Statistics and Forecasting Analysis

This assignment focuses on applying various statistical techniques to real-world scenarios. Students will conduct hypothesis tests using ANOVA and t-tests to analyze differences in test scores and accident occurrences. Furthermore, they will utilize moving averages and exponential smoothing to forecast closing prices of a stock based on historical data.

Running Head: BUSINESS DATA ANALYSIS
Business Data Analysis
Name of the Student
Name of the University
Author Note

1BUSINESS DATA ANALYSIS
TUTORIAL
Part A
Answer 1
Task 1
Figure 1 shows that the count of Rock Road * Bus Lane Beside Park fluctuates a lot each
day.
1/1/2015
1/6/2015
1/11/2015
1/16/2015
1/21/2015
1/26/2015
1/31/2015
2/5/2015
2/10/2015
2/15/2015
2/20/2015
2/25/2015
3/2/2015
3/7/2015
3/12/2015
3/17/2015
3/22/2015
3/27/2015
0
100
200
300
400
500
600
700
800
Rock Road *Bus Lane Beside Park
Days
Count
Figure 1

2BUSINESS DATA ANALYSIS
Task 2
Figure 2
From the chart it can be seen that in countries like Cork, Dublin, Limerick and Tipperary
have shown very high fatality rates compared to other countries.
Task 3
Data has been selected on the GDP growth of Singapore. The data has been collected
from World Bank. With the help of a line graph in figure 3, the trend of the data is described.
The GDP of Singapore had a dramatic fall in 2008 and 2009 and after that there has been
recovery in the GDP in 2010. It again started to fall from 2011 gradually.

3BUSINESS DATA ANALYSIS
2007 2008 2009 2010 2011 2012 2013 2014 2015 2016
-2
0
2
4
6
8
10
12
14
16
18
Real GDP Growth Rate of Singapore
Year
GDP Growth
Figure 3
Answer 2
Problem 1:
a) The null hypothesis (H0) and the alternate hypothesis (HA) has been stated as
follows:
Ho: μ ≤ 0.75
HA: μ >0.75
The test statistic for this test can be given by the following formula:
t= X−μ
σ
√ n
Here, X is the sample mean, μ is the mean value that has been hypothesized, σ is the standard
deviation of the sample and n is the sample size. In this study,
Sample mean ( X ) = 0.68

4BUSINESS DATA ANALYSIS
Hypothesized mean ( μ) = 0.75
Standard deviation of the sample ( σ )
Sample size (n) = 400
Degrees of freedom (n-1) = 399
The null hypothesis will be rejected if the tabulated value of t statistic is less than the observed
value of absolute t.
b) In stratified random sampling, the different strata are created from the population
considering the variables based on which the study will be conducted. Sampling is then
done from each stratum. Sampling error is reduced in this case.
In cluster sampling, different clusters are formed from the whole population and
as samples some clusters are selected from all the clusters formed.
c) Smaller
Greater
d) 1.96
e) -1.29 and 1.29
f) Z tests are used when sample size is greater than 30 and standard deviation is
known. T test is used when sample size is less than 30.
g) To test whether the proportion wishing to travel to Europe has fallen below 44%,
one sample z-test has to be conducted.
Here, the sample proportion ( p1) = (400/1000) = 0.4
The population proportion (p) = 0.44
Sample size (n) = 100

5BUSINESS DATA ANALYSIS
The standard deviation of the sample (s) = √ p1 (1−p1)
n = √ 0.4(1−0.4)
100 =0.049
The null (H0) and the alternate ( H A) hypothesis can be given as:
H0 : p ≥ 0.44
H A : p<0.44
The test statistic for the test can be given as:
z= p1− p
s = 0.4−0.44
0.049 =−0.816
The tabulated value of z for 1% level of significance is 2.58 which is more than
the absolute value of calculated z statistic. Thus, null hypothesis is accepted. the
proportion wishing to travel to Europe has not fallen below 44%
Problem 2:
The null hypothesis (H0) and the alternate hypothesis (HA) has been stated as follows:
Ho: μ =0.75
HA: μ ≠0.75
The test statistic for this test can be given by the following formula:
t= X−μ
σ
√ n
Here, X is the sample mean, μ is the mean value that has been hypothesized, σ is the standard
deviation of the sample and n is the sample size. In this study,
Sample mean ( X ) = 0.68

