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STATISTICS ASSIGNMENT GROUP HI6007 1
Statistics Assignment Group Hi6007
Holmes Institute
Faculty of Higher Education
STATISTICS ASSIGNMENT GROUP HI6007 2
Question 1
Table 1: F-test
Source of Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
Between treatments
90
3
__30___?
__5___?
Within treatments (Error)
120
20
__6___?
Total __210___? __23___?
a)
Mean of squares (Between treatments) = 90 ÷3
= 30
Mean of squares (Within treatments) = 120 ÷20
= 6
Total sum of squares = 90 + 120 = 120
Total sum of squares = 3 + 20 = 23
b) F- value= (Mean square between treatments) ÷ (Mean square within
treatments)
F- value = 30 ÷ 6
= 5
c) What has been the total number of observations?
Number of observations = Degrees of freedom + 1
= (3 + 20) + 1
= 27 observations.
Question 2
Develop a linear trend expression and project the sales (the number of cars sold) for time period t = 11.
Let the number of cars sold be represented x, then, the linear trend will be expressed as,
btay
, where a is a constant, t is the time period in years and x is the number of cars cold in
STATISTICS ASSIGNMENT GROUP HI6007 3
a thousand units in a particular period and y is the total number of cars sold in a given time
period (Lun 2017). Therefore, the estimated linear equation will be represented as;
Table 2: Trend line estimation
ti
yi
tti
yyi
(
tti
)(
yyi
)
)
2
(
tti
tiy 181818.39136
^
1
195
-4.5
-156.5
704.25
20.25
175.1818
2
200
-3.5
-151.5
530.25
12.25
214.3636
3
250
-2.5
-101.5
253.75
6.25
253.5455
4
270
-1.5
-81.5
122.25
2.25
292.7273
5
320
-0.5
-31.5
15.75
0.25
331.9091
6
380
0.5
28.5
14.25
0.25
371.0909
7
440
1.5
88.5
132.75
2.25
410.2727
8
460
2.5
108.5
271.25
6.25
449.4545
9
500
3.5
148.5
519.75
12.25
488.6364
10
500
4.5
148.5
668.25
20.25
527.8182
∑
ti
=55
∑
yi
=3515
∑
tti
=0
∑
yyi
=0
∑(
tti
)
(
yyi
)=3232.50
∑
)
2
(
tti
=82.50
∑
^
y
=3515
5.5
10
55
n
ti
t
,
n
yi
y
where n is the number of observations (the sample)
5.351
10
3515
n
yi
y
btiayi
-Actual trend line equation
ty
^
- Actual trend line equation.
a in the actual trend line is the constant that does not have any effect on the number of
cars sold as time changes.
STATISTICS ASSIGNMENT GROUP HI6007 4
2
)(
))((
tti
yyitti
b
=
181818.39
50.82
50.3232
tbya
=
)5.5(181818.395.351
=
5.2155.351
136a
tiy 181818.39136
^
The estimated equation determines how the number of cars sold relates with time in
years.
yi
is the observed y while
^
y
is the estimated y.
According to the table generated above, for every value of t, there is an estimated y. The
summation of the actual number of vehicles sold and summation of the number of vehicles
estimated to be sold are the same. However, these values vary each year.
tiy 181818.39136
When t = 11
y
= 136 + 136 + 39.181818(11)
y
= 136 + 490.9999
= 566.9999
= 566.9999 × 1000
y
= 566999.9
STATISTICS ASSIGNMENT GROUP HI6007 5
y
= 567000
Therefore, the number of vehicles sold for the time period t = 11, will be 567000 vehicles as
shown in the trend line below. A plot of the number of vehicles sold against time in years.
At the 11
th
year, 567 cars will be sold.
Question 3
a) F-test
Table 3: F-Test Two-Sample for Variances
Variable 1
Variable 2
Mean
35
4.285714286
Variance
11.66666667
7.238095238
Observations
7
7
df
6
6
F
1.611842105
P(F<=f) one-tail
0.288286332
F Critical one-tail
8.46612534
195
200
250
270
320
380
440
460
500 500
567
0
100
200
300
400
500
600
1 2 3 4 5 6 7 8 9 10 11
Cars sold
STATISTICS ASSIGNMENT GROUP HI6007 6
The calculated F-value is 1.611 is < the critical value (8.466) at .01 significance level. Thus,
we fail reject the null hypothesis and conclude that price and the number of flash drives sold are
not related. This is further confirmed by the p-value (0.288)
a) T-test
Table 4: T-test: Paired Two Sample for Means
Variable 1
Variable 2
Mean
35
4.285714286
Variance
11.66666667
7.238095238
Observations
7
7
Pearson Correlation
-0.924982219
Hypothesized Mean Difference
0
Df
6
t Stat
13.56167756
P(T<=t) one-tail
4.98633E-06
t Critical one-tail
3.142668403
P(T<=t) two-tail
9.97267E-06
t Critical two-tail
3.707428021
The calculated t Stat is > than the t Critical. Thus, the null hypothesis that the price and
number of flash disks sold are not related is rejected. Therefore, it can be concluded that the
price number of flash drives sold are related. This is further confirmed by p-value (<.01) which
is significant at 1%.
