Probability and Statistics Archives - OnlineEngineeringNotes https://onlineengineeringnotes.com/category/probability-and-statistics/ A Complete Guide to future Engineers Thu, 17 Jun 2021 06:51:59 +0000 en-US hourly 1 https://wordpress.org/?v=6.5.2 Sampling and Estimation: Central limit theorem and its application and concept of point and interval estimation and confidence interval for population mean and proportion https://onlineengineeringnotes.com/2021/06/17/sampling-and-estimation-central-limit-theorem-and-its-application-and-concept-of-point-and-interval-estimation-and-confidence-interval-for-population-mean-and-proportion/ https://onlineengineeringnotes.com/2021/06/17/sampling-and-estimation-central-limit-theorem-and-its-application-and-concept-of-point-and-interval-estimation-and-confidence-interval-for-population-mean-and-proportion/#respond Thu, 17 Jun 2021 06:51:54 +0000 https://onlineengineeringnotes.com/?p=934 Population (Universe) Population means aggregate of all possible units under investigation. It need not be human population. It may be population of plants, population of insects, population of fruits, etc. Finite population When the number of observation can be counted and is definite, it is known as finite population. • No. of plants in a ... Read more

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Population (Universe)

Population means aggregate of all possible units under investigation. It need not be human

population. It may be population of plants, population of insects, population of fruits, etc.

Finite population

When the number of observation can be counted and is definite, it is known as finite population.

• No. of plants in a plot.

• No. of farmers in a village.

• All the fields under a specified crop.

Infinite population

When the number of units in a population is innumerably large, that we cannot count all of them, it is known as infinite population.

• The plant population in a region.

• The population of insects in a region.

Frame

A list of all units of a population is known as frame.

Parameter

A summary measure that describes any given characteristic of the population is known as parameter (in another word constant drawn from population is parameter). Population are described in terms of certain measures like mean, standard deviation etc. These measures of the population are called parameter and are usually denoted by Greek letters. For example, population mean is denoted by μ, standard deviation by σ and variance by σ2.

Sample

A portion or small number of unit of the total population is known as sample.

• All the farmers in a village(population) and a few farmers(sample)

• All plants in a plot is a population of plants.

• A small number of plants selected out of that population is a sample of plants.

Statistic

A summary measure that describes the characteristic of the sample is known as

Statistic (in other word constant drawn from sample observation is statistic). Thus sample mean , sample standard deviation (s) etc is statistic. The statistic is usually denoted by roman letter.

The statistic is a random variable because it varies from sample to sample.

Sampling

The method of selecting samples from a population is known as sampling.

Sampling technique

There are two ways in which the information is collected during statistical survey.

They are

1. Census survey

2. Sample survey

Census

It is also known as population survey and complete enumeration survey. Under census survey the information are collected from each and every unit of the population or universe.

Sample survey

A sample is a part of the population. Information are collected from only a few units of a population and not from all the units. Such a survey is known as sample survey. Sampling technique is universal in nature, consciously or unconsciously it is adopted in every day life.

For eg.

1. A handful of rice is examined before buying a sack.

2. We taste one or two fruits before buying a bunch of grapes.

3. To measure root length of plants only a portion of plants are selected from a plot.

Need for sampling

The sampling methods have been extensively used for a variety of purposes and

in great diversity of situations. In practice it may not be possible to collected information on all units of a population due to various reasons such as

1. Lack of resources in terms of money, personnel and equipment.

2. The experimentation may be destructive in nature. Eg- finding out the germination percentage of seed material or in evaluating the efficiency of an insecticide the experimentation is destructive.

3. The data may be wasteful if they are not collected within a time limit. The census survey will take longer time as compared to the sample survey. Hence for getting quick results sampling is preferred. Moreover, a sample survey will be less costly than complete enumeration.

4. Sampling remains the only way when population contains infinitely many number of units.

5. Greater accuracy.

Sampling methods

The various methods of sampling can be grouped under;

  1. Probability sampling or random sampling

Probability samplingis a technique in which every unit in the population has a chance

(non-zero probability) of being selected in the sample, and this chance can be accurately determined. Sample statistics thus produced, such as sample mean or standard deviation, are unbiased estimates of population parameters, as long as the sampled units are weighted according to their probability of selection. All probability sampling has two attributes in common: (1) every unit in the population has a known non-zero probability of being sampled, and (2) the sampling procedure involves random selection at some point.

  • Non-probability sampling or non-random sampling

Nonprobability samplingis a sampling technique in which some units of the population have zero chance of selection or where the probability of selection cannot be accurately determined. Typically, units are selected based on certain non-random criteria, such as quota or convenience. Because selection is non-random, nonprobability sampling does not allow the estimation of sampling errors, and may be subjected to a sampling bias. Therefore, information from a sample cannot be generalized back to the population.

Probability sampling or Random sampling

Under this method, every unit of the population at any stage has equal chance (or)

each unit is drawn with known probability. It helps to estimate the mean, variance etc of

the population. Under probability sampling there are two procedures

1. Sampling with replacement (SWR)

2. Sampling without replacement (SWOR)

When the successive draws are made with placing back the units selected in the

preceding draws, it is known as sampling with replacement. When such replacement is

not made it is known as sampling without replacement. When the population is finite sampling with replacement is adopted otherwise SWOR is adopted.

There are many kinds of random sampling technique. Some of them are.

1. Simple Random Sampling

2. Systematic Random Sampling

3. Stratified Random Sampling

4. Cluster Sampling

Simple Random sampling (SRS)

There are two methods in SRS

1. Lottery method

2. Random no. table method

Lottery method

This is most popular method and simplest method. In this method all the items of

the universe are numbered on separate slips of paper of same size, shape and color. They

are folded and mixed up in a drum or a box or a container. A blindfold selection is made.

Required number of slips is selected for the desired sample size. The selection of items

thus depends on chance.

For example, if we want to select 5 plants out of 50 plants in a plot, we number the 50 plants first. We write the numbers from 1-50 on slips of the same size, role them and mix them. Then we make a blindfold selection of 5 plants. This method is also called unrestricted random sampling because units are selected from the population without any restriction. This method is mostly used in lottery draws. If the population is infinite, this method is inapplicable. There is a lot of possibility of personal prejudice if the size and shape of the slips are not identical.

Random number table method

As the lottery method cannot be used when the population is infinite, the alternative method is using of table of random numbers. There are several standard tables of random numbers. But the credit for this technique goes to Prof. LHC. Tippet (1927). The random number table consists of 10,400 four-figured numbers. There are various other random numbers. They are fishers and

Yates (19380 comprising of 15,000 digits arranged in twos. Kendall and B.B Smith

(1939) consisting of 1, 00,000 numbers grouped in 25,000 sets of 4 digits random

numbers, Rand corporation (1955) consisting of 2, 00,000 random numbers of 5 digits

each etc.,

Merits

1. There is less chance for personal bias.

2. Sampling error can be measured.

3. This method is economical as it saves time, money and labor.

Demerits

1. It cannot be applied if the population is heterogeneous.

2. This requires a complete list of the population but such up-to-date lists are not available in many enquires.

3. If the size of the sample is small, then it will not be a representative of the

population.

