Parametric t Test

Conventional statistical procedures are also called parametric tests. In a parametric test a sample statistic is obtained to estimate the population parameter. Because this estimation process involves a sample, a sampling distribution, and a population, certain parametric assumptions are required to ensure all components are compatible with each other. 

For example, in Analysis of Variance (ANOVA) there are three assumptions:

• Observations are independent.
• The sample data have a normal distribution.
• Scores in different groups have homogeneous variances.

In a repeated measure design, it is assumed that the data structure conforms to the compound symmetry. A regression model assumes the absence of collinearity, the absence of auto correlation, random residuals, linearity...etc. In structural equation modeling, the data should be multivariate normal.

Why are they important? Take ANOVA as an example. ANOVA is a procedure of comparing means in terms of variance with reference to a normal distribution. The inventor of ANOVA, Sir R. A. Fisher (1935) clearly explained the relationship among the mean, the variance, and the normal distribution: "The normal distribution has only two characteristics, its mean and its variance. The mean determines the bias of our estimate, and the variance determines its precision." (p.42) It is generally known that the estimation is more precise as the variance becomes smaller and smaller.

Put it in another way: the purpose of ANOVA is to extract precise information out of bias, or to filter signal out of noise. When the data are skewed (non-normal), the means can no longer reflect the central location and thus the signal is biased. When the variances are unequal, not every group has the same level of noise and thus the comparison is invalid. More importantly, the purpose of parametric test is to make inferences from the sample statistic to the population parameter through sampling distributions.

When the assumptions are not met in the sample data, the statistic may not be a good estimation to the parameter. It is incorrect to say that the population is assumed to be normal and equal in variance, therefore the researcher demands the same properties in the sample. Actually, the population is infinite and unknown. It may or may not possess those attributes. The required assumptions are imposed on the data because those attributes are found in sampling distributions. However, very often the acquired data do not meet these assumptions. There are several alternatives to rectify this situation.

Free Sample Trinomials

Trinomials:

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.(source : WIKIPEDIA)

The trinomial must be one of the following form .

Examples for free sample trinomial:

1. 9x + 3y + 5z , where x , y, z are variables.

2. xy + x + 2y, Where x, y are variables.

3. x2+x-8, where x is variable.

4. ax + by + c = 0 , Where a,b,c are constants and x,y are variables.

Our  tutor vista website provide opportunity to learn about sample trinomials with free of cost. In this article we are going to see some sample problems on solving and factoring trinomials.

Free sample problems on trinomials:

Problem 1:

Square the following trinomial,

x+3y-2

Solution:

Given, x+3y-2

We need to find the square for the given trinomial,

That is (x+3y-2)2

We can apply the following formukla to find the square for the above trinomial,

( a + b + c)2 = a2+b2+c2+2ab+2bc+2ca

(x+3y-2)2  = x2+(3y)2+22+2(x)(3y)+2(3y)(-2)+2(-2)(x)

= x2+9y2+4+6xy-12y-4x

= x2+ 9y2+ 6xy - 4x -12y + 4

Answer: (x+3y-2)2 = x2+ 9y2+ 6xy - 4x -12y + 4

Problem 2:

Factor the trinomail x2 + x – 156 .

Solution:

Given , x2 + x – 156 .

- 156 (product)

/     \

-12     13

\    /

1 (sum)

= x2 - 12x + 13x -156

= x ( x - 12 ) + 13 ( x -12 )

= (x+13) (x-12)

Answer: (x+13) and  (x-12) are the factors of the given trinomial.

Problem 3:

Solve the trinomial 2x2 - 2x = 12.

Solution:

Given, 2x2 - 2x = 12.

Subtract 12 on both sides,

2x2 - 2x - 12 = 12 - 12

2x2 - 2x - 12 = 0

2(x2 - x - 6 ) =0

Divide by 2 on both sides,

x2 - x - 6 = 0

-6

/  \

-3  2

\  /

-1

x2 -3x + 2x - 6 = 0

x(x - 3) + 2(x-3) = 0

(x+2) (x-3) = 0

(x+2) = 0

x = -2

(x-3) = 0

x =3

Answer: x = 3 , -2

Practice problems on free sample trinomials:

Problems:

1.Find the roots of the trinomial x2 - 5x = -6

2.Factor the trinomial 2x2 + 12 x - 14

Answer Key:

1. x = -1 , x = 6

2. ( x - 1) and (x + 7)

Examples Functions in Math

Functions in math deals with finding unknown variable from the given expression with the help of known values. In algebraic expression the variable are represented in alphabetic letters. Functions in math, the numbers are consider as constants. Algebraic expression deals with real number, complex number, and polynomials. In algebraic expression several identities to find the x values by using this we can easily find the algebraic expression of the particular function. The example function in math  may include the function of p(x), q(x),… to find the x value of the functions.

