Relationship Between Students’ Attendance And Chocolate Sales

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

1. Frequency distribution table

classes Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	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

1. Constructing a histogram

 

1. Mean

Median = 670

Mode = 338

1. 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;

1. 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.

1. 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.

1. 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

1. 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

1. Probability of external and scientific

P(E) = 66/193

P(ES) = 54/89

P(E) and P(ES) = 66/193 * 54/89 = 3564/17177

1. P(H|S) = P(H and S)/P(H)

P(H) and P(s)/P(H)

=(127/193*35/89)/(127/193) = 35/89

1. 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

1. 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. 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

1. P(1) = 1/10

P(<=2) = 2*1/10 = 1/5

= 1/10 + 1/5 = 3/10

   

1. Normal distribution

1. 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