Stratified Sampling:

When the population is heterogeneous with respect to the characteristic in which

we are interested, we adopt stratified sampling. When the heterogeneous population is divided into homogenous sub-population, the sub-populations are called strata. From each stratum a separate sample is selected using simple random sampling. This sampling method is known as stratified sampling. We may stratify by size of farm, type of crop, soil type, etc.

The number of units to be selected may be uniform in all strata (or) may vary from stratum to stratum.

There are four types of allocation of strata

1. Equal allocation

2. Proportional allocation

3. Neyman’s allocation

4. Optimum allocation

If the number of units to be selected is uniform in all strata it is known as equal allocation of samples.

If the number of units to be selected from a stratum is proportional to the size of the stratum, it is known as proportional allocation of samples.

When the cost per unit varies from stratum to stratum, it is known as optimum allocation.

When the costs for different strata are equal, it is known as Neyman’s allocation.

Merits

1. It is more representative.

2. It ensures greater accuracy.

3. It is easy to administrate as the universe is sub-divided.

Demerits

1. To divide the population into homogeneous strata, it requires more money, time

and statistical experience which is a difficult one.

2. If proper stratification is not done, the sample will have an effect of bias.

Systematic sampling:

The systematic sampling technique is operationally more convenient than simple random sampling. It also ensures, at the same time that each unit has an equal probability of inclusion in the sample. In this method of sampling, the first unit is selected with the help of random numbers, and the remaining units are selected automatically according to a predetermined pattern. This method is known as systematic sampling.

Suppose the N units in the population are numbered 1 to N in some order. Suppose further that N is expressible as a product of two integers n and k, so that N = nk.

To draw a sample of size n,

 – select a random number between 1 and k.

 – Suppose it is i.

– Select the first unit, whose serial number is i.

– Select every kth unit after ith unit.

 – The sample will contain i, i+k, i+2k, i+3k, ……… i+(n-1) k.

For example: A researcher wants to select a systematic random sample of 10 people from a population of 100. If she or he has a list of all 100 people, he would assign each person a number from 1 to 100. The researcher then picks a random number from 1 to 10. let us say selected number is 6, which is taken as starting number (or random start). He or she would then select every tenth person for the sample (because the sampling interval(k) = 100/10=10). The final sample would contain those individuals who were assigned the following numbers: 6,16,26,36,46,56,66,76,86,96.

Merits

(a) This is a simple method of selecting a sample.

(b) It reduces the field cost.

(c) Inferential statistics may be used.

(d) Sample may be comprehensive and representative of population.

(e) Observations of the sample may be used for drawing conclusions and generalizations.

Demerits

(a) This is not free from error, since there is subjectivity due to different ways of systematic list by different individuals. Knowledge of population is essential.

(b) Information of each individual is essential.

(c) This method can’t ensure the representativeness.

(d) There is a risk in drawing conclusions from the observations of the sample.

Cluster sampling:

Cluster sampling is a sampling technique where the entire population is divided into groups, or clusters and a random sample of these clusters are selected. All observations in the selected clusters are clusters are included in the sample. Here, in order to increase the precision of the estimates, the population should be partitioned into clusters in such a way that variation within clusters should be as large as possible, while the variation between clusters should be as small as possible.

            Cluster sampling is typically used when the researcher cannot get a complete list of the members of the population they wish to study but can get a complete list of groups or clusters of the population. It is also used when a random sample would produce a list of subjects so widely scattered that surveying them would prove to be far too expensive.

For example: the list of all the agricultural farms in a village or district may not be easily available but the list of villages and the districts are available. In each case farm is sampling unit and every village or the district is the cluster.

Merits

(a) It may be a good representative of the population.

(b) It is an easy method.

(c) It is an economical method.

(d) It is practicable and highly applicable in education.

(e) Observations can be used for inferential purpose.

Demerits

(a) Cluster sampling is not free from error.

(b) It is not comprehensive.

Sampling distribution of mean:

Mean

The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Therefore, if a population has a mean μ, then the mean of the sampling distribution of the mean is also μ. The symbol  is used to refer to the mean of the sampling distribution of the mean. Therefore, the formula for the mean of the sampling distribution of the mean can be written as:

Variance

The variance of the sampling distribution of the mean is computed as follows:

That is, the variance of the sampling distribution of the mean is the population variance divided by ‘n’, the sample size. Thus, the larger the sample size, the smaller the variance of the sampling distribution of the mean.

The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is therefore the square root of the variance of the sampling distribution of the mean and can be written as:

Sampling distribution of Proportion:

The Central limit theorem(CLT):

Example:

A population consists of five members 2,3,6,8 and 11.

  • Draw all possible samples of size 2 that can be drawn from this population without replacement.
  • Find mean and variance of the population.
  • Find the mean of sampling distribution of means and show that it is equal to the population mean.
  • Find the variance of sampling distribution of means and also verify with the formula
  • Find the standard error of the mean.

Solution:

Here, N = 5, n = 2

Possible number of samples of size 2 which can be drawn from the population of size 5 without replacement is,

Possible sample are:

{(2,3), (2,6), (2,8), (2,11), (3,6), (3,8), (3,11), (6,8), (6,11), (8,11)}

Calculation of population mean and variance.

Example:

Consider a population of 4 units with values 3,6,2,1.

  • Write down all possible samples of size of 2 that can be drawn with replacement for this population.
  • Find mean and variance of the population.
  • Find mean of the sampling distribution of means and show that it is equal to the population mean.
  • Find the variance of the sampling distribution of means , also verify that it agrees with the formula,

where = population variance and n = sample size.

  • Find the standard error of the mean.

Solution:

Here population consists of 4 values 3,6,2,1. Thus sample size of 2 can be drawn with replacement from the population of size 4 in N2 = 42 =16 ways.

Thus possible samples are: {(3,3), (3,6), (3,2), (3,1), (6,3), (2,3), (1,3), (6,6), (6,2), (6,1), (2,6), (1,6), (2,2), (2,1), (1,2), (1,1)}

Mean and variance of the population.

Estimation:

 It is a process of obtaining the characteristics of a population from the information contained in the sample. Samples are used instead of the whole population for various reasons; financial or time constraints, checking the items involves a destructive process or the whole population is not accessible. The population characteristic (Parameter) is thus estimated from the sample characteristic (Statistic). An estimate is a value calculated from sample observations while an estimator is the Statistic whose value can be used to guess the value of the population parameter.