Examples function in math:

Q(y) = 4y2+12y + 40. In this equation we need to find the variable of Q(2) functions in math her y is 2.

Problems using examples functions in math

Examples functions in math

Problem 1: Examples functions in math using p(y) = y2 +2y +4

p(y) = y2 +2y +4 find the f(6).

Solution :

Given the function of p(y) there is y value is given function in math

p(y) = y2 +2y +4 find the p(6)

The value of x is 2 is given

p(6) = 62 +2*6 +4

p(6) = 36 +12 +4 In this step 6 square is 36 it is calculate and 2*6 is 12 be added

p(6) = 52.

The functions in math p(6) = y2 +2y +4 find the p(6) is 52.

Problem 2 Examples functions in math using q(x) = x2 +2x +40 find the q(3).

q(x) = x2 +2x +40 find the q(3).

Solution :

Given the examples functions in math of q(x) there is x value is given functions in math

q(x) = x2 +2x +40 find the f(3)

The value of x is 2 is given

q(3) = 32 +2*3 +40

q(3) = 9 + 6 +40 In this step 3 square is 9 it is calculate with 2*3 is 6 be added to 40 to find the example function in math q(3).

q(3) = 55.

The examples functions in math q(3) = x2 +2x +40 find the q(3) is 55

Example functions in math using cubic equation

Problems1: examples functions in math using f(x) = x3 +2x2 + 2x + 4 to find the f(3).

F(x) = x3 +2x2 + 2x + 4 find the functions in math f(3).

Solution

Given the function in math of  f(x) there is x value is given as 3. Find example function in math.

f(x) = x3 +2x2 + 2x + 4 find the f(3).

Here the value of x is given as 3

f(3) = 33 + 2*32 + 2*3 +4

f(3) = 27 +18+ +6 +4 In this step 3cube is calculated  as 27 and  3square is 9.

f(3) = 55.

The math functions of f(x) = x3 +2x2 + 2x + 4 find the f(3) = 55.

Problems 2: Examples functions in math using q(x) = x3 +2x2 + 4x + 25 to find the q(3).

q(x) = x3 +2x2 + 4x + 25 find the functions in math q(3).

q(x) = x3 +2x2 + 4x + 25 find the math function in q(3).

Solution Given the examples functions in math of  q(x) there is x value is given as 3. Find example function in math.

q(x) = x3 +2x2 + 2x + 25 find the f(3).

Here the value of x is given as 3

q(3) = 33 + 2*32 + 2*3 +25

q(3) = 27 +18+ 6 + 25 In this step 3cube is calculated  as 27 and  3 square is 9 multiplied with 2 and add 25 to find the example function in math of q(3).

q(3) = 76 .

The examples functions in math of q(x) = x3 +2x2 + 2x + 25  is the q(3) = 76.

Length of Line Segment

The line segment is the straight line and it has two points. They are starting point and ending point. The starting point is in the starting place of the line and the ending point is in the ending place of the line. The length of line segment is the distance between the starting point and ending point of a line.

Diagram and formula - Length of line segment:

The formula to find the length of the line segment with two points (x1,y1) and (x2, y2) is

Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Example problems - Length of line segment:

Find the length of line segment and the points are (1,1) and (2, 2).

Solution:

Given , (1,1) and (2, 2).

Let us take (1 ,1 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-1)^2 + (2-1)^2)`

=`sqrt(1^2 + 1^2)`

= `sqrt(1 + 1)`

= `sqrt(2)`

Find the length of line segment and the points are (1,1) and (3, 3).

Solution:

Given , (1,1) and (3, 3).

Let us take (1 ,1 ) as (x1, y1) and (3, 3) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((3-1)^2 + (3-1)^2)`

= `sqrt(2^2 + 2^2)`

= `sqrt(4+4)`

= `sqrt(8)` .

= `sqrt(4 * 2)`

= `sqrt(4)` `xx` `sqrt(2)`

= 2 `xx` `sqrt(2)` .

Find the length of line segment and the points are (3,3) and (2, 2).

Solution:

Given , (3,3) and (2, 2).

Let us take (3 ,3 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-3)^2 +(2-3)^2)`

= `sqrt((-1)^2 + (-1)^2)`

= `sqrt(1 + 1)`

= `sqrt(2)` .

Find the length of line segment and the points are (2,2) and (5, 5).

Solution:

Given , (2,2) and (5, 5).

Let us take (2 ,2 ) as (x1, y1) and (5, 5) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((5-2)^2 + (5-2)^2)`

= `sqrt(3^2 + 3^2)`

= `sqrt(9+9)`

= `sqrt(18)`

= `sqrt(9 * 2)`

= `sqrt(9)` `xx` `sqrt(2)`

= 3 `xx` `sqrt(2)`

Mean Difference Standard Deviation

Definition of mean difference:

Mean difference is defined as the measure of the difference between the given data set and mean value.For finding the mean difference first have to find the mean,

Formula for finding the mean,

`barx = (sumx) / n`

Using the mean value mean difference have to be found Formula for mean difference is,

`x-barx`

Definition of Standard Deviation:

Standard Deviation is the determination of describing the variability and spread of the Data set in the given total values in data set. It is used to take the measurement for the average of numbers in the given Data set. Standard Deviation given by the square root for the summation of the total squared mean difference and it is divided by the total number of values minus one.

Formula for standard deviation,

S =` sqrt(((sum(x - barx))) / (n-1))`

Steps for calculating mean difference and standard deviation:

Get the mean for the given n numbers in the given data set.
Get mean difference of each given numbers in the Data set from the mean.
Take Square for all each deviations. It is called as the squared mean deviation.
Calculate the summation for standard  mean deviations.
Now apply the Standard Deviation formula for finding the Standard deviation form the mean.


Mean difference standard Deviation - Example Problems:

Mean difference standard Deviation - Problem1:

Calculate the mean difference and standard deviation in the following data set.56, 52, 54, 57, 58.

Solution:

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (56+ 52+ 54+ 57+ 58) / 5`        

`barx = 277 / 5`

`barx = 55.4`

Mean difference is given below

x                             `(x - barx)`

56                         56 - 55.4 =   0.6

52                         52 - 55.4 =  -3.4

54                        54 - 55.4 =   -1.4

57                        57 - 55.4 =    1.6

58                        58 - 55.4 =    2.6


Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.6)^2+(-3.4)^2+(-1.4)^2+(1.6)^2+(2.6)^2 )/ (5-1))`

S = `sqrt(23.2 / 4)`

S = `sqrt(5.8)`

S = 2.40831892

Mean difference standard Deviation - Problem 2:

Calculate the mean difference and standard Deviation for the given data set. 23, 25, 24, 26.

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (23+ 25+ 24+ 26) /4`        

`barx = 98 / 4`

`barx = 24.5`

Mean difference is given below

x                                  `(x-barx)`

23                        23 - 24.5 =   0.5

25                        25 - 24.5 = - 0.5

24                        24 - 24.5 =   1.5

26                        26 - 24.5 = - 1.5

Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.5)^2+(-0.5)^2+(1.5)^2+(-1.5)^2)/ (4-1))`

S = `sqrt(5 /3)`

S = `sqrt(1.66666667)`



S = 1.29099445

Finding Orthocenter

Orthocentre is point of intersection of all the three altitudes of a triangle.

The altitude of a triangle is the line perpendicular from one vertex to the opposite side.

Steps to find the orthocentre of triangle ABC algebrically :-

1) Find the equation of any two altitudes of the triangle

The slope of the line joining the opposite side BC  is calculated ,

Then,  the perpendicular slope is then calculated.

slope of a line is `(y2 - y1 )/(x2-x1)` =  m ,

perpendicular slope is  `(-1)/(m)`

Given are vertex A and slope of line,

we plug the values in y = mx +b we get the y intercept b

The equation of the altitude is y = mx + b

Same way we find the equation of the other altitude

2) Find point of intersection of the two altitudes is the orthocentre of the triangle.

