Data collection
Class interval = range/number of classes
Class interval = 7560/10 = 756
Relative frequency = frequency/total frequency
For instance the relative frequency for 169-925 class = 37/60 = 0.616667
- Frequency distribution table
classes |
freq |
midpoint |
rel. freq |
cum. Freq |
169-925 |
37 |
547 |
0.616667 |
37 |
926-1682 |
17 |
1304 |
0.283333 |
54 |
1683-2439 |
2 |
2061 |
0.033333 |
56 |
2440-3196 |
3 |
2818 |
0.05 |
59 |
3197-3953 |
0 |
3575 |
0 |
59 |
3954-4710 |
0 |
4332 |
0 |
59 |
4711-5467 |
0 |
5089 |
0 |
59 |
5468-6224 |
0 |
5846 |
0 |
59 |
6225-6981 |
0 |
6603 |
0 |
59 |
6982-7738 |
1 |
7360 |
0.016667 |
60 |
- Constructing a histogram
- Mean
Median = 670
Mode = 338
- The above is a sample not a population since a sample is the subset or the proportion of the population while population is a set of objects or elements under study/investigation.
Total sample n = 7
Attendance (x) |
(x – 484.5714) |
(x-484.57114)2 |
472 |
-12.5714 |
158.0408 |
413 |
-71.5714 |
5122.469 |
503 |
18.42857 |
339.6122 |
612 |
127.4286 |
16238.04 |
399 |
-85.5714 |
7322.469 |
538 |
53.42857 |
2854.612 |
455 |
-29.5714 |
874.4694 |
32909.7124 |
Therefore the standard deviation of the attendance will be;
- Correlation coefficient
Mean of (x) = mean of (y) = 47618/7 = 6802.571429
Attendance (x) |
Bars (y) |
xy |
X2 |
Y2 |
|
472 |
6 916 |
3264352 |
222784 |
47831056 |
|
413 |
5 884 |
2430092 |
170569 |
34621456 |
|
503 |
7 223 |
3633169 |
253009 |
52171729 |
|
612 |
8 158 |
4992696 |
374544 |
66552964 |
|
399 |
6 014 |
2876391 |
159201 |
36168196 |
|
538 |
7 209 |
3878442 |
289444 |
51969681 |
|
455 |
6 214 |
2827370 |
207025 |
38613796 |
|
∑(x) =3392 |
47618 |
23902512 |
1676576 |
327928878 |
The correlation coefficient of r=0.968 showed that there was a strong positive correlation between the weekly students’ attendance and the number of chocolate bars sold at the supermarket.
- Linear regression equation is given by
y=a+bx
Where y is number of chocolate bars sold (dependent variable), x is weekly students’ attendance (independent variable), a is the y-intercept and b is the slope or gradient of the linear equation.
Attendance (x) |
Bars (y) |
xy |
X2 |
Y2 |
472 |
6 916 |
3264352 |
222784 |
47831056 |
413 |
5 884 |
2430092 |
170569 |
34621456 |
503 |
7 223 |
3633169 |
253009 |
52171729 |
612 |
8 158 |
4992696 |
374544 |
66552964 |
399 |
6 014 |
2876391 |
159201 |
36168196 |
538 |
7 209 |
3878442 |
289444 |
51969681 |
455 |
6 214 |
2827370 |
207025 |
38613796 |
∑(x) =3392 |
47618 |
23902512 |
1676576 |
327928878 |
From the above calculations,
Number of chocolate bars sold = 1628.689 + 10.67723*students’ weekly attendance.
The weekly attendance of the students affected the number of chocolate sold at the supermarket in that case, the students’ weekly attendance is an independent variable while number of chocolate sold is a dependent variable. That is to mean, when the Holmes are closed and there are no students, the number of chocolate sold will reduce as in the following example;
When (x) the students’ weekly attendance is zero (0), the number of chocolate sold will be as below;
Y = 1628.689 + 10.67723*0
Y= 1628.689 which will approximately be 1629 chocolate bars.
When the number of students increase by 10, the number of chocolate that will be sold by the supermarket will be;
Y = 1628.689 + 10.67723*10
Y = 1628.689 + 106.7723 = 1735.4613 which will be approximately 1735 chocolate bars. The number of chocolate bars that will be sold will increase by (1735 – 1629) = 106 chocolates.
- The coefficient of the determinant i.e. the coefficient of the number of chocolate bars is the constant value by which when the number of students’ weekly attendance will be increasing since the coefficient is 10.7 approximately 11, it then means that each student from Holmes will be buying at least 11 chocolate bars in their weekly attendance.
Scientific training |
Grassroots training |
Total |
|
Recruited from Holmes students |
35 (35/127) |
92 (92/127) |
127 |
External recruitment |
54 (54/66) |
12 (12/66) |
66 |
Total |
89 |
104 |
193 |
- P(H) = probability of student from Holmes
P(E) = probability of the external players
P(H) = number of players from Holmes/total number of players
P(H) = 127/193
P(G) is probability of grassroots training
P(G) = number of players from Holmes/total players in grassroots training
P(G) = 92/104 = 23/26
P(H) or P(G) = 127/193 + 23/26 = 2921/5018
- Probability of external and scientific
P(E) = 66/193
P(ES) = 54/89
P(E) and P(ES) = 66/193 * 54/89 = 3564/17177
- P(H|S) = P(H and S)/P(H)
P(H) and P(s)/P(H)
=(127/193*35/89)/(127/193) = 35/89
- Independent probability
The probability that Holmes will be going for training is 35/89 and that of recruitment is 54/89 for scientific training and for the grassroots training Holmes is 92/104 and external 12/104. Training is not independent of recruitment.
SegProd |
X |
Y |
Z |
A 0.55 |
0.2 |
||
B 0.3 |
0.35 |
||
C 0.1 |
0.60 |
||
D 0.05 |
0.9 |
- Probability of segment (A) = 0.55
P(X) = 0.2
P(AX) = 0.55*0.2 = 0.11
The probability that a person comes from segment A and prefers product X over Y and Z is 0.11
- 1 – (0.3*0.35 + 0.1*0.6 + 0.05*0.9)
1 – 0.21 = 0.79
Therefore the probability that a random consumer preference is product X is 0.79
- P(1) = 1/10
P(<=2) = 2*1/10 = 1/5
= 1/10 + 1/5 = 3/10
- Normal distribution
- The t-distribution will still be used to test for the assistance findings against mine since I need to make use of the sample size and compare the means
H0: µ>=1100000
H1: µ<1100000
T-test