Characteristics of good estimator:

 A good estimator must be

  1.  Unbiased
  2.  Efficient
  3. Consistent and
  4. Sufficient.
  • Unbiasedness: An estimator (tn) is said to be unbiased if its expected value is equal to the parameter. i.e.  E(tn) =θ, where tn is an estimate.
  • Consistency: An estimator is consistent if, as the sample size (n) increases, the estimate approaches the true population value. This implies that as the sample size increases, the variance of the estimator tends to zero.
  • Efficiency:  This implies having the minimum variance among the group of unbiased estimators. i.e., An estimator tn and tn’ are two consistent estimators the, tn is more efficient than tn’ if V(tn)< V(tn’).
  • Sufficiency: An estimator is said to be sufficiency for a parameter, if it contains all the information in the sample regarding the parameter.

Example:1

A population consists of value 1,2,3,6,8. Prove that the sample mean of sample size 3 is unbiased estimator of population mean. (without replacement)

Solution:

Here population consists of 5 values 1,2,3,6 and 8. So population mean of five observation is,

These 10 ways are tabulated in figure as,

So we can say that sample mean of size 3 is unbiased estimator of population mean.

Types of estimation:

There are two types of estimation

  1. Point estimation: If a single value is estimated from the sample observation as an unbiased estimate of population parameter then, the method is point estimation.
  2. Interval estimation: The interval (C1, C2) within which the unknown value of population parameter is expected to lie with confidence probability (1-α) is confidence interval and the method of finding (C1, C2) using confidence probability (1-α) is interval estimation.

where,      

C1 = Lower confidence limit (LCL).

C2 = Upper confidence limit (UCL).

 θ= Parameter and

(1-α) = Confidence probability.

The confidence level describes the uncertainty associated with a sampling method. Suppose we used the same sampling method to select different samples and to compute a different interval estimate for each sample. Some interval estimates would include the true population parameter and some would not. A 90% confidence level means that we would expect 90% of the interval estimates to include the population parameter; a 95% confidence level means that 95% of the intervals would include the parameter; and so on.

Confidence interval for Population mean (µ ) for large sample (

a. When the sample size is large (i.e. n ≥ 30), the (1-α) x100% confidence interval for µ is, 

Where α is the level of significance, mostly taken as 5% (0.05) or 1% (0.01), is the standardized value of the given level of confidence {(1-α) x100%} obtained from tables of Normal distribution. The values for 5% and 1% level are:

Determination of sample size:

            If population standard deviation (σ )and maximum size of error (E)  are known then we can estimate the sample size for the survey as;

When the sample size is small (n< 30 and σ is unknown), the (1-α) x100% confidence interval for µ is assumed to have t-distribution with (n-1) degree of freedom.

Example:

In an experiment to estimate the heart beat rate per minute for a certain male population, a random sample of 49 men were selected and the average heart beat was found to be 130 per minute. if the population variance is 100 pulse rate per minute, find (i)95% (ii) 99% confidence interval for the population mean (µ).

Solution:

Since n is large (i.e. n>30), we assume normal distribution.

Example:

A consumer group wishes to estimate the average electric bills for the month of July for single –family homes in a large city. Based on studies conducted in other cities the standard deviation is assumed to be Rs. 250. The group wants to estimate the average bill for July to within Rs. 50 of the true average with 99% confidence.

  1. What sample size is needed?
  2. If 95% confidence is desired, what sample size (n) is needed?

Solution:

Here, standard deviation  =Rs. 250, maximum size of error (E) =50

Example:4

From a population of 540, a sample of 60 individuals is taken. From this sample, the mean is found to be 6.2 and standard deviation 1.368.

  1. Find the standard error of the mean.
  2. Construct 96% confidence for the mean.

Solution:

Example:

The weights of 10 pieces of potatoes from a farm are 9.8, 9.9, 10.3, 10.1, 10.4, 10.3, 10.2, 9.7, 10.1 and 9.8 grams. Find the (i) 95% and (ii) 99% confidence interval for the Population mean of all potatoes in the farm.

Solution:

Since ‘n’ is small and population standard deviation (σ) is unknown (i.e. n< 30 and σ unknown), we assume the student t distribution.

Now, calculation of sample mean and sample s.d.

xx2
9.896.04
9.998.01
10.3106.09
10.1102.01
10.4108.16
10.3106.09
10.2104.04
9.794.09
10.1102.01
9.896.04
100.61012.58

Confidence interval for population proportion (P)

Example:

A survey of 500 people shopping at a shopping mall, selected at random, showed that 350 of them used cash and 150 of them used credit cards. Construct a 95% confidence interval estimate of the proportion of all the persons at the mall, who use cash for shopping.

Solution:

Here, sample size (n)= 500, out of which 350 used cash for shopping.

So sample proportion of cash users (p) =  and q = (1-p) =1-0.7 = 0.3

Then 95% CI for population proportion P is

Reference : Probability and statistics for engineers by Toya Naryan Paudel & Pradeep Kunwar, Sukunda Pustak Bhawan.

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Simple Linear Regression and Correlation: Concept of simple regression analysis, estimation of regression coefficient by using square estimation method and Standard error, coefficient of determination https://onlineengineeringnotes.com/2021/05/31/simple-linear-regression-and-correlation-concept-of-simple-regression-analysis-estimation-of-regression-coefficient-by-using-square-estimation-method-and-standard-error-coefficient-of-determination/ https://onlineengineeringnotes.com/2021/05/31/simple-linear-regression-and-correlation-concept-of-simple-regression-analysis-estimation-of-regression-coefficient-by-using-square-estimation-method-and-standard-error-coefficient-of-determination/#respond Mon, 31 May 2021 07:56:13 +0000 https://onlineengineeringnotes.com/?p=648 CORRELATION In many natural systems, changes in one attribute are accompanied by changes in another attribute and that a definite relation exists between the two. In other words, there is a correlation between the two variables. For instance, several soil properties like nitrogen content, organic carbon content or pH are correlated and exhibit simultaneous variation. ... Read more

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CORRELATION

In many natural systems, changes in one attribute are accompanied by changes in another attribute and that a definite relation exists between the two. In other words, there is a correlation between the two variables. For instance, several soil properties like nitrogen content, organic carbon content or pH are correlated and exhibit simultaneous variation. Strong correlation is found to occur between several morphometric features of a tree. In such instances, an investigator may be interested in measuring the strength of the relationship. Having made a set of paired observations (xi,yi); i = 1, …, n, from n, independent sampling units, a measure of the linear relationship between two variables can be obtained by the following quantity called Pearson’s product moment correlation coefficient or simply correlation coefficient.

Definition: If the change in one variable affects a change in the other variable, the two variables are said to be correlated and the degree of association ship (or extent of the relationship) is known as correlation.

Types of correlation:

a). Positive correlation:  If the two variables deviate in the same direction, i.e., if the increase (or decrease) in one variable results in a corresponding increase (or decrease) in the other variable, correlation is said to be direct or positive.