6BUSINESS DATA ANALYSIS
Hypothesized mean ( μ) = 0.75
Standard deviation of the sample ( σ )
Sample size (n) = 400
Degrees of freedom (n-1) = 399
The null hypothesis will be rejected if the tabulated value of t statistic is less than the observed
value of absolute t, at 0.025 level of significance at each of the two tails.
Problem 3:
Confidence Interval ¿ x ± z0.05
s
√ n
¿ 29.8 ±1.96 0.41
√ 30
¿ 29.8 ±1.96 × 0.075
¿ 29.8 ± 0.147
¿( 29.65 ,29.95)
There are no values outside this range, thus the process is in control
Part B
Section A
1. 13 packets of beans were sampled.
a) Median = 498
b) Mean = 497.15
c) Variance = 29.64

7BUSINESS DATA ANALYSIS
d) Null Hypothesis: μ <500
Alternate hypothesis: μ ≥500
e) z= X −μ
σ
√ n
= 497.15−500
5.44
√ 13
=−1.89
f) The probability of committing a type 1 error at α = 0.05 is 1.96
g) The absolute value of z statistic is less than the probability of type 1 error. Thus
null hypothesis is accepted. The average weights of the packets are less than 500
grams.
2. Statistics course is available online and in traditional classroom.
a) Null Hypothesis: μonline =μclassroom
Alternate Hypothesis: μonline ≠ μclassroom
b) The test statistic for the test can be given as:
T = Y online−Y classroom
√ sonline
2
Nonline
+ sclassroom
2
N classroom
= 63.3−58.33
√ 33.57
10 + 73.75
9
=1.46
The degree of freedom is given by:
v=
( sonline
2
N online
+ sclassroom
2
Nclassroom ) 2
( sonline
2
N online )
2
( Nonline −1 ) +
( sclassroom
2
Nclassroom )
2
( N classroom−1 )
= ( 33.57
10 + 73.75
9 )
2
( 33.57
10 )
2
( 10−1 ) + ( 73.75
9 )
2
( 9−1 )
=14
c) The probability of committing a type 1 error at 0.05 level of significance with 14
degree of freedom is 2.145

8BUSINESS DATA ANALYSIS
d) The test statistic is less than the probability of committing a type 1 error. Thus,
the null hypothesis is accepted. Thus, there is no difference in the exam results at
the end of the course between the two groups.
Section B
3.
a) From the ANOVA table it can be seen that sig. value is 0.000 which is less than
the level of significance. Thus, it can be said that there is significant difference of
noise pollution created by small, medium and large sized cars.
b) The null hypothesis in this case is:
Null Hypothesis: There is no significant difference in the premiums paid per six
months by these households with these three companies
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups
1020
6 2
510
3
1.09718
3
0.37462
2
4.25649
5
Within Groups
4185
9 9
465
1
Total
5206
5
1
1
The ANOVA table shows that p-value is more than the level of
significance (0.05). Thus, null hypothesis is accepted.
4.
a) Vsb
b) Bchjs

9BUSINESS DATA ANALYSIS
5.
a) Simple linear regression model will be calculated
i. Correlation coefficient = 0.144
ii. Slope of a line best fit = 0.003
iii. The Y intercept = 3.82
iv. The value of Y when X = 22 is 3.89
The correlation coefficient shows that the study hours and grade points
have a very weak relation. With one-unit increase in study hours, grade points
increase by 0.003 times. The y-intercept indicates the grade points in the
absence of study hours.
b) A simple linear regression model predicts the value of Y with changes in the
value of X.

10BUSINESS DATA ANALYSIS
2015-2016
Answer 1
a) Null Hypothesis:There is no significant difference between the salaries of male and
female professors
Alternate Hypothesis:There is no significant difference between the salaries of male and
female professors
b) The test statistic to test the hypothesis is 2.636
c) The probability of committing type 1 error is 2.101
d) The test statistic is more than the probability of committing a type 1 error. Thus, null
hypothesis is rejected.
Answer 2
(a) Null Hypothesis:There is no significant difference between the salaries of male and
female professors
Alternate Hypothesis:There is no significant difference between the salaries of male and
female professors
(b) The test statistic to test the hypothesis is 4.049
(c) The probability of committing type 1 error is 2.262
(d) The test statistic is more than the probability of committing a type 1 error. Thus, null
hypothesis is rejected.
Answer 3
(a) In case of an ANOVA test, the null and alternate hypothesis can be given as:

11BUSINESS DATA ANALYSIS
Null Hypothesis:There is no significant difference between the rental units at different
time points
Alternate Hypothesis:There is no significant difference between the rental units at
different time points
From the ANOVA table given in the question, it can be seen that the p-value
(Sig.) is more than the 55 level of significance (0.05). Thus, the Null hypothesis is
accepted. There is no significant difference in the rental units over different time points.
(b) The dataset contains the test scores from three statistics classes of three different
teachers.
a) Null Hypothesis: There is no significant difference in test scores of three
different classes
Alternate Hypothesis:There are significant differences in test scores of three
different classes
b) The test statistic can be given as 4.84
c) The probability of committing a type 1 error is 4.459
d) The critical value of F statistic is less than the observed value of the F statistic.
Thus, null hypothesis is rejected.
Answer 4
a) To perform this test, chi-square test of association has been conducted.
i. Null Hypothesis: There is no association between the observed and the expected
values
Alternate Hypothesis: There is association between the observed and the expected
values

