Estimating Area And Analysis Of Cricket Chirps

Question and Steps to Help Answer

Q1)

Site 1Area of Complete squares = 138 cm²

Area of Incomplete squares = 40/2 = 20cm²

Total area = 158 m²

But the scale = 1: 40 000

Actual area = 158 * 40000 *40000 = 2.528 *10¹¹ cm²

But,

1Km² = 100000cm * 100000 cm

1Km² = 10000000000 cm²

2.528 *10¹¹ cm²

= 25.28 km²

Site 2Area of Complete squares = 119 cm²

Area of Incomplete squares = 70/2 = 35cm²

Total area = 154 m²

But the scale = 1: 40 000

Actual area = 154 * 40000 *40000 = 2.464 *10¹¹ cm²

But,

1Km² = 100000cm * 100000 cm

1Km² = 10000000000 cm²

2.464 *10¹¹ cm²

= 24.64 km²

Evaluation

The calculation of the area of the sites involves subdividing the area using equal squares of dimension of 1 cm, where the incomplete squares covered will be divided by two and added to the total number of complete squares, the total will then be multiply by area of one square since they are similar to determine the total surface area covered, after which it will be multiplied by the scale to determine the actual area then it will be converted to square kilometers.

Assumption

1. The square are equal and have a dimensions of 1 cm
2. The incomplete squares are made complete by dividing their total number by 2 

Relevance to real world

The surveyors and architectures will use this method in drawing their plan and also interpreting it, to easily help to determine the surface area of a particular land or area to be covered with a building respectively. 

Q2)

Table  2.1: Chirp rate of crickets at different temperatures.

	Chirps per second
Temperature (oC)	Species A	Species B
20.8	15.4	16.0
20.9	14.7	12.7
22	16.0	16.0
24	15.5	15.0
24.6	14.4	18.9
26.4	15.0	12.8
27	17.1	17.1
27	16.0	16.3
27.8	17.1	18.3
28.1	17.2	19.1
28.5	16.2	15.9
28.6	17.0	17.8
29.5	18.1	14.0
31.4	18.5	21.0
34.1	19.8	20.7

	Chirps per second
Temperature (oC)  [X]	Species A     [Y]	X2	XY
20.8	15.4	432.64	320.32
20.9	14.7	436.81	307.23
22	16	484	352
24	15.5	576	372
24.6	14.4	605.16	354.24
26.4	15	696.96	396
27	17.1	729	461.7
27	16	729	432
27.8	17.1	772.84	475.38
28.1	17.2	789.61	483.32
28.5	16.2	812.25	461.7
28.6	17	817.96	486.2
29.5	18.1	870.25	533.95
31.4	18.5	985.96	580.9
34.1	19.8	1162.81	675.18
400.7	248	10901.25	6692.12

 Normal equation

₁₀

₁₀

248 = 400.7₁ + 15₀

6692.12 = 10901.25₁ + 400.7₀

Solving by elimination method

₁ = 0.3408

Substituting into the equation

248 = 400.7₁ + 15₀

₀ = 7.4292

Equation  in the form of y =mx + c

y = 0.3408x + 7.4292

For Species A, calculate the chirp rate at 45 degrees Celsius

y = 0.3408x + 7.4292

Substituting x = 45 degree

y = 0.3408*45 + 7.4292

= 22.8⁰ 

	Chirps per second
Temperature (oC)  [X]	Species B	X2	XY
20.8	16	432.64	332.8
20.9	12.7	436.81	265.43
22	16	484	352
24	15	576	360
24.6	18.9	605.16	464.94
26.4	12.8	696.96	337.92
27	17.1	729	461.7
27	16.3	729	440.1
27.8	18.3	772.84	508.74
28.1	19.1	789.61	536.71
28.5	15.9	812.25	453.15
28.6	17.8	817.96	509.08
29.5	14	870.25	413
31.4	21	985.96	659.4
34.1	20.7	1162.81	705.87
Total	Total	Total	Total
400.7	251.6	10901.25	6800.84

 251.6 = 400.7₁ + 15₀

6800.84 = 10901.25₁ + 400.7₀

Solving by elimination method

₁ = 0.4045

Substituting into the equation

251.6 = 400.7₁ + 15₀

₀ = 5.969

Equation

y = 0.4045x + 5.969 

Determine at what temperature the species would stop chirping.

y = 0.4045x + 5.969

at y = 0

0 = 0.4045x + 5.969

0.4045x = -5.969

X = -14.8⁰C

The test is correct since it is expected at negative temperature the cricket will stop chirping just as what has been shown in calculation

Site 1 and Site 2 calculations

Evaluation

A regression model involves the following variable

• The unknown parameters denoted by which represent a scalar or a vector
• The dependent variable x
• The dependent variable y

To carry out regression analysis involves finding solution for the known parameter

In our case we involve in determining linear regression , where the model specification of y is a linear combination of of the parameters.