Ex:     (i) Heights and weights

(ii) Household income and expenditure

(iii) Amount of rainfall and yield of crops

(iv) Prices and supply of commodities

(v) Feed and milk yield of an animal

(vi) Soluble nitrogen and total chlorophyll in the leaves of paddy.

b) Negative correlation: If the two variables constantly deviate in the opposite direction i.e.,  if  increase  (or  decrease)  in  one  variable  results  in  corresponding                                                                         decrease   (or increase) in the other variable, correlation is said to be inverse or negative.

Ex:     (i) Price and demand of a goods

(ii) Volume and pressure of perfect gas

(iii) Sales of woolen garments and the day temperature

(iv) Yield of crop and plant infestation

c) No or Zero Correlation: If there is no relationship between  the two variables such that the value of one variable change and the other variable remain constant is called no or zero correlation.

Methods of studying correlation:

1. Scatter Diagram                                                              

2. Karl Pearson‟s Coefficient of Correlation                     

3. Spearman‟s Rank   Correlation                                        

4.  Regression Lines

1. Scatter Diagram 

It is the simplest way of the diagrammatic representation of bivariate data. Thus for the bivariate distribution (xi,yi); i = j = 1,2,…n,          If  the  values  of  the variables X and Y be plotted along the X-axis and Y-axis respectively in the xy-plane, the diagram of dots so obtained is known as scatter diagram.  From the scatter diagram, if the points are very close to each other, we should expect a fairly good amount of correlation between the variables and if the points are widely scattered, a poor correlation is expected. This method, however, is not suitable if the number of observations is fairly large.

If the plotted points show an upward trend of a straight line, then we say that both the variables are positively correlated.

When the plotted points shows a downward trend of a straight line then we say that both the variables are negatively correlated.

If the plotted points spread on whole of the graph sheet, then we say that both the variables are not correlated.

2. Karl Pearson‟s Coefficient of Correlation   

Prof. Karl Pearson, a British Biometrician suggested a measure of correlation between two variables. It is known as Karl Pearson‟s coefficient of correlation. It is useful for   measuring the degree and direction of linear relationship between the two variables X and Y. It is usually denoted by rxy or r(x,y) or ‘r’.

a. Direct method

b. When deviations are taken from an assumed mean.

Where,

n = number of items;  dx = x-A, dy = y-B                                                                    

A = assumed value of x and B = assumed value of y

c. By changing origin

mean value of Y series.

If U = X-A and V = Y-B

Where,          A= Assumed mean value of X series.

                     B = Assumed

d. By changing origin and scale.

Where,          A = Assumed mean in value of X series

                     B = Assumed mean in value of Y series.

                     h = common factor or width of class interval of X series.

                     k = Common factor or width of class interval of Y series.

  • The correlation coefficient never exceeds unity.  It always lies between –1 and +1.                                         (i.e. –1≤ r  ≤1)
  • If r = +1 then we say that there is a perfect positive correlation between x and y.
  • If r = -1 then we say that there is a perfect negative correlation between x and y .
  • If r = 0 then the two variables x and y are called uncorrelated variables.
  • Correlation coefficient (r) is the geometric mean between two regression coefficients. i.e. , where bxy and byx are regression coefficient of X on Y and Y on X respectively.
  • Correlation coefficient is independent of change of origin and scale.
Interpretation of correlation coefficient
If the correlation coefficient lies between,
±1 ,                      : Then correlation is perfect. 
±0.75 to ±1        : Then correlation is very high (significant).
±0.50 to ±0.75  : Then the correlation is high.
±0.25 to ±0.50  : Then the correlation is low.
0 to ± 0.25,    : Then the correlation is very low (insignificant)
0,                  : No correlation.

Test for significance of correlation coefficient

If “r” is the observed correlation coefficient in a sample of ‘n’ pairs of observations from a bivariate normal population, then Prof. Fisher proved that under the null hypothesis

H0 : r = 0

the variables x, y follows a bivariate normal distribution.   If the population correlation coefficient of x and y is denoted by ρ, then it is often of interest to test whether ρ is zero or different from zero, on the basis of observed correlation coefficient “r”. Thus if “r” is the sample correlation coefficient based on a sample of

“n” observations, then the appropriate test statistic for testing the null hypothesis                             

   H0 : r  = 0 against the alternative hypothesis H1: r ≠ 0 is

“t” follows Student’s t – distribution with (n-2) d.f.

If calculated value of |t| > is greater than critical value of t (n-2) d.f. at specified level of significance, then the null hypothesis is rejected. That is, there may be significant coefficient correlation between two variables. Otherwise, the null hypothesis is accepted.

Example: Consider the data on pH and organic carbon content measured from soil

samples collected from 12 pits taken in natural forests, given in table.

Table : Values of pH and organic carbon content observed in soil samples

collected from natural forest.

Soil pit123456789101112
pH(x)5.76.15.25.75.65.15.85.55.45.95.35.4
Organic carbon(%) (y)2.102.171.971.392.261.291.171.142.091.010.891.60

Find the correlation coefficient between values of pH and organic carbon content. Also test the significance of observed correlation coefficient at 5% level of significance.

Solution:

Direct method;

Calculation of correlation

Soil pitpH(x)Organic carbon (%) (y)(x2)(y2)(x y)
15.72.132.494.4111.97
26.12.1737.214.708913.24
35.21.9727.043.880910.24
45.71.3932.491.93217.92
55.62.2631.365.107612.66
65.11.2926.011.66416.58
75.81.1733.641.36896.79
85.51.1430.251.29966.27
95.42.0929.164.368111.29
105.91.0134.811.02015.96
115.30.8928.090.79214.72
125.41.629.162.568.64
Total66.719.08371.7133.1124106.28

Then, correlation coefficient is,

Here, r = 0.138, which means there is positive and very low correlation or insignificant correlation between ph and organic carbon content.

Test of significance of observed correlation coefficient.

            Null hypothesis,                      H0: r = 0

            Alternative hypothesis,           H1: r ≠0

            Under null hypothesis test statistic is ;

The critical value of t is 2.23, for 10 degrees of freedom at the probability level, a = 0.05. Since the computed t value is less than the critical value, we conclude that the pH and organic carbon content measured from soil samples are not significantly correlated.

Rank correlation

Before learning about Spearman’s correlation it is important to understand Pearson’s correlation which is a statistical measure of the strength of a linear relationship between paired data. Its calculation and subsequent significance testing of it requires the following data assumptions to hold:

 · interval or ratio level;

 · linearly related;

 · bivariate normally distributed.

 If your data does not meet the above assumptions, then use Spearman’s rank correlation.

Monotonic function

            To understand Spearman’s correlation, it is necessary to know what a monotonic function is. A monotonic function is one that either never increases or never decreases as its independent variable increases. The following graphs illustrate monotonic function.

· Monotonically increasing – as the ‘x’ variable increases the ‘y’ variable never decreases;

· Monotonically decreasing – as the ‘x’ variable increases the ‘y’ variable never increases;

· Not monotonic – as the ‘x’ variable increases the ‘y’ variable sometimes decreases and sometimes increases.