Question 4.
Table 5: F-statistics
Source of Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
Between treatments
____3,200_
__4___
800.00
_0.9730____
Within treatments (Error)
__7,400___
__9___
__822.22___
Total 10,600 _13____
Since the levels of the factors were five, then the degrees of freedom between treatments
were 4. Considering the number of experimental units, it means that the degrees of freedom
STATISTICS ASSIGNMENT GROUP HI6007 7
within treatments will be nine since the total number of degrees of freedom are supposed to be
13. According to Kenny et al. (2015) as degrees of freedom will be calculated as (n – 1)
= 14 -1 = 13.
Since, the degrees of freedom between treatments are 4, the degrees of freedom within treatments
will be calculated as follows, 13 – 4 = 9
Sum of squares = degrees of freedom × Mean sum of squares.
= 800 × 4
= 3200
Total sum of squares = Sum of squares between treatments + Sum of squares within
treatments. Let the sum of squares within groups be x,
10,600= x +3200
x
= 10600 – 3200
x
= 7400
F value = mean of square between treatments divided by mean square within treatments.
= 800 ÷ 822.22
= 0.9730
Question 5
STATISTICS ASSIGNMENT GROUP HI6007 8
Table 6: Oneway ANOVA table
Number of obs. = 11 R-squared = 0.8988
Root MSE = 2.08167 Adj. R-squared = 0.8735
Source
Partial SS
df
MS
F
Prob. > F
Model
307.87879
2
153.93939
35.52
0.0001
Store
307.87879
2
153.93939
35.52
0.0001
Residual
34.666667
8
4.3333333
Total
342.54545
10
34.254545
The results indicate significant (at better than the 1% level) differences in the average sales
of the three stores. However, it cannot be ascertained whether the difference is between only two
of the stores or all three stores. To unravel this, unpaired t-test is performed. Tables 2, 3 and 4
present the t-test results.
Table 7: Two-sample t test with equal variances between store 1 and store 2
Group
Obs.
Mean
Std. Err.
Std. Dev.
[95% Conf. Interval]
Store 1
5
45
0.83666
1.870829
42.67706
47.32294
Store 2
3
36.33333
1.452966
2.516611
30.08172
42.58494
Combined
8
41.75
1.729471
4.891684
37.66045
45.83955
Diff
8.666667
1.539601
4.899399
12.43393
diff = mean (Store 1) – mean (Store 2) t = 5.6292
H
o
: diff = 0 degrees of freedom = 6
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(T < t) = 0.9993 Pr(|T| > |t|) = 0.0013 Pr(T > t) = 0.0007
Results in Table 2 indicates that average sales of store 1 and store 2 are significantly
different at 5% significance level (Ha: diff != 0; p-value = 0.0013). Furthermore, results in Table
3 indicates that average sales of store 1 and store 3 are significantly different at 5% significance
level (Ha: diff != 0; p-value = 0.0001). Lastly, results in Table 4 indicates that average sales of
store 2 and store 3 are not significantly different at 5% significance level (Ha: diff ! = 0; p-value
= 0.0001).
STATISTICS ASSIGNMENT GROUP HI6007 9
Table 8: Two-sample t test with equal variances between store 1 and store 3
Group
Obs.
Mean
Std. Err.
Std. Dev.
[95% Conf. Interval]
Store 1
5
45
0.83666
1.870829
42.67706
47.32294
Store 3
3
33
1.154701
2
28.03172
37.96828
Combined
8
40.5
2.283481
6.45866
35.10043
45.89957
Diff
12
1.398412
8.57821
15.42179
diff = mean (Store 1) – mean (Store 3) t = 8.5812
H
o
: diff = 0 degrees of freedom = 6
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(T < t) = 0.9999 Pr(|T| > |t|) = 0.0001 Pr(T > t) = 0.0001
Table 9: Two-sample t test with equal variances between store 2 and store 3
Group
Obs.
Mean
Std. Err.
Std. Dev.