12BUSINESS DATA ANALYSIS
ii. The test statistic for this test has been obtained as 14.767
iii. The critical value of the chi square test with two degrees of freedom is 5.99
(Obtained from chi-square table). The value of the test statistic is more than the
critical value. Thus, the null hypothesis is rejected. There is association between
the observed and the expected frequencies.
b) Time series
i. The estimate for the number of road deaths in 2012 is 212.
Year
Road
Deaths
3-point
MA
exp
smoothing
2000 400 #N/A
2001 411 400
2002 376 406.6
2003 335
395.66666
7 388.24
2004 374 374 356.296
2005 396
361.66666
7 366.9184
2006 365
368.33333
3 384.36736
2007 338
378.33333
3 372.746944
2008 279
366.33333
3 351.8987776
2009 238
327.33333
3 308.159511
2010 212 285 266.0638044
2011 186 243 233.6255218
2012 212 205.0502087
Answer 5
a) The simple linear regression model can be given as:
i. The correlation coefficient is 0.97
ii. The slope of a line best fit for this data is 0.67
iii. The Y intercept is 1.01

13BUSINESS DATA ANALYSIS
iv. The value of Y when X = 55 is 38.09
b) These two variables have a very strong negative association with each other. The increase
in one variable implies the decrease in the other.

14BUSINESS DATA ANALYSIS
2016 – 2017
Answer 1
In this problem, the difference in the weights before and after the programme is to be measured.
Paired t-test will be used for this.
a) Null Hypothesis: There is no significant difference in the weights before and after the
programme
Alternate Hypothesis: There is significant difference in the weights before and after the
programme.
b) The test has been run in excel and the test statistic has been found to be 2.88
c) Probability of committing a type 1 error is 2.201
d) The value of the test statistic is ore than the probability of type 1 error. Thus, null
hypothesis is rejected.
Answer 2
To test whether the scores of the students in the classes of two different statistics lecturers are
same of different, two sample t-test for unequal variances will be conducted. The test is
conducted in Excel.
a) Null Hypothesis: There is no significant difference in the mean scores of the students of
the two different classes
Alternate Hypothesis: There are significant differences in the mean scores of the students
of the two different classes
b) The test statistic is given by 0.424
c) An appropriate probability of committing type 1 error is 0.05

15BUSINESS DATA ANALYSIS
d) The value of the test statistic obtained is less than the type 1 error. Thus, the null
hypothesis is accepted. There is no significant difference in the test scores of students
going to different statistics lecturers.
Answer 3
To determine whether the mean time required to complete a certain task differ for the three levels
of employee training, ANOVA test has to be conducted. The test has been conducted in Excel.
a) Null Hypothesis: There is no significant difference in the mean time required to complete
a task for three levels of employee training.
Alternate Hypothesis: There are significant differences in the mean time required to
complete a task for three levels of employee training.
b) The test statistic is given by 5.446
c) The probability of committing a type one error is 0.05
d) The value of the test statistic is more than the critical value of F (3.885). Thus, the null
hypothesis is rejected. There is no difference in the mean time required to complete a task
for three levels of employee training.
Answer 4
a) To test whether there is an equal number of an accident taking place each day, one
sample t-test has to be conducted. The mean number of accidents taking place each day is
12.71 (Approx). Let the expected number of accidents happening each day is 13
i. The null hypothesis (H0) and the alternate hypothesis (HA) has been stated as
follows:
Ho: μ = 13

16BUSINESS DATA ANALYSIS
HA: μ ≠ 13
ii. The test statistic for this test can be given by the following formula:
t= X−μ
σ
√ n
=12.71−13
3.09
√ 7
=−0.25
iii. The tabulated value of t for 6 degrees of freedom is 2.447 which is more than the
absolute value of the test statistic. Thus, the null hypothesis is accepted. There are
on an average of 13 accidents taking place each day.
b) Moving average and exponential smoothing
Date
Closing
Price 3 pt MA
Exp
Smoothing
10/27/201
6 129.69 #N/A
10/28/201
6 131.29 129.69
10/31/201
6 130.99 130.49
11/01/201
6 129.5
130.656
7 130.74
11/02/201
6 127.17
130.593
3 130.12
11/03/201
6 120 129.22 128.645
11/04/201
6 120.75
125.556
7 124.3225
11/07/201
6 122.15 122.64 122.53625
11/08/201
6
120.966
7 122.343125