3. Spearman’s rank correlation:

When the data is non- normal or qualitative, e.g., intelligence, beauty, honesty, taste etc. Karl Pearson’s coefficient of correlation cannot be used. In such cases we arrange the variables or individuals in order of merit and proficiency and assign ranks. The coefficient of rank correlation as given by Spearman is

Where, rs = Spearman’s rank correlation coefficient

d = difference of corresponding ranks (i.e., d = xi-yi or d = Rx-Ry )

Note:

If ranks are repeated or tied then above formula of Spearman’s rank correlation coefficient become,

Where denotes the number of times an item is repeated.

Example :

Calculate the rank correlation coefficient from the following data.

X8078757568676859
Y1213141414161517

Solution:

Let Rx be the rank of X series and Ry be the rank of Y series.

Calculation of Rank correlation coefficient

XYRxRyd = Rx –Ryd2
801218-749
781327-525
75143.55-1.52.25
75143.55-1.52.25
68145.550.50.25
671672525
68155.532.56.25
591781749
Total    159

Here,  in X series two items 75 and 68 repeated 2-times each.

So, m1 = 2, m2 = 2

Again in Y series one item 14 is repeated 3 times.

So, m3 = 3 Thus rank correlation coefficient is ;

Here rs = – 0.93, which shows that there is very high but negative relationship between two variables X and Y.

REGRESSION

Correlation coefficient measures the extent of interrelation between two variables which are simultaneously changing with mutually extended effects. In certain cases, changes in one variable are brought about by changes in a related variable but there need not be any mutual dependence. In other words, one variable is considered to be dependent on the other variable changes, in which are governed by extraneous factors. Relationship between variables of this kind is known as regression. When such relationships are expressed mathematically, it will enable us to predict the value of one variable from the knowledge of the other. For instance, the photosynthetic and transpiration rates of trees are found to depend on atmospheric conditions like

temperature or humidity but it is unusual to expect a reverse relationship. However, in many cases, it so happens that the declaration of certain variables as independent is made only in a statistical sense although when reverse effects are conceivable in such cases. For instance, in a volume prediction equation, tree volume is taken to be dependent on dbh although the dbh cannot be considered as independent of the effects of tree volume in a physical sense. For this reason, independent variables in the context of regression are sometimes referred as regressor or preditor or explanatory variables and the dependent variable is called the regressand or explained variable.

            The dependent variable is usually denoted by y and the independent variable by x.When only two variables are involved in regression, the functional relationship is known as simple regression. If the relationship between the two variables is linear, it is known as simple linear regression, otherwise it is known as nonlinear regression. When one variable is dependent on two or more independent variables, the functional relationship between the dependent and the set of independent variables is known as multiple regression.

Simple linear regression

The simple linear regression of y on x in the population is expressible as

y = a + bx + e

Where y = dependent variable

 x = independent variable

 α = y- intercept (expected value of y when x assumes the value zero)

 β = regression coefficient of y on x or slope of the line. (it measures the average change in the value of Y when unit change in the value of X)

ɛ = residual or error.

The slope of a linear regression line may be positive, negative or zero depending on the relation between y and x. In practical applications, the values of a and b are to be estimated from observations made on y and x variables from a sample. For instance, to estimate the parameters of a regression equation proposed between atmospheric temperature and transpiration rate of trees, a number of paired observations are made on transpiration rate and temperature at different times of the day from a number of trees. Let such pairs of values be designated as (xi, yi); i = 1, 2, . . ., n where n is the number of independent pairs of observations. The values of a and b are estimated using the method of least squares such that the sum of squares of the difference between the observed and expected value is minimum. In the estimation process, the following assumptions are made viz., (i) The x values are non-random or fixed (ii) For any given x, the variance of y is the same (iii) The y values observed at different values of x are completely independent. Appropriate changes will need to be made in the analysis when some of these assumptions are not met by the data. For the purpose of testing hypothesis of parameters, an additional assumption of normality of errors will be required.

Estimating the values of a and b using least square method.

  1. Regression equation of Y on X.

It is the line which gives the best estimates for the values of Y for any specified values of X.

The regression equation of Y on X is given as

Y = α +βX ………………(1)

The values of two constants α and β can be determined by solving following two normal equations (applying principle of least square)

Now, substituting the value of α and β in equation (1), we get the required estimated regression equation of Y on X as,

Alternative method of estimating a and b.  

 Let y = a + bx + e be the simple linear regression equation of y on x

Then a and b are estimated directly as,

Testing the significance of the regression coefficient

Once the regression function parameters have been estimated, the next step in

regression analysis is to test the statistical significance of the regression function. It is

usual to set the null hypothesis against the alternative hypothesis, H1 : b ¹

0 or (H1 : b < 0 or H1 : b > 0, depending on the anticipated nature of relation). For the

testing, we may use the analysis of variance procedure.

Testing procedure:

  • Set up hypothesis as;

Null hypothesis, H0: b = 0

Alternative hypothesis, H1: b ¹ 0 or (H1: b < 0 or H1: b > 0, depending on the anticipated nature of relation)

  • Construct an outline of analysis of variance table as follows.
Source of Variation (SV)Degree of freedom (d.f.)Sum of square (SS)Mean sum of square (MSS=SS/d.f.)F-ratio
Due to regression Error1 n-2SSR SSEMSR MSEMSR/MSE
Totaln-1TSS  
  • Compute the different sums of squares as follows

a. Total sum of square

b.  Sum of square due to regression

c. Sum of squares due to deviation from regression

  • Enter the values of sums of squares in the analysis of variance table and perform the rest of the calculations.
  • Compare the computed value of F with tabular value at (1,n-2) degrees of freedom. If the calculated value is greater than the tabular value of F  at (1,n-2) degrees of freedom at α% level of significance then F value is significant otherwise insignificance. If the computed F value is significant, then we can state that the regression coefficient, b is significantly different from 0.

Coefficient of determination:

The coefficient of determination (r2) is a measure of the variation of the dependent variable that is explained by the regression line and the independent variable. It is often used as a measure of the correctness of a model, that is, how well a regression model will fit the data. A ‘good’ model is a model where the regression sum of squares is close to the total sum of squares, SSR TSS . A “bad” model is a model where the residual sum of squares is close to the total sum of squares, SSE TSS . The coefficient of determination represents the proportion of the total variability explained by the model.            

i.e.,      

Here,

Total sum of square    = Regression sum of square +Error sum of square

                                    TSS     = RSS   +   ESS

Of course, it is usually easier to find the coefficient of determination by squaring correlation coefficient (r) and converting it to a percentage. Therefore, if r = 0.90, then r 2 = 0.81, which is equivalentto 81%. This result means that 81% of the variation in the dependent variable isaccounted for by the variations in the independent variable. The rest of the variation,0.19, or 19%, is unexplained. This value is called the coefficient of non-determination andis found by subtracting the coefficient of determination from 1. As the value of r approaches 0, r2 decreases more rapidly. For example, if r = 0.6, then r 2 = 0.36, whichmeans that only 36% of the variation in the dependent variable can be attributed to thevariation in the independent variable. Another way to arrive at the value for r2 is to square the correlation coefficient.

The coefficient of determination can have values 0 ≤ r2 ≤ 1. The “good” model means that r2 is close to one.

Standard error  of the estimate:

The standard error of estimate is the measure of variation of an observation made around the estimated regression line. Simply, it is used to check the accuracy of predictions made with the regression line.

Like, a standard deviation which measures the variation in the set of data from its mean, the standard error of estimate measures the variation in the actual values of Y from the estimated values of Y (predicted) on the regression line. The smaller the value of a standard error of estimate the closer are the dots to the regression line and better is the estimate based on the equation of the line. If the standard error is zero, then there is no variation corresponding to the computed line and the correlation will be perfect.

The standard error of estimate of ‘Y’ for given ‘X’ (i.e. Y on X) is denoted by Syx or Se and given by;


Where; Y= value of dependent variable.

            = estimated value of Y.

            n = number of observations in the sample.

Here, we use (n-2) instead of ‘n’ because of the fact that two degree of freedom are lost in estimating the regression line as the value of two parameters ‘β0’ and ‘β1’ are to be determined from the data.

Above formula is not convenient as it requires to calculate the estimated value of Y i.e. Thus, more convenient and easy formula is given below:

Where;

X and Y are independent and dependent variables respectively.
β0 = Y- intercept.

β1 = slope of the line.

n = number of observations in the sample.

Properties of regression lines and coefficients:

  1. Two regression lines pass through the mean point .
  2. Two regression lines coincide when the variables X and Y are perfectly correlated.

i.e. (r = ±1).

  • The two regression lines are right angle to each other if X and Y are independent to each other. (i.e. when r = 0)
  • The two regression lines must have same sign. Also, the correlation coefficient and regression coefficients must have the same sign.
  • Correlation coefficient is the geometric means between the two regression coefficients.
  •  The product of two regression coefficients must be less than or equal to 1.
  • Regression coefficient is independent of change of origin but not scale.

Example:

The following observations are from a morphometric study of cottonwood trees. The widths of 12 leaves from a single tree were measured (in mm) while fresh and after drying.

Fresh leaf width (X)9011555110761008495849510090
Dry leaf width (Y)88109521057195789077919686
  1. Estimate the parameters of simple linear regression using least square method and using the model predict the dry leaf width on the basis of 95 mm fresh leaf width.
  2. Find the linear relationship between fresh leaf and dry leaf width.
  3. Calculate the coefficient of determination and interpret it in practical manner.
  4. Find the standard error (Se) of the estimate.
  5. Test the significance of regression coefficient (β) at 5% level of significance.

Solution:

  • Let y= α+ β1 x be the simple linear regression equation of Y on X.

Then its normal equations are;

Fresh leaf (x)Dry leaf width (y)x2y2xy
9088810077447920
115109132251188112535
5552302527042860
110105121001102511550
7671577650415396
100951000090259500
8478705660846552
9590902581008550
100771000059297700
9591902582818645
100961000092169600
9086810073967740
111010381054329242698548

Substituting the sum values in above two normal equations we get;

1038 = 12α +1110 β   ………………………..(c)

98548=1110α + 105432 β   …………………..(d)

Solving equations (c) and (d), we get

α = 1.515              β = 0.919

Therefore fitted regression equation of Y on X is;

                        Y= 1.515 + 0.919 X  …………………..(e)

Thus estimated width of dry leaf having 95 mm fresh leaf is ,

Y = 1.515+ 0.919 x 95 = 88.82 mm

  • Correlation coefficient between dry and fresh leaf is ;

Here r = 0.929, which means there is positive and very high correlation between width of dry leaf and fresh leaf.

  • Here r = 0.939 so r2 = 0.88, which means about 88% variation in width of dry leaf is due to fresh leaf and remaining 12 % variation is due to other variables.
  • Standard error of the estimate of Y on X is,
  • Test of significance of observed regression coefficient (β):

Null hypothesis                (H0): β = 0

Alternative hypothesis     (H1): β ≠ 0

Computing the different sums of squares as follows.

Total sum of square(TSS)

Sum of square due to regression

Sum of squares due to deviation from regression

TABLE

Source of Variation (SV)Degree of freedom (d.f.)Sum of square (SS)Mean sum of square (MSS=SS/d.f.)F-ratio
Due to regression   Error1   12-2=10SSR=2327.199   SSE=311.801MSR=2327.199   MSE= 31.1801= 74.637
Totaln-1=11TSS= 2639  

Thus test statistics F= 74.637.

Critical value of F at α= 5% level of significance for (1,10) degree of freedom is 4.96.

Decision: Here calculated value of F (74.637) is greater than critical value of F at 5% level of significance for (1,10) degree of freedom. So accept alternative hypothesis and reject null hypothesis and conclude that regression coefficient (β) is significant.

Example:1

A physical measurement, such as intensity of light of a particular wavelength transmitted through a solution, can often be calibrated to give the concentration of particular substance in the solution. ‘9’ pairs of values of intensity (X) and concentration ‘Y’ were obtained and can be summarized as follows:

  1. Find the regression equation of Y on X and estimate the value of Y when X=3.
  2. Find the coefficient of determination and interpret it.
  3. Compute the standard error of the estimate.

Solution:

(b) Since, square of correlation coefficient is coefficient of determination. So let us find correlation coefficient between X and Y as,

Coefficient of determination (r2) = (0.9910)2 =0.982

This gives, more than 98% of the variation in Y has able to explain by fitted regression model.

(c) Standard error of the estimate of Y on X is,

Example:2

In a study between the amount of rainfall and the quantity of air pollution removed the following data were collected:

Daily rainfall:             4.3      4.5      5.9      5.6      6.1            5.2      3.8      2.1

(in .01 cm)

Pollution removed:     12.6     12.1    11.6    11.8    11.4            11.8    13.2    14.1

(mg/m3)

Find the regression line of y on x.

Solution:

Let Y= β0 + β1X , be the simple linear regression equation of Y( Pollution removed) on X (daily rainfall).

Then its normal equations are;

Now,

 Substituting the sum values in above two normal equations we get;

98.6     =      8 β0     +   37.5 β1 …………………..(c)

453.82 = 37.5 β0     +   188.01 β1……………(d)

Solving equations (c) and (d), we get

           β0= 15.532 and β1 = -0.684

Then, fitted or estimated regression line of Y on X is,

Example:

In a partially destroyed laboratory record of an analysis of correlation data, the following results only are legible:

Variable of X = 9

What were,     i. The mean values of X and Y.

                        ii. The correlation coefficient between X and Y, and

                        iii. The standard deviation of Y?

Solution:

Example:

Average Temperature and Precipitation Temperatures (in degrees Fahrenheit) and precipitation (in inches) are as follows:

Avg. daily temp. x 86 81 8389 80 7464
Avg. monthly precip. y3.41.8 3.5 3.6 3.7 1.5 0.2
  1. Find y when x =70F.
  2. Find the correlation coefficient and interpret it in practical manner.
  3. Find the coefficient of determination and interpret it.
  4. Find the standard error of the estimate.

Solution:

Avg. daily temp. xAvg. monthly precip. y
863.4739611.56292.4
811.865613.24145.8
833.5688912.25290.5
893.6792112.96320.4
803.7640013.69296
741.554762.25111
640.240960.0412.8
55717.74473955.991468.9

Now substituting the values of various sum and solving it we get,Now substituting the values of various sum and solving it we get,

Now correlation coefficient between X and Y is,

 x y 
863.43.476-0.0760.005776
811.82.751-0.9510.904401
833.53.0410.4590.210681
893.63.911-0.3110.096721
803.72.6061.0941.196836
741.51.736-0.2360.055696
640.20.286-0.0860.007396
55717.771.771-54.0712.477507

Reference : Probability and statistics for engineers by Toya Naryan Paudel & Pradeep Kunwar, Sukunda Pustak Bhawan.

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Theory of Probability https://onlineengineeringnotes.com/2021/05/20/theory-of-probability/ https://onlineengineeringnotes.com/2021/05/20/theory-of-probability/#respond Thu, 20 May 2021 13:05:57 +0000 https://onlineengineeringnotes.com/?p=520 Probability: The probability theory provides a means of getting an idea of the likelihood of occurrence of different events resulting from a random experiment in terms of quantitative measures ranging between zero and one. The probability is zero for an impossible event and one for an event which is certain to occur. Some terminologies: Random ... Read more

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Probability:

The probability theory provides a means of getting an idea of the likelihood of occurrence of different events resulting from a random experiment in terms of quantitative measures ranging between zero and one. The probability is zero for an impossible event and one for an event which is certain to occur.

Some terminologies:

Random experiment:

An experiment is called a random experiment if outcome of experiment can’t be predicted with certainty ( or result is not unique), but may be any one of the various possible outcomes.

For example, flipping a coin, throwing a balance dice etc. where one cannot predict the outcome of events with certainty.

Sample space:

The set of all possible outcomes of random experiment is sample space. It is denoted by S. For example, while throwing a balance dice, the possible outcomes are 1, 2,3,4,5,6. So sample space is S= {1, 2,3,4,5,6}.

Trial and events:

 performing the random experiment is trial and its outcomes are events. For example, flipping a fair coin is a trial where as its outcomes are events.

Simple and Compound Events:

Simple events.

 In the simple events we think about the probability of the happening or not-happening of the simple events. Whenever we are tossing the coin we are considering the occurrence of the events of head and tail. In another example, if in a bag there are 10 white balls and 6 red balls and whenever we are trying to find out the probability of drawing a red ball, is included in simple events.

Compound events:

But on the other hand when we consider the joint occurrence of two or more events, it becomes compound events. Unlike simple events here more than one event are taken into consideration.

Equally Likely Events:

Events are said to be equally likely, when there is equal chance of occurring. If one event is not occurred like other events then events are not considered as equally likely. Or in other words events are said to be equally likely when one event does not occur more often than the others.

Mutually Exclusive Events:

The events are said to be mutually exclusive when they are not occurred simultaneously. Among the events, if one event will remain present in a trial other events will not appear. In other words, occurrence of one precludes the occurrence of all the others.

Independent and Dependent Events:

Two or more events are said to be independent when the occurrence of one trial does not affect the other. It indicates the fact that if trial made one by one, one trial is not affected by the other trial. And also one trial never describes anything about the other trials.

Dependent events are those in which the occurrence and non-occurrence of one event in a trial may affect the occurrence of the other trials. Here the events are mutually dependent on each other.

Different Schools of Thought on the Concept of Probability:

There are different schools of thought on the concept of probability:

  • Mathematical or prior or classical approach of probability

The definition of probability has been given by a French mathematician named “Laplace”. According to him “ if there are ‘n’ exhaustive, mutually exclusive and equally likely outcomes of an experiment and m of them are favorable to an event A. the probability of happening of event A is denoted by P(A).

            This definition was given by James Bernoulli.

Unfavorable no of events for the occurrence of an event A is n-m, so the probability of not happening of an event A is defined as the ratio of n-m/n

  • Statistical or empirical or relative frequency approach:

Suppose that an event A occurs ‘m’ times in ‘n’ repetitions of a random experiment. In the limiting case when n become sufficiently large, it more or less settle to a number which is called probability of happening of an event A.

  • Axiomatic or modern approach of probability:

Let S be a sample space of random experiment and A be any event in S, then P(.) is probability function define in S and the function P(A) satisfying the following axioms.

i. P(A) ≥ 0, (non- negativity)

ii.         P(S) = 1, (total probability)

iii.        if A1, A2, ……. An are n mutually exclusive (mutually disjoint) events define in sample space S then,

            P (A1UA2U……UAn) = P(A1) + P(A2) +……..+ P(An)

      

Theorems of probability (or law):

  1. Additive theorem of Probability.

Statement:

 if A and B are two non-mutually exclusive events define in sample space ‘S’ then,                                                                                         

Proof:

      Let A and B are two non- mutually exclusive events define in sample space S then from above Venn- diagram we can write,

Note:

If events are mutually exclusive, then an intersection part between events is zero. Then,

2. Multiplicative theorem of Probability (And law)

Case 1: when events are dependent.

If A and B are dependent events then,

P(A and B) = P(A B) = P(A).P(B/A) ; P(A) > 0 = P(B).P(A/B) ; P(B) > 0

Where P(A/B) and P(B/A) are conditional probabilities of an event A given that, B has been occurred and B given A has been occurred respectively.

Proof:

 Suppose sample space contains ‘N’ number of sample points. Out of which  number of sample points belongs to the occurrence of an event A, nB number of sample points belongs to the occurrence of an event B and nAB number of sample points belongs to the compound events of A ∩B. Then from the definition of probability we can write,

Now, for the conditional probability P(A/B), sample points in the event B acts as sample space for the occurrence of an event A or (A∩B). thus,

Independence

Events A and B are said to be independent to each other if the occurrence or non-occurrence of one event doesn’t affect the probability of happening of other event. In other words, if A and B are two independent events, then the probability that they will both occur together is equal to the product of the probability of event A and the probability of event B.

i.e. P (A and B) = P(A∩B) = P(A). P(B)

Thus, if events are independent, then conditional probability becomes unconditional.

i.e. P(A/B) = P(A) and P(B/A) = P(B)

Example:

A coin is tossed three times. Find the probability of getting (i)two heads (ii) no head (iii) at least one head (iv) at most one head (v) head, tail and head in that order.

Solution:

 For a fair or unbiased coin P(H)=P(T)=1/2

Example:

In a company, out of 20 candidates 14 men and 6 women apply fort wo vacancies. What is the probability that, i. both  men are selected, ii) both women are selected, iii) one man and women are selected in the vacancies.

Solution:

Total no of Candidates(n)= 20

2 candidates can be selected from 20 candidates in 20C2 ways which is exhaustive no. of outcomes, i.e,

Example:

Three groups of children contain 3 boys and 1 girl, 2 boys and 2 girls and 1 boy and 3 girls respectively. One child is selected at random from each group. Show that probability of selecting 1 boy and 2 girls is 13/32.

Solution:

Example:

A Problem of statistics is given to three students A, B and C whose chance of solving it are  and  respectively. Find the probability that,

  1. All of them can solve the problem.
  2. None of them can solve.
  3. Problem will be solved.
  4. A and B can solve the problem but C cannot.
  5. Only one of them can solve.

Solution:

Example:

Past records show that 4 of 135 parts are defective in length, 3 of 141 are defective in width, and 2 of 347 are defective in both. Use these figures to estimate probabilities of the individual events assuming that defects occur independently in length and width.

a) What is the probability that a part produced under the same conditions will be defective in length or width or both?

b) What is the probability that a part will have neither defect?

Solution:

Conditional probability:

      If A and B are two dependent events, i.e. occurrence of one affects the probability of others and vice versa. Then conditional probability of an event A given that an event B has been already occurred is denoted by P(A/B), and given by,

            

Similarly , conditional probability of an event B given that A has been already occurred is denoted by P(B/A) and given by,

               

Example:

Suppose that the population of a certain city is 40 percent male and 60 percent female. Suppose also that 50 percent of male and 30 percent of the female smokers. (i) What is the probability that the person selected is male given that it is smoker? (ii) What is the probability that that person selected is female given that she is smoker.

Solution:

For the sake of convenience, let us construct a table from the given information as;

Now,

SmokerNon-smokerTotal
Male Female20 1820 3240 60
Total3852100

Example:

The probabilities of the monthly snowfall exceeding 10 cm at a particular location in the months of December, January and February are 0.20, 0.40 and 0.60, respectively. For a particular winter:

a) What is the probability of not receiving 10 cm of snowfall in any of the months of December, January and February in a particular winter?

b) What is the probability of receiving at least 10 cm snowfall in a month, in at least two of the three months of that winter?

c) Given that the snowfall exceeded 10 cm in each of only two months, what is the probability that the two months were consecutive?

Solution:

From the given information’s, the probabilities of the monthly snowfall exceeding 10 cm at a particular location in the months of December, January and February are 0.20, 0.40 and 0.60, respectively.

Baye’s theorem for conditional probability

Statement:

Let E1, E2 and E3 be mutually exclusive disjoint events of sample space S with P(Ei) ≠ 0, i = 1, 2, 3 then for any arbitrary event say ‘A’ which is subset of ‘S’ such that P(A) > 0, we have
                   

Proof:

Let E1, E2 and E3 are three mutually exclusive events define in sample space ‘S’ and ‘A’ be any arbitrary event. Then relationship between  mutually exclusive events and an arbitrary event ‘A’ can be shown in venn-diagram as,

Remarks:

  1. The probabilities P(E1),P(E2),…….,P(En) are called prior probabilities because they exist before performing the experiment.
  2. The probabilities P(A/E1), P(A/E2),…..P(A/En) are called likelihood probabilities because that indicate how often A occur for each and every E1,E2,……,En given.
  3. The probabilities P(Ei/A); i=1,2,3….n are called posteriori probabilities because they are determined after the result of the experiment are known.

Example:

It is known that of the articles produced by a factory, 20% come from Machine A, 30% from Machine B, and 50% from Machine C. The percentages of satisfactory articles among those produced are 95% for A, 85% for B and 90% for C. An article is chosen at random.

a) What is the probability that it is satisfactory?

b) Assuming that the article is satisfactory, what is the probability that it was produced by Machine A?

Solution:

Here from given information’s (Prior probabilities),

20% articles come from machine A; P(A)= 20/100 = 0.20

30% articles come from machine B; P(B) = 30/100 = 0.30

50% articles come from machine C; P(C)= 50/100 = 0.50

Example:

At a particular day of covid-19 pandemics,  2000 frontline works  from Kathmandu district are randomly selected and tested for PCR,1500 frontline workers from Bhaktapur and 500 from Lalitpur are randomly selected and tested for PCR. In Kathmandu, Bhaktapur and Lalitpur 25, 7 and 10 are seems to be PCR positive.               
1.what is the probability of PCR positive patients?
2. if  PCR positive patients are sent in a quarantine and a doctor thoroughly check a PCR positive patient. What is the probability that infected patient is from Kathmandu district?

Solution:

Example:

Box A contains 18 items of which 3 are defective and box B contains 12 non-defectives and 6 defective items. One of the box is selected at random and an item is drawn from it.

1. find the probability that it is defective.

2. if the item is defective, what is the probability that the defective item came from box A?

Solution:

Here, P(selecting box A at random) =P(A) = 1/2= 0.5

And  P(selecting Box B at random ) = P(B) = 1/2= 0.5

                  Let b = defective item

Then likelihood probabilities are ;

                  P(b/A) = 3/18 and P(b/ B) = 6/18

1. P(defective item) = P(A).P(b/A) + P(B).P(b/B)

= 0.5*3/18 +0.5*6/18

 =1/4 =0.25

2. using bayes theorem,

Example:

Three road construction firms, X, Y and Z, bid for a certain contract. From past experience, it is estimated that the probability that X will be awarded the contract is 0.40, while for Y and Z the probabilities are 0.35 and 0.25. If X does receive the contract, the probability that the work will be satisfactorily completed on time is 0.75. For Y and Z these probabilities are 0.80 and 0.70.

i) What is the probability that Y will be awarded the contract and complete the work satisfactorily?

ii) What is the probability that the work will be completed satisfactorily?

iii) It turns out that the work was done satisfactorily. What is the probability that Y was awarded the contract?

Solution:

From given information ,

   P(X) = 0.40, P(Y) = 0.35 and P(Z) = 0.25                               [  P(X)+P(Y)+P(Z)=1]

Let us define a arbitrary event say (A) then,

  A = work will be completed satisfactorily.

Again fron likelihood probabilities,

P(A/X) =0.75, P(B/Y) = 0.80 and P(A/Z) = 0.70

  • P(Y will awarded the contract and complete the work satisfactorily) = P(Y∩A)

= P(Y).P(A/Y)

= 0.40 x 0.75

= 0.30

  • P( work will be completed satisfactorily)

 P(A) = P(X).P(A/X) + P(Y).P(A/Y) + P(Z).P(A/Z)

            = 0.40×0.75 + 0.35×0.80 + 0.25×0.70

            = 0.755

  • Now using Bayes theorem,

                                                     

Reference : Probability and statistics for engineers by Toya Naryan Paudel & Pradeep Kunwar, Sukunda Pustak Bhawan.

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