Steps to find the orthocentre of a triangle graphically:

1) First we plot all the points of the triangle

2) We draw two altitudes of the trianlge

3) The point of intersection of the altitudes is the Orthocentre

Examples of Solving Orthocentre

Find the orthocentre of the of the triangle ABC whose vertices are A(-2,-1) , B ( -1,-4) and C (0,-5) .

Solution:    {i) Equation of altitude AD which is perpendicular to BC

x 2 = -1, y2 = -4 , x 1 = 0 , y1 = -5

slope of a line BC is  `(-4 +5)/(-1-0)` =  `(1)/(-1)`  =  -1

slope of line AD is `(-1)/(-1)` =  1

Line AD passes through A(-2,-1) and slope m = 1,

Equation of line is y = 1x +b

we plug in x = -2, y = -1,  we get,

-1 = -2 + b

-1 +2  = b

b = 1

Equation of AD is   y = x + 1 ------------------> 1
(ii) Equation of Altitude BE perpendicular to AC

x2 = 0, y2 = -5, x1 = -2, y1 = -1

slope of AC =  `(-5+1)/(0+2)` = `(-4)/(2)`  =  -2

Perpendicular slope is  `(-1)/(-2)`  =  `1/(1/2)`  = slope of BE

Equation of BE is y = `(1)/(2)` x + b

B (-1,-4)  Plug in x = -1, y = -4              ,

-4   = -1/2 +b

-4 + 1/2 = b ,  b = 7/2

Altitude BE is        y = `(1)/(2)` x + `(7)/(2)`  ---------------> 2

(iii) Point of intersection of AD , BE

Solve 1 and 2,    x +1  = `(1)/(2)` x + `(7)/(2)`

we get (-9,-8)

Solving Orthocentre of Different Types of Triangles using Graphs

Orthocentre of a right angled triangle is the vertex which is the right angle.

Orthocentre of a an obtuse angled triangle is  outside the triangle.

For an equilateral triangle , the orthocenter lies on the perpendicular bisector of each side of the triangle.

Some graphs to illustrate the above facts:-

Does Geometry use Integers

The geometry generally use the integers and the variables. The answer for the question "Does geometry use integers?" is yes, the geometry uses the integers.For specific purpose of using the geometry with integers, the separate topic available as algebraic geometry. The integers are used for to represent the co-ordinates in vertex, some equations of the line in the 2d geometry. The examples and practice problems are given below for the question "Does geometry use integers or not?".

some examples to explain "does geometry use integers"

• Consider the given points  (4,4), (2,3), (8,3), (6, 2 ), (5, 1) and plot them in the graph. And also we have to denote the quadrant in which each of the point lies.

Solution:

The plotted points in the coordinate plane are shown in the graph. All the co-ordinates of the points are in positive so all will be in the first quadrant itself. And also that  the co-ordinates of the points are not having the zero term, they are all having only the non-zero values. Therefore the points are not lie on the x-axis and the y-axis.

• Consider the points (−3, −4) and (−9, 11) and find the horizontal and the vertical distances between them.

Solution:

For the points (−3, −4) and (−9, 11) The horizontal distance between the two points is a distance between the point corresponding to x coordinates−3 and −9 on the number line x- axis; i.e., (−3) − (−9) = 9 − 3 = 6. and the vertical distances between the two points is the distance between the points corresponding to y co-ordinates −4 and 11 on the number line y axis; i.e., (11) − (−4) = 15.

some more problems to explain "does geometry use integers"

• Consider the line passing through (5,6) and (15,9) and state whether the line is rising up or falling down find the slope.

Solution:

Take (5,6) as (x₁, y₁) and (15, 9) as (x₂, y₂). Then the slope of the line is

m = y₂-y₁ / x₂-x₁

  = `(9-6)/(15-5)`

  = `3/10`

The slope is a positive(+ve) number and so the line is rising up. Here the geometry is used to determine(find) the direction of the lines using the integers.

• Consider the line passing through (−16, 29) and (40, −6) and state whether the line is rising up or falling down and find the slope.
  

Solution:

The slope of the line is

m = y₂-y₁ / x₂-x₁

  = `(-6-29)/(40-(-16))`

  = `-35/56` 

  = `-5/8`

Here m is a negative number, the line will be falling down

In this problem the geometry is used to determine the direction of the lines using the integers.