Statistics and Business Intelligence

STATISTICS AND BUSINESS INTELLIGENCE 1
Statistics and Business Intelligence
Name
Course
Professor’s Name
University
City
Date
]

STATISTICS AND BUSINESS INTELLIGENCE 2
KC7021-Individual Assignment
Question 1
Prepare the following questionnaire in SPSS, so that its data can be stored.
Attached is a screen shot of the created SPSS file. Included in the attachment
Question 2
Participants N Mean Std. Deviation Variance Mode
Exercise class N1=43 μ1=1.8Kgs (S1)=1.04 (S12)=1.08kgs M1=1.6 kgs
Gym-only
workouts
N2=57 μ2=2.3Kgs (S2)=1.34 (S22)=1.80kgs M2=1.8 kgs
Hypothesis can be written as;
Null Hypothesis H0: (μ1- μ2) = 0
Alternative Hypothesis is H1: (μ1- μ2) > 0
The test statistic is:
t =( X1− X2 )−( μ1−μ2 )
√ s p
2 (1
n1
+1
n2
)
d . f .=n1+ n2−2
The pooled variation:
s p
2 =(n1−1) s1
2+(n2−1)s2
2
n1 +n2−2
= (43-1)*1.08 + (57-1)*1.80/ (43+57-2) = 1.49

STATISTICS AND BUSINESS INTELLIGENCE 3
t value computed is 2.028
Since, t = 1.21
From the t-test tables;
At α=5%, t = 2.028>1.645, therefore we reject null hypothesis.
The table above shows that the exercise class participants had a weight loss (M=1.8,
SD=1.04) while that of gym only workout participants was (M=2.3, SD=1.34). From the analysis
above, the staff at the gym who only workouts had a higher mean than the exercise class
participants. The t-test also reveals that the t computed value is higher than the tabulated value of
t (from the t-table). This, therefore, implies that the null hypothesis that suggests that there exist
no difference in weights loss between the participants who attended the gym only workouts and
the exercise class. This, therefore, implies that the exercise class has a smaller mean weight loss
than the gym only group. This suggests that participants who were in the gym only workouts
class lost statistically significant weight (p<.05) (Gupta and Kapoor,2019) compared to the
exercise class only participants. The mode of the weight lost by gym only workouts participants
was relatively higher than the exercise-only group.
Question 3
.Referring to the Venn diagram of this situation in the Figure below, the state in words the
events represented by the following regions
a. region 5
Considering that;
M- the event that they will experience mechanical problems
T - the event that they will receive a ticket for committing a traffic violation

STATISTICS AND BUSINESS INTELLIGENCE 4
V-the event that they will arrive at a campsite with no vacancies
Region 5= (V ∪ T ¿− M
It is the event the family will arrive at a campsite with no vacancies and receives a ticket
for committing a traffic violation less the probability they had a mechanical problem.
b. region 3
Region 1 is the intersection of M, T and V. M∩T ∩V .
This can be said to be the event that the family arrives at the campsite with no vacancies
given that they received a ticket for violation and had a mechanical problem.
Region 3 =T- (M∩T ∩V )
Is the difference between events that the family receives a ticket for committing a traffic
violation given that they had no vacancies at the campsite, they received a ticket for
violation and had a mechanical problem.
c. regions 1 and 2 together
=V-M; This is the event that the family arrives at a campsite with no vacancies given that
they experienced mechanical problems.
d. regions 4 and 7 together
=V+ Region 5
But Region 5= (V ∪ T ¿− M
Therefore Region 4 and 7= V+ (V ∪ T ¿− M
This is the event that the family will arrive at a campsite with no vacancies and does not
receive a ticket for committing a traffic violation given that they had a mechanical
problem.