Assumptions

There is one independent variable x and two parameters ₀ and ₁ giving a straight line

Y = ₀ + ₁x

Given a random sample from a population we estimate the population parameters and obtain the sample linear regression model

 , this equation is known as the least squares line of regression of y on x

Relevance

In statistical modeling, regression analysis is a set of statistical processes for estimating the relationships among variables. It includes many techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables (or ‘predictors’). More specifically, regression analysis helps one understand how the typical value of the dependent variable (or ‘criterion variable’) changes when any one of the independent variables is varied, while the other independent variables are held fixed.

Most commonly, regression analysis estimates the conditional expectation of the dependent variable given the independent variables – that is, the average value of the dependent variable when the independent variables are fixed. Less commonly, the focus is on a quantile, or other location parameter of the conditional distribution of the dependent variable given the independent variables. In all cases, a function of the independent variables called the regression function is to be estimated. In regression analysis, it is also of interest to characterize the variation of the dependent variable around the prediction of the regression function using a probability distribution. A related but distinct approach is Necessary Condition Analysis^[1](NCA), which estimates the maximum (rather than average) value of the dependent variable for a given value of the independent variable (ceiling line rather than central line) in order to identify what value of the independent variable is necessary but not sufficient for a given value of the dependent variable.

Q3)

Total number of medals (including gold medals) won by Australian and British Olympic Athletes since World War II.

	AUSTRALIA	GREAT BRITAIN
Olympics	Gold	TOTAL	Gold	TOTAL
1948 London	2	13	3	23
1952 Helsinki	6	11	1	11
1956 Melbourne	13	35	6	24
1960 Rome	8	22	2	20
1964 Tokyo	6	18	4	18
1968 Mexico	5	17	5	13
1972 Munich	8	17	4	18
1976 Montreal	0	5	3	13
1980 Moscow	2	9	5	21
1984 Los Angeles	4	24	5	37
1988 Seoul	3	14	5	24
1992 Barcelona	7	27	5	20
1996 Atlanta	9	41	1	15
2000 Sydney	16	58	11	28
2004 Athens	17	50	9	30
2008 Beijing	14	46	19	47
2012 London	8	35	29	65
2016 Rio De Janiero	8	29	27	67

Australia data

	AUSTRALIA
Olympics	TOTAL
1976 Montreal	5
1980 Moscow	9
1952 Helsinki	11
1948 London	13
1988 Seoul	14
1968 Mexico	17
1972 Munich	17
1964 Tokyo	18
1960 Rome	22
1984 Los Angeles	24
1992 Barcelona	27
2016 Rio De Janiero	29
1956 Melbourne	35
2012 London	35
1996 Atlanta	41
2008 Beijing	46
2004 Athens	50
2000 Sydney	58

 Calculate a five number summary for Australia

The total has been arranged in ascending order as shown on the table

Minimum number = 5

Maximum number = 58

Median number = average of the 9^th number and 10^th number

Median = (22 + 24)/2

= 23

Quartile

Q1 can be thought of as a median in the lower half of the data, and Q3 can be thought of as a median for the upper half of data

5,9,11,13,14,17,17,18,22,24,27,29,35,35,41,46,50,58

Lower quartile (Q1) = (13+14)/2 = 13.5

Upper quartile = (35+41)/2 = 38

Great Britain data 

	GREAT BRITAIN
Olympics	TOTAL
1952 Helsinki	11
1968 Mexico	13
1976 Montreal	13
1996 Atlanta	15
1964 Tokyo	18
1972 Munich	18
1960 Rome	20
1992 Barcelona	20
1980 Moscow	21
1948 London	23
1956 Melbourne	24
1988 Seoul	24
2000 Sydney	28
2004 Athens	30
1984 Los Angeles	37
2008 Beijing	47
2012 London	65
2016 Rio De Janiero	67

Calculate a five number summary for Great Britain

The total has been arranged in ascending order as shown on the table

Minimum number = 11

Maximum number = 67

Median number = average of the 9^th number and 10^th number

Median = (21 + 24)/2

= 22.5

Quartile

Q1 can be thought of as a median in the lower half of the data, and Q3 can be thought of as a median for the upper half of data

11,13,13,15,18,18,20,20,21,23,24,24,28,30,37,47,65,67

Lower quartile (Q1) = (15+18)/2 = 16.5

Upper quartile = (30+37)/2 = 33.5 

Mean of Australian

= 494/18

= 27.44

Mean of Great Britain

= 471/18

= 26.67 

It is clear that Australian gotten more medals when the means was compared similarly to the median which shows that the Great Britain had less

Evaluation

A five number summary involve determination of minimum value from the data, maximum value from the data, median value of the data, lower quartile value of the data and finally upper quartile value of the data.

Where the minimum value is the smallest number of the data, while the maximum value is the highest value of the data, median value is the center values of the data, while the upper quartile is the median of the upper values after the median of all the data while the lower quartile is the median of the lower values before the median of the whole data.

 Assumptions

All the data should be first be arranged in ascending order, failure in which all the five number summary will be wrong.

Relevance

• The five-number summary provides a concise summary of the distributionof the observations.
• The five-number summary gives information about the location (from the median), spread (from the quartiles) and range (from the sample minimum and maximum) of the observations.