[95% Conf. Interval]
Store 2
3
36.33333
1.452966
2.516611
30.08172
42.58494
Store 3
3
33
1.154701
2
28.03172
37.96828
Combined
6
34.66667
1.115547
2.73252
31.79906
37.53427
diff
3.333333
1.855921
-1.819531
8.486197
diff = mean (Store 1) – mean (Store 3) t = 1.7961
H
o
: diff = 0 degrees of freedom = 4
Ha: diff < 0 Ha: diff ! = 0 Ha: diff > 0
Pr(T < t) = 0.9265 Pr(|T| > |t|) = 0.1469 Pr(T > t) = 0.0735
Question 6
a) Hypotheses
i. H
o
: There is no significant difference in the average sales from the five store, that
is,
54321
H
a
: There is significant difference in the average sales from the five store, that is
54321
ii. H
o
: There is no significant difference in the average sales from three boxes, that
is,
321
STATISTICS ASSIGNMENT GROUP HI6007 10
H
a
: There is significant differences in the average sales from the three boxes, that
is
321
b) Anova table
Table 10: Twoway ANOVA table
Number of obs. = 15 R-squared = 0.9260
Root MSE = 15.8572 Adj. R-squared = 0.8706
Source
Partial SS
df
MS
F
Prob. > F
Model
25188.8
6
4198.1333
16.70
0.0004
Store
721.6
4
180.4
0.72
0.6033
Box
23932.483
2
11966.242
47.59
0.0000
Residual
2011.6
8
251.45
Total
27200.4
14
1942.8857
c) Results from the factorial ANOVA indicates that the overall model that was used to fit the
data is statistically significant (F = 16.70; p = 0.0004). Store is not statistically significant
(F = 0.72; p = 0.6033), implying that average sales in the five different stores is not
significantly different. In contrast, the average sales of boxes are statistically significant (F
= 47.59; p = 0.0000), suggesting that the average sales differ among the three boxes.
Question 7
The data is unpaired because only one method is used to measure the tyres for tread ware.
Therefore, two sample t-test can be used to test the mean mileage for the three brands of tyres.
Letting
1
,
2
and
3
represent the mean sample mean for Brand A, Brand B and Brand
C tyres, respectively.
2
2
2
1
2
1
21
n
s
n
s
xx
t
STATISTICS ASSIGNMENT GROUP HI6007 11
Null hypothesis 1:
210
:
H
0.6 -
1.667
1-
9
4
9
3
03837
22
t
Null hypothesis 2:
310
:
H
3.333
20.1
4
9
2
9
3
03337
22
t
Null hypothesis 3:
320
:
H
3.354
1.491
5
9
2
9
4
03338
22
t
The
9,5.0
t
critical value from the t-table is 1.833. The t-calculated values for a test between
Brands A and B, Brands A and C, and Brands B and C, are -0.6, 3.333, and 3.354, respectively.
The calculated it value for the difference in the average mileage for Brands A and B does not
exceed the t-critical value, hence we fail to reject the null hypothesis that average mileage for wear
characteristics of Brand A and Brand B are equal. Conversely, the calculated t-values for the
differences between the average mileage for Brands A and C and Brands B and C exceeds the t-
critical of 1.833, so the null hypothesis that the difference in the average mileage between the
Brands of tyres is reject and alternative hypothesis that the difference between average mileage of
Brands A and C and Brands B and C are statistically different is accepted.
STATISTICS ASSIGNMENT GROUP HI6007 12
Question 8
Table 11: Moving avaerage
Day
Tips
Moving average
1
18
2
22
3
17
19
4
18
19
5
28
21
6
20
22
7
12
20
1
st
set = (18 + 22 + 17)/3 = 19
2
nd
set = (22 + 17 + 18)/3 = 19
3
rd
set = (17 + 18 + 28)/3 = 21
4
th
set = (18 + 28 + 20)/3 = 22
5
th
set = (28 + 20 + 12)/3 = 20
a. Compute the mean square error for the forecasts.
Let days be represented by x while tips y.
- actual trend equation.
xy
^^^
- ,estimated trend equation
x
yx
var
,cov
^
, Where covx,y, is the covariance of x and y and varx is the variance of x
^
= -11/28
= -0.393
bxay
STATISTICS ASSIGNMENT GROUP HI6007 13
Table 12: Coefficient of determination
xi
yi
xxi
yyi
)
(
2
x
xi
(
xxi
)(
yyi
)
^
y
)(
^
yyi
(
^
yyi
)^2
)
(
2
y
yi
yy
^
)(
^
y
y
1
18
-3
-1.29
9
3.87
20.47
-2.47
6.10
1.66
1.18
1.39
2
22
-2
2.71
4
-5.42
20.08
1.92
3.69
7.34
0.79
0.62
3
17
-1
-2.29
1
2.29
19.68
-2.68
7.18
5.24
0.39
0.15
4
18
0
-1.29
0
0
19.29
-1.29
1.66
1.66
0
0
5
28
1
8.71
1
8.71
18.88
9.12
83.17
75.86
-0.41
0.17
6
20
2
0.71
4
1.42
18.50
1.50
2.25
0.50
-0.79
0.62
7
12
3
7.29
9
-21.87
18.11
-6.11
37.33
53.14
-1.18
1.39
28
135
0
-0.03
28
-11
135.01
-0.01
141.38
145.40
-0.02
4.34
xy
^
=
y
--0.393 (4)
= 19.29 + 0.393 (4)
= 19.29 + 1.572
^
= 20.862
^
y
= 20.862 – 0.393
x
2
)(
yyiTSS
= 145.40
SSE = 141.38
MSE = TSS – SSE
STATISTICS ASSIGNMENT GROUP HI6007 14
= 145.40 – 141.38
= 4.02
OR
MSE =
2
^
)(
yy
= 4.34
b. Compute the mean absolute deviation for the forecasts.
Mean absolute deviation = √S^2
√S^2 =
pn
SSE
Where, p is the number of independent variables regressed on y. Hence p, = 2
5
38.141
27
38.141
3175.5
276.28
Question 9
Table 13: F-statistics
Source of
Variation
Degrees of
Freedom
Sum of
Squares
Mean F
Square
Regression
4
283,940.60
0.456690
Error
18
621,735.14
Total
a. Compute the coefficient of determination and fully interpret its meaning.
Coefficient of determination is also known as
R
2
, it explains the variations caused in the
model by the explanatory variables (Miljkovic 2017).
R
2
= Residual sum of squares divided by the total sum of squares.
STATISTICS ASSIGNMENT GROUP HI6007 15
F value = SSR/SSE
= 283940.60 ÷ 621735.14
= 0.456690
TSS = SSR +SSE
TSS = 283940.60 + 621735.14
= 905675.74
R
2
=
74.905675
60.283940
SST
SSR
R
2
= 0.3135
It means that 31.35% of the variations in the model are explained by the explanatory
variables while 68.65% of the variations are explained by other factors outside the model hence
the error term.
b. Is the regression model significant? Explain what your answer implies. Let α = .05.
The model is not significant. This implies that at 5% significance level, the model does not
fit the data well. The explanatory variable does not significantly account for the variations in the
model.
c. What has been the sample size for this analysis?
The sample size for analysis = the total degrees of freedom +1
= (4 + 18) +1
=23
STATISTICS ASSIGNMENT GROUP HI6007 16
Question 10
a) Equation that can be used to predict the price of a stock
)1.57(- )0.02(- 118.51
ˆ
21
xxy
Note: The intercept (constant) and slope estimates have been rounded to two decimal places
b) Interpretation the coefficients of the estimated regression equation
- An additional share of stocks sold decreases the price of Rawlston Inc. stock by 0.02 units.
- A unit increase in the volume exchange (
2
x
) on the New York stock Exchange
reduces/decrease the price of the Rawlston Inc. stock by 1.57 units.
c) The volume of exchange (
2
x
) on the New York Stock exchange was statistically significant in
influencing the price Rawlston Inc. stock at all confidence levels. It is significant at α = .05
because 0.0018 < 0.05. In contrast, the number of shares of the company's stocks sold (
1
x
) is
insignificant since 0.6176 > 0.05
d)
000,000)1.5726(16,- 500)0.0163(94,- 118.51
ˆ
y
25120000- 1890- 118.51
ˆ
y
202,163,52
ˆ
y
STATISTICS ASSIGNMENT GROUP HI6007 17
References
Kenny, D. A., Kaniskan, B., & McCoach, D. B. 2015. "The performance of RMSEA in models
with small degrees of freedom." Sociological Methods & Research 44(3), 486-507.
Lun, A. T., & Smyth, G. K. 2017. "No counts, no variance: allowing for loss of degrees of
freedom when assessing biological variability from RNA-seq data." Statistical
Applications in Genetics and Molecular Biology. 34-46.
Miljkovic, T., & Orr, M. 2017. "An evaluation of the reconstructed coefficient of determination
and potential adjustments. ." Communications in Statistics-Simulation and Computation,
1-14.
SUBJECT: STATISTICS
WORD COUNT: 2430
DEADLINE: 22/01/2018
REFERENCE STYLE: APA
COUNTRY: AUSTRALIA

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