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BUSINESS STATISTICS (PART-25) UNIT- VIII APPROXIMATIONS AND INTERVAL ESTIMATION 1. INTRODUCTION Hello viewers, in my last three lectures we have discussed about the Binomial, Poisson & the Normal distributions. These are the basic distributions which we make use of, in many applications. The first two distributions namely the Binomial & Poisson are the discrete distributions whereas; the Normal distribution is a continuous distribution. We have also discussed the applications of these distributions in these lectures. Now here the question arises: Is there any relationship between these three distributions? And can we also estimate the unknown parameters of these distributions? The answer of these questions is YES. Therefore, in this lecture we shall first discuss at length the relationship between the Binomial, Poisson & Normal distributions. Then we shall discuss interval estimation of population mean of a normal and a nonnormal population. 2. POISSON APPROXIMATION DISTRIBUTION TO BINOMIAL Now here we first discuss the Poisson approximation to Binomial distribution. We know that the Binomial distribution function is given by: f ( x) ncx p x q nx where, x 0,1,2,..., n 0 p, q 1, q 1 p Whereas the Poisson distribution function is given by: e x g ( x) x! x 0,1,... And λ, which is the parameter of the distribution, is a positive constant. Now, if in case of the Binomial distribution the n is very large i.e. 𝑛 → ∞ and, If p is very small .Mathematically ,speaking that, p→ 0, such that np = λ, a constant or rather it is a finite constant then f(x) g(x). That is, Binomial distribution function the Poisson distribution function Now let us look at this example which makes this approximate relation more clear. We know that, if the distribution of a random variable X is Binomial with parameters n and p then f ( x) n c x p x q n x Then for n=10 & p =0.2 f (0) 10C0 (0.2) 0 (0.8)10 0.1074 That is, this gives me the probability that x takes value 0. Now here if I take np then it is equal to 2. Now using the Poisson distribution function, if we calculate the same probability, i.e., probability of x takes value 0 then this comes out to be 0.1353. Now, in this table we consider the probabilities which are calculated using the Binomial & Poisson distribution functions. In the first column we take different values of ‘n’ in Binomial distribution function. And in the next column we take the value of ‘p’. And the third column gives me the value of ‘np’, which we know is λ here. And the fourth column gives me the P(x=0) and the last column gives the P(x=5). Now, in the first case when n=10, I take the value p=0.2. We have just seen that for this the P(x=0) is 0.1704. And when we use Binomial distribution function for (x=5) then this probability comes out to be 0.0881. Now we increase the value of ‘n’ from 10 to 20 and reduce the value of ‘p’ from 0.2 to 0.1 such that this product remains the same i.e. 2 (np=2). For this P(x=0) is 0.1216 and P(x=5) is 0.0898. This way when we increase the value of ‘n’ from 20 to 50 (say) and reduce the value of ‘p’ now from 0.1 to 0.04 such that the product remains the same i.e. 2. For this the P(x=0) is 0.1299 and P(x=5) is 0.0902. Now, when we increase the value of ‘n’ from 50 to 100 and the value of ‘p’ reduce from 0.4 to 0.2 such that the product remains the same i.e. 2, then p(x=0) is 0.1326 and P(x=5) is 0.0902. All these probabilities we have calculated by using the Binomial distribution function. Now we calculate these probabilities using the Poisson distribution function, where we take λ=2 as a product of n and p. We remember here that the λ we keep constant. That is, the value of λ is always 2 in this example. So here when we use Poisson distribution function the p(x=0) is 0.1353 and the p(x=5) is 0.0902. So we can see that these two values are very close to Binomial distribution probabilities for n=100 & p=0.02. Now there is another justification for this approximation. We know that the mean and variance of Poisson distribution with parameter λ, is equal to λ. That is, the mean and variance of Poisson distribution are equal. Whereas the mean of Binomial distribution is np and the variance is npq. So, when we take as p→0 in Binomial distribution, then automatically q→1. And when we do this then the mean and variances of Binomial distribution will be same as like of Poisson distribution. 3. NORMAL APPROXIMATION DISTRIBUTION TO BINOMIAL Earlier we discuss the Poisson approximation to Binomial distribution. Now, we are discussing Normal approximation to Binomial distribution. Now here is the example of electric fuses produced by a factory. Here we consider a sample of 1000 fuses and this sample size we denote by ‘n’. In this example there are 27 defective fusses and 973 non- defective fusses. Also the P(observing a defective when a single fuse is tested)=0.02 That is p=0.02 in Binomial distribution. Now the question is: What is P(x ≥ 27)=??? Here x is the number of defective fuses. So, P(x 27) 1000C2 7 (0.02) 2 7 (0.098)9 7 3 ...... (0.02)1 0 0 0 We can see that the calculation of this probability is becomes very tedious when we use the Binomial distribution. This can be simplified if we use the Normal approximation or the Normal probability distribution function to calculate this probability. So, in Normal distribution when we take µ=np then the value of µ is 20. And the value of σ= √npq, i.e., 4.43. Now we make a transformation on x variable. And this transformation is given by z. That is When we substitute the values of µ & σ, I get the value of z as 1.47. Now P (x ≥ 27) = P (z ≥ 1.47). Now we are in a position to make use of the Normal table to calculate this probability. And using the normal table the P (z ≥ 1.47) = 0.0708. Now we consider another example of Binomial distribution; where n=15 and the value of p is 0.04. So if we use Normal approximation the value of µ = np = 6 and the value of σ = √npq =1.897. Now here in this table calculