Unit 7
Sampling
Population
The population is the collection or group of observations under study.
The total number of observations in a population is known as population size and it is denoted by N.
Types of population
 finite population – the population contains finite numbers of observations is known as finite population
 Infinite population it contains infinite number of observations.
 Real population the population which comprises the items which are all present physically is known as real population.
 Hypothetical population if the population consists the items which are not physically present but their existence can be imagined, is known as hypothetical population.
Sample –
To get the information from all the elements of a large population may be time consuming and difficult. And also, if the elements of population are destroyed under investigation then getting the information from all the units is not make a sense. For example, to test the blood, doctors take very small amount of blood. Or to test the quality of certain sweet we take a small piece of sweet in such situations, a small part of population is selected from the population which is called a sample
Complete survey
When each and every element of the population is investigated or studied for the characteristics under study then we call it complete survey or census.
Sample Survey
When only a part or a small number of elements of population are investigated or studied for the characteristics under study then we call it sample survey or sample enumeration Simple Random Sampling or Random Sampling
The simplest and most common method of sampling is simple random sampling. In simple random sampling, the sample is drawn in such a way that each element or unit of the population has an equal and independent chance of being included in the sample. If a sample is drawn by this method then it is known as a simple random sample or random sample
Simple Random Sampling without Replacement (SRSWOR)
In simple random sampling, if the elements or units are selected or drawn one by one in such a way that an element or unit drawn at a time is not replaced back to the population before the subsequent draws is called SRSWOR.
Suppose we draw a sample from a population, the size of sample is n and the size of population is N, then total number of possible samples is
Simple Random Sampling with Replacement (SRSWR)
In simple random sampling, if the elements or units are selected or drawn one by one in such a way that a unit drawn at a time is replaced back to the population before the subsequent draw is called SRSWR.
Suppose we draw a sample from a population, the size of sample is n and the size of population is N, then total number of possible sample is .
Parameter
A parameter is a function of population values which is used to represent the certain characteristic of the population. For example, population mean, population variance, population coefficient of variation, population correlation coefficient, etc. are all parameters. Population parameter mean usually denoted by μ and population variance denoted by
Sample mean and sample variance
Let be a random sample of size n taken from a population whose pmf or pdf function f(x,
Then the sample mean is defined by
And sample variance 
Statistic
Any quantity which does not contain any unknown parameter and calculated from sample values is known as statistic.
Suppose is a random sample of size n taken from a population with mean μ and variance then the sample mean
Is a statistic.
Estimator and estimate
if a statistic is used to estimate an unknown population parameter then it is known as estimator and the value of the estimator based on observed value of the sample is known as estimate of parameter.
Standard error
The standard deviation of the sampling distribution is known as standard error.
When we increase the sample size then standard increases.
It plays an important role in large sample theory.
Note The reciprocal of the standard error is called ‘precision’.
If the size of the sample is less than 30, it is considered as small sample otherwise it is called large sample.
The sampling distribution of large samples is assumed to be normal.
Key takeaways
 Population
The population is the collection or group of observations under study.
2. Simple Random Sampling without Replacement (SRSWOR)
If the elements or units are selected or drawn one by one in such a way that an element or unit drawn at a time is not replaced back to the population before the subsequent draws is called SRSWOR.
3. Simple Random Sampling with Replacement (SRSWR)
If the elements or units are selected or drawn one by one in such a way that a unit drawn at a time is replaced back to the population before the subsequent draw is called SRSWR.
4. Statistic
Any quantity which does not contain any unknown parameter and calculated from sample values is known as statistic.
Hypothesis
A hypothesis is a statement or a claim or an assumption about the value of a population parameter.
Similarly, in case of two or more populations a hypothesis is comparative statement or a claim or an assumption about the values of population parameters.
For example
If a customer of a car wants to test whether the claim of car of a certain brand gives the average of 30km/hr is true or false.
Simple and composite hypotheses
If a hypothesis specifies only one value or exact value of the population parameter then it is known as simple hypothesis. And if a hypothesis specifies not just one value but a range of values that the population parameter may assume is called a composite hypothesis.
Null and alternative hypothesis
The hypothesis which is to be tested as called the null hypothesis.
The hypothesis which complements to the null hypothesis is called alternative hypothesis.
In the example of car, the claim is and its complement is .
The null and alternative hypothesis can be formulated as
And 
Critical region
Let be a random sample drawn from a population having unknown population parameter .
The collection of all possible values of is called sample space and a particular value represent a point in that space.
In order to test a hypothesis, the entire sample space is partitioned into two disjoint subspaces, say, and S – . If calculated value of the test statistic lies in , then we reject the null hypothesis and if it lies in then we do not reject the null hypothesis. The region is called a “rejection region or critical region” and the region is called a “nonrejection region”.
Therefore, we can say that
“A region in the sample space in which if the calculated value of the test statistic lies, we reject the null hypothesis then it is called critical region or rejection region.”
The region of rejection is called critical region.
The critical region lies in one or two tails on the probability curve of sampling distribution of the test statistic it depends on the alternative hypothesis.
Therefore, there are three cases
CASE1: if the alternative hypothesis is right sided such as then the entire critical region of size lies on right tail of the probability curve. CASE2: if the alternative hypothesis is left sided such as then the entire critical region of size lies on left tail of the probability curve.
CASE3: if the alternative hypothesis is two sided such as then the entire critical region of size lies on both tail of the probability curve
Type1 and Type2 error Type1 error The decision relating to rejection of null hypo. When it is true is called type1 error. The probability of type1 error is called size of the test, it is denoted by and defined as Note is the probability of correct decision. Type2 error The decision relating to nonrejection of null hypo. When it is false is called type1 error. It is denoted by and defined as

Decision  true  true 
Reject  Type1 error  Correct decision 
Do not reject  Correct decision  Type2 error 
One tailed and two tailed tests
A test of testing the null hypothesis is said to be twotailed test if the alternative hypothesis is twotailed whereas if the alternative hypothesis is onetailed then a test of testing the null hypothesis is said to be onetailed test. For example, if our null and alternative hypothesis are then the test for testing the null hypothesis is twotailed test because the alternative hypothesis is twotailed. If the null and alternative hypotheses are
then the test for testing the null hypothesis is righttailed test because the alternative hypothesis is righttailed. Similarly, if the null and alternative hypotheses are then the test for testing the null hypothesis is lefttailed test because the alternative hypothesis is lefttailed

Procedure for testing a hypothesis
Step1: first we set up null hypothesis and alternative hypothesis .
Step2: after setting the null and alternative hypothesis, we establish a
criteria for rejection or nonrejection of null hypothesis, that is,
decide the level of significance (), at which we want to test our
hypothesis. Generally, it is taken as 5% or 1% (α = 0.05 or 0.01).
step3: The third step is to choose an appropriate test statistic under H0 for
testing the null hypothesis as given below

Now after doing this, specify the sampling distribution of the test statistic preferably in the standard form like Z (standard normal), , t, F or any other wellknown in literature
Step4: Calculate the value of the test statistic described in Step III on the basis of observed sample observations.
Step5: Obtain the critical (or cutoff) value(s) in the sampling distribution of the test statistic and construct rejection (critical) region of size .
Generally, critical values for various levels of significance are putted in the form of a table for various standard sampling distributions of test statistic such as Ztable, table, ttable, etc
Step6: After that, compare the calculated value of test statistic obtained from Step IV, with the critical value(s) obtained in Step V and locates the position of the calculated test statistic, that is, it lies in rejection region or nonrejection region.
Step7: in testing the hypothesis we have to reach at a conclusion, it is performed as below
First If calculated value of test statistic lies in rejection region at level of significance then we reject null hypothesis. It means that the sample data provide us sufficient evidence against the null hypothesis and there is a significant difference between hypothesized value and observed value of the parameter
Second If calculated value of test statistic lies in nonrejection region at level of significance then we do not reject null hypothesis. Its means that the sample data fails to provide us sufficient evidence against the null hypothesis and the difference between hypothesized value and observed value of the parameter due to fluctuation of sample
Procedure of testing of hypothesis for large samples
The sample size more than 30 is considered as large sample size. So that for large samples, we follow the following procedure to test the hypothesis.
Step1: first we set up the null and alternative hypothesis.
Step2: After setting the null and alternative hypotheses, we have to choose level of significance. Generally, it is taken as 5% or 1% (α = 0.05 or 0.01). And accordingly, rejection and nonrejection regions will be decided.
Step3: Third step is to determine an appropriate test statistic, say, Z in case of large samples. Suppose Tn is the sample statistic such as sample
mean, sample proportion, sample variance, etc. for the parameter
then for testing the null hypothesis, test statistic is given by
Step4: the test statistic Z will be assumed to be approximately normally distributed with mean 0 and variance 1 as
By putting the values in above formula, we calculate test statistic Z.
Suppose z be the calculated value of Z statistic
Step5: After that, we obtain the critical (cutoff or tabulated) value(s) in the sampling distribution of the test statistic Z corresponding to assumed in Step II. we construct rejection (critical) region of size α in the probability curve of the sampling distribution of test statistic Z.
Step6: Take the decision about the null hypothesis based on the calculated and critical values of test statistic obtained in Step IV and Step V.
Since critical value depends upon the nature of the test that it is one tailed test or twotailed test so following cases arise
Case1 onetailed test when
(righttailed test)
In this case, the rejection (critical) region falls under the right tail of the probability curve of the sampling distribution of test statistic Z.
Suppose is the critical value at level of significance so entire region greater than or equal to is the rejection region and less than
is the nonrejection region
If z (calculated value ) ≥ (tabulated value), that means the calculated value of test statistic Z lies in the rejection region, then we reject the null hypothesis H0 at level of significance. Therefore, we conclude that sample data provides us sufficient evidence against the null hypothesis and there is a significant difference between hypothesized or specified value and observed value of the parameter.
If z < that means the calculated value of test statistic Z lies in nonrejection region, then we do not reject the null hypothesis H0 at level of significance. Therefore, we conclude that the sample data fails to provide us sufficient evidence against the null hypothesis and the difference between hypothesized value and observed value of the parameter due to fluctuation of sample.
So, the population parameter
Case2: when
(lefttailed test)
the rejection (critical) region falls under the left tail of the probability curve of the sampling distribution of test statistic Z.
Suppose  is the critical value at level of significance then entire region less than or equal to  is the rejection region and greater than is the nonrejection region
If z ≤, that means the calculated value of test statistic Z lies in the rejection region, then we reject the null hypothesis H0 at level of significance.
If z >, that means the calculated value of test statistic Z lies in the nonrejection region, then we do not reject the null hypothesis H0 at level of significance.
In case of two tailed test
In this case, the rejection region falls under both tails of the probability curve of sampling distribution of the test statistic Z. Half the area (α) i.e. α/2 will lies under left tail and other half under the right tail. Suppose and are the two critical values at the lefttailed and righttailed respectively. Therefore, entire region less than or equal to and greater than or equal to are the rejection regions and between is the nonrejection region.
If Z that means the calculated value of test statistic Z lies in the rejection region, then we reject the null hypothesis H0 at level of significance.
If that means the calculated value of test statistic Z lies in the nonrejection region, then we do not reject the null hypothesis H0 at level of significance.
Testing of hypothesis for population mean using ZTesnt
For testing the null hypothesis, the test statistic Z is given as
The sampling distribution of the test statistics depends upon variance
So that there are two cases
Case1: when is known 
The test statistic follows the normal distribution with mean 0 and variance unity when the sample size is the large as the population under study is normal or nonnormal. If the sample size is small then test statistic Z follows the normal distribution only when population under study is normal. Thus,
Case1: when is unknown –
We estimate the value of by using the value of sample variance
Then the test statistic becomes

After that, we calculate the value of test statistic as may be the case ( is known or unknown) and compare it with the critical value at prefixed level of significance α.
Example: A company of pens claims that a certain pen manufactured by him has a mean writinglife at least 460 A4 size pages. A purchasing agent selects a sample of 100 pens and put them on the test. The mean writinglife of the sample found 453 A4 size pages with standard deviation 25 A4 size pages. Should the purchasing agent reject the manufacturer’s claim at 1% level of significance?
Sol.
It is given that
Specified value of population mean = = 460, Sample size = 100 Sample mean = 453 Sample standard deviation = S = 25 
The null and alternative hypothesis will be
Also the alternative hypothesis lefttailed so that the test is left tailed test.
Here, we want to test the hypothesis regarding population mean when population SD is unknown. So we should used ttest for if writinglife of pen follows normal distribution. But it is not the case. Since sample size is n = 100 (n > 30) large so we go for Ztest. The test statistic of Ztest is given by

We get the critical value of left tailed Z test at 1% level of significance is
Since calculated value of test statistic Z (= ‒2.8) is less than the critical value
(= −2.33), that means calculated value of test statistic Z lies in rejection region so we reject the null hypothesis. Since the null hypothesis is the claim so we reject the manufacturer’s claim at 1% level of significance.
Example: A big company uses thousands of CFL lights every year. The brand that the company has been using in the past has average life of 1200 hours. A new brand is offered to the company at a price lower than they are paying for the old brand. Consequently, a sample of 100 CFL light of new brand is tested which yields an average life of 1220 hours with standard deviation 90 hours. Should the company accept the new brand at 5% level of significance?
Sol.
Here we have
The company may accept the new CFL light when average life of
CFL light is greater than 1200 hours. So the company wants to test that the new brand CFL light has an average life greater than 1200 hours. So our claim is > 1200 and its complement is ≤ 1200. Since complement contains the equality sign so we can take the complement as the null hypothesis and the claim as the alternative hypothesis. Thus,
Since the alternative hypothesis is righttailed so the test is righttailed test.
Here, we want to test the hypothesis regarding population mean when population SD is unknown, so we should use ttest if the distribution of life of bulbs known to be normal. But it is not the case. Since the sample size is large (n > 30) so we can go for Ztest instead of ttest.
Therefore, test statistic is given by
The critical values for righttailed test at 5% level of significance is
1.645
Since calculated value of test statistic Z (= 2.22) is greater than critical value (= 1.645), that means it lies in rejection region so we reject the null hypothesis and support the alternative hypothesis i.e. we support our claim at 5% level of significance
Thus, we conclude that sample does not provide us sufficient evidence against the claim so we may assume that the company accepts the new brand of bulbs
Significance test of difference between sample means
Given two independent examples and with means standard derivations from a normal population with the same variance, we have to test the hypothesis that the population means are same For this, we calculate

It can be shown that the variate t defined by (1) follows the t distribution with degrees of freedom.
If the calculated value the difference between the sample means is said to be significant at 5% level of significance. If , the difference is said to be significant at 1% level of significance. If the data is said to be consistent with the hypothesis that . 
Cor. If the two samples are of same size and the data are paired, then t is defined by
=difference of the ith member of the sample d=mean of the differences = and the member of d.f.=n1. 
Example
Eleven students were given a test in statistics. They were given a month’s further tuition and the second test of equal difficulty was held at the end of this. Do the marks give evidence that the students have benefitted by extra coaching?
Boys  1  2  3  4  5  6  7  8  9  10  11 
Marks I test  23  20  19  21  18  20  18  17  23  16  19 
Marks II test  24  19  22  18  20  22  20  20  23  20  17 
Sol. We compute the mean and the S.D. of the difference between the marks of the two tests as under:
Assuming that the students have not been benefitted by extra coaching, it implies that the mean of the difference between the marks of the two tests is zero i.e. Then, nearly and df v=111=10 
Students  
1  23  24  1  0  0 
2  20  19  1  2  4 
3  19  22  3  2  4 
4  21  18  3  4  16 
5  18  20  2  1  1 
6  20  22  2  1  1 
7  18  20  2  1  1 
8  17  20  3  2  4 
9  23  23    1  1 
10  16  20  4  3  9 
11  19  17  2  3  9 




we find that (for v=10) =2.228. As the calculated value of , the value of t is not significant at 5% level of significance i.e. the test provides no evidence that the students have benefitted by extra coaching.
Example:
From a random sample of 10 pigs fed on diet A, the increase in weight in certain period were 10,6,16,17,13,12,8,14,15,9 lbs. For another random sample of 12 pigs fed on diet B, the increase in the same period were 7,13,22,15,12,14,18,8,21,23,10,17 lbs. Test whether diets A and B differ significantly as regards their effect on increases in weight?
Sol. We calculate the means and standard derivations of the samples as follows
 Diet A 

 Diet B 

10  2  4  7  8  64 
6  6  36  13  2  4 
16  4  16  22  7  49 
17  5  25  15  0  0 
13  1  1  12  3  9 
12  0  0  14  1  1 
8  4  16  18  3  9 
14  2  4  8  7  49 
15  3  9  21  6  36 
9  3  9  23  8  64 


 10  5  25 


 17  2  4 






120 

 180  0  314 
Assuming that the samples do not differ in weight so far as the two diets are concerned i.e. For v=20, we find =2.09 The calculated value of Hence the difference between the samples means is not significant i.e. thew two diets do not differ significantly as regards their effects on increase in weight.

Testing of hypothesis for difference of two population means using ZTest
Let there be two populations, say, populationI and populationII under study.
Also let denote the means and variances of populationI and populationII respectively where both are unknown but may be known or unknown. We will consider all possible cases here. For testing the hypothesis about the difference of two population means, we draw a random sample of large size n1 from populationI and a random sample of large size n2 from populationII. Let be the means of the samples selected from populationI and II respectively.
These two populations may or may not be normal but according to the central limit theorem, the sampling distribution of difference of two large sample means asymptotically normally distributed with mean and variance
And We know that the standard error = Here, we want to test the hypothesis about the difference of two population means so we can take the null hypothesis as Or And the alternative hypothesis is Or The test statistic Z is given by
Or Since under null hypothesis we assume that , therefore, we get Now, the sampling distribution of the test statistic depends upon that both are known or unknown. Therefore, four cases arise
Case1: When are known and In this case, the test statistic follows normal distribution with mean 0 and variance unity when the sample sizes are large as both the populations under study are normal or nonnormal. But when sample sizes are small then test statistic Z follows normal distribution only when populations under study are normal, that is,
Case2: When are known and In this case, the test statistic also follows the normal distribution as described in case I, that is, Case3: When are unknown and In this case, σ2 is estimated by value of pooled sample variance where,
and test statistic follows tdistribution with (n1 + n2 − 2) degrees of freedom as the sample sizes are large or small provided populations under study follow normal distribution. But when the populations are under study are not normal and sample sizes n1 and n2are large (> 30) then by central limit theorem, test statistic approximately normally distributed with mean 0 and variance unity, that is, Case4: When are unknown and In this case, are estimated by the values of the sample variances respectively and the exact distribution of test statistic is difficult to derive. But when sample sizes n1 and n2 are large (> 30) then central limit theorem, the test statistic approximately normally distributed with mean 0 and variance unity, that is, After that, we calculate the value of test statistic and compare it with the critical value at prefixed level of significance α.

Example: A college conducts both face to face and distance mode classes for a particular course indented both to be identical. A sample of 50 students of facetoface mode yields examination results mean and SD respectively as
and other sample of 100 distancemode students yields mean and SD of their examination results in the same course respectively as:
Are both educational methods statistically equal at 5% level?
Sol. here we have Here we wish to test that both educational methods are statistically equal. If denote the average marks of face to face and distance mode students respectively then our claim is and its complement is ≠ . Since the claim contains the equality sign so we can take the claim as the null hypothesis and complement as the alternative hypothesis. Thus,
Since the alternative hypothesis is twotailed so the test is twotailed test. We want to test the null hypothesis regarding two population means when standard deviations of both populations are unknown. So we should go for ttest if population of difference is known to be normal. But it is not the case. Since sample sizes are large (n1, and n2 > 30) so we go for Ztest. For testing the null hypothesis, the test statistic Z is given by The critical (tabulated) values for twotailed test at 5% level of significance are Since calculated value of Z ( = 2.23) is greater than the critical values (= ±1.96), that means it lies in rejection region so we reject the null hypothesis i.e. we reject the claim at 5% level of significance

Level of significance
The probability of type1 error is called level of significance of a test. It is also called the size of the test or size of the critical region. Denoted by .
Basically, it is prefixed as 5% or 1% level of significance.
If the calculated value of the test statistics lies in the critical region then we reject the null hypothesis.
The level of significance relates to the trueness of the conclusion. If null hypothesis do not reject at level 5% then a person will be sure “concluding about the null hypothesis” is true with 95% assurance but even it may false with 5% chance.
Confidence limits
Let be a random sample of size n drawn from a population having pdf (pmf) . Let and (here be two statistic such that the probability that the random interval [] including the true value of population parameter , that is Here does not depends on . Then the random interval [] is called as (1 – 100 % confidence interval for unknown population parameter and (1 – is known as confidence coefficient. The length of interval can be defined as Length = Upper confidence – Lower confidence limit 
Key takeaways
The decision relating to rejection of null hypo. When it is true is called type1 error. The probability of type1 error is called size of the test, it is denoted by and defined as Note is the probability of correct decision.
2. Type2 error The decision relating to nonrejection of null hypo. When it is false is called type1 error. It is denoted by and defined as
3.
4.
5. For testing the null hypothesis, the test statistic Z is given as

General procedure of ttest for testing hypothesis
Let X1, X2,…, Xn be a random sample of small size n (< 30) selected from a normal population, having parameter of interest, say,
which is actually unknown but its hypothetical value then
Step1: First of all, we setup null and alternative hypotheses
Step2: After setting the null and alternative hypotheses our next step is to decide a criteria for rejection or nonrejection of null hypothesis i.e. decide the level of significance at which we want to test our null hypothesis. We generally take = 5 % or 1%.
Step3: The third step is to determine an appropriate test statistic, say, t for testing the null hypothesis. Suppose Tn is the sample statistic (may be sample mean, sample correlation coefficient, etc. depending upon ) for the parameter then teststatistic t is given by
Step4: As we know, ttest is based on tdistribution and tdistribution is described with the help of its degrees of freedom, therefore, test statistic t follows tdistribution with specified degrees of freedom as the case may be.
By putting the values of Tn, E(Tn) and SE(Tn) in above formula, we calculate the value of test statistic t. Let tcal be the calculated value of test statistic t after putting these values.
Step5: After that, we obtain the critical (cutoff or tabulated) value(s) in the sampling distribution of the test statistic t corresponding to assumed in Step II. The critical values for ttest are corresponding to different level of significance (α). After that, we construct rejection (critical) region of size in the probability curve of the sampling distribution of test statistic t.
Step6: Take the decision about the null hypothesis based on calculated and critical value(s) of test statistic obtained in Step IV and Step V respectively.
Critical values depend upon the nature of test.
The following cases arises
In case of one tailed test
Case1:
[Righttailed test]
In this case, the rejection (critical) region falls under the right tail of the probability curve of the sampling distribution of test statistic t.
Suppose is the critical value at level of significance then entire region greater than or equal to is the rejection region and less than is the nonrejection region.
If ≥ that means calculated value of test statistic t lies in the rejection (critical) region, then we reject the null hypothesis at level of significance.
If < that means calculated value of test statistic t lies in non rejection region, then we do not reject the null hypothesis at level of significance.
Case2:
[Lefttailed test]
In this case, the rejection (critical) region falls under the left tail of the probability curve of the sampling distribution of test statistic t.
Suppose  is the critical value at level of significance then entire region less than or equal to  is the rejection region and greater than  is the nonrejection region.
If ≤ − that means calculated value of test statistic t lies in the rejection (critical) region, then we reject the null hypothesis at level of significance.
If > −, that means calculated value of test statistic t lies in the nonrejection region, then we do not reject the null hypothesis at level of significance.
In case of two tailed test
In this case, the rejection region falls under both tails of the probability curve of sampling distribution of the test statistic t. half the area (α) i.e. α/2 will lies under left tail and other half under the right tail. Suppose , and are the two critical values at the left tailed and righttailed respectively. Therefore, entire region less than or equal to and greater than or equal to are the rejection regions and between and is the non rejection region.
If ≥ or ≤ , that means calculated value of test statistic t lies in the rejection(critical) region, then we reject the null hypothesis at level of significance.
And if  < < , that means calculated value of test statistic t lies in the nonrejection region, then we do not reject the null hypothesis at level of significance.
Testing of hypothesis for population mean using tTest
There are the following assumptions of the ttest
 Sample observations are random and independent.
 Population variance is unknown
 The characteristic under study follows normal distribution.
For testing the null hypothesis, the test statistic t is given by
Example: A tube manufacturer claims that the average life of a particular category
of his tube is 18000 km when used under normal driving conditions. A random sample of 16 tube was tested. The mean and SD of life of the tube in the sample were 20000 km and 6000 km respectively.
Assuming that the life of the tube is normally distributed, test the claim of the manufacturer at 1% level of significance using appropriate test.
Sol.
Here we have We want to test that manufacturer’s claim is true that the average life () of tube is 18000 km. So claim is μ = 18000 and its complement is μ ≠ 18000. Since the claim contains the equality sign so we can take the claim as the null hypothesis and complement as the alternative hypothesis. Thus, Here, population SD is unknown and population under study is given to be normal. So here can use ttest For testing the null hypothesis, the test statistic t is given by The critical value of test statistic t for twotailed test corresponding (n1) = 15 df at 1% level of significance are Since calculated value of test statistic t (= 1.33) is less than the critical (tabulated) value (= 2.947) and greater that critical value (= − 2.947), that means calculated value of test statistic lies in nonrejection region, so we do not reject the null hypothesis. we conclude that sample fails to provide sufficient evidence against the claim so we may assume that manufacturer’s claim is true. 
When a fair coin is tossed 80 times, we expect from the theoretical considerations that heads will appear 40 times and tail 40 times. But this never happens in practice that is the results obtained in an experiment do not agree exactly with the theoretical results. The magnitude of discrepancy between observations and theory is given by the quantity (pronounced as chi squares). If the observed and theoretical frequencies completely agree. As the value of increases, the discrepancy between the observed and theoretical frequencies increases.
Definition.
If and be the corresponding set of expected (theoretical) frequencies, then is defined by the relation 
Chi – square distribution
If be n independent normal variates with mean zero and s.d. unity, then it can be shown that is a random variate having distribution with ndf.
The equation of the curve is
(1) Properties of distribution
have been tabulated for various values of P and for values of v from 1 to 30. (Table V Appendix 2) ,the curve approximates to the normal curve and we should refer to normal distribution tables for significant values of . IV. Since the equation of curve does not involve any parameters of the population, this distribution does not dependent on the form of the population. V. Mean = and variance =

Goodness of fit
The values of is used to test whether the deviations of the observed frequencies from the expected frequencies are significant or not. It is also used to test how will a set of observations fit given distribution therefore provides a test of goodness of fit and may be used to examine the validity of some hypothesis about an observed frequency distribution. As a test of goodness of fit, it can be used to study the correspondence between the theory and fact.
This is a nonparametric distribution free test since in this we make no assumptions about the distribution of the parent population.
Procedure to test significance and goodness of fit
(i) Set up a null hypothesis and calculate (ii) Find the df and read the corresponding values of at a prescribed significance level from table V. (iii) From table, we can also find the probability P corresponding to the calculated values of for the given d.f. (iv) If P<0.05, the observed value of is significant at 5% level of significance If P<0.01 the value is significant at 1% level. If P>0.05, it is a good faith and the value is not significant. 
Example. A set of five similar coins is tossed 320 times and the result is
Number of heads  0  1  2  3  4  5 
Frequency  6  27  72  112  71  32 
Solution.
For v = 5, we have P, probability of getting a head=1/2;q, probability of getting a tail=1/2. Hence the theoretical frequencies of getting 0,1,2,3,4,5 heads are the successive terms of the binomial expansion Thus the theoretical frequencies are 10, 50, 100, 100, 50, 10. Hence, Since the calculated value of is much greater than the hypothesis that the data follow the binomial law is rejected. 
Example. Fit a Poisson distribution to the following data and test for its goodness of fit at level of significance 0.05.
x  0  1  2  3  4 
f  419  352  154  56  19 
Solution.
Mean m = Hence, the theoretical frequency are 
X  0  1  2  3  4  Total 
F  404.9 (406.2)  366  165.4  49.8  11..3 (12.6)  997.4 
Hence, Since the mean of the theoretical distribution has been estimated from the given data and the totals have been made to agree, there are two constraints so that the number of degrees of freedom v = 5 2=3 For v = 3, we have Since the calculated value of the agreement between the fact and theory is good and hence the Poisson distribution can be fitted to the data. 
Example. In experiments of pea breeding, the following frequencies of seeds were obtained
Round and yellow  Wrinkled and yellow  Round and green  Wrinkled and green  Total 
316  101  108  32  556 
Theory predicts that the frequencies should be in proportions 9:3:3:1. Examine the correspondence between theory and experiment.
Solution.
The corresponding frequencies are Hence, For v = 3, we have Since the calculated value of is much less than there is a very high degree of agreement between theory and experiment. 
Key takeaways
2. 
References
 E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
 P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
 S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
 W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
 N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
 B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
 T. Veerarajan, “Engineering Mathematics”, Tata McgrawHill, New Delhi, 2010
 Higher engineering mathematics, HK Dass