number of tyres of each brand * (Average
mileage of each brand – Average mileage of all brands)2].
Therefore, SSB = 10 * (37 – 36)2 + 10 * (38 – 36)2 + 10 * (33 – 36)2 = 140.
Error Sum of Squares (SSE) = Σ [Variance of each brand * (Number of tyres – 1)] = 3 * (10 – 1)
+ 4 * (10 – 1) + 2 * (10 – 1) = 81
Therefore, degrees of freedom for between groups = Number of brands – 1 = 3 – 1 = 2.
Total degrees of freedom = total number of tyres – 1 = 30 – 1 = 29
Degrees of freedom for errors = Total degrees of freedom – Degrees of freedom between groups
= 29 – 2 = 27
Source of Variation Sum of
Square
s
Degrees of
Freedom
Mean Square F
Between groups SSB =
140
(k – 1) = (3 – 1)
= 2
MSB = (SSB / k
– 1) = (140 / 2)
= 70
(MSB / MSE) =
(70 / 3) = 23.33
Within treatments (Errors) SSE =
81
(N – k) = (29 –
2) = 27
MSE = (SSE / N
– k) = (81 / 27)

7HI6007 STATISTICS ASSIGNMENT
= 3
Total TSS =
221
(N – 1) = (30 –
1) = 29
The critical value of F at α = 0.05 with 2 and 27 degrees of freedom is 3.35. The
calculated value of F is higher than the significance value. Thus, there is significant difference
between the average mileage of the tyres.
Answer 8
Table 8.1: Calculations of Moving Average, MSE and MAD
Day Tips 3-Day Moving Average Error Error^2
1 18
2 22
3 17
4 18 19 1 1
5 28 19 9 81
6 20 21 1 1
7 12 22 10 100
Total 21 183
a) The third column in table 8.1 gives the three day moving averages for the time series.
b) The mean square error for the forecasts = (183 / 4) = 45.75
c) The mean absolute deviation for the forecasts = (21 / 4) = 5.25
Answer 9
Source of Variation Degrees of Freedom Sum of Squares Mean Square F
Regression 4 283940.6 70985.15 2.055108
Error 18 621735.14 34540.84111
Total 22 905675.74
a) The coefficient of determination is given by the formula: 70985.15/905675.74 = 0.0784.
This indicates that only 7.84 percent of the variation in sales can be explained by price
per unit, competitor’s price, advertising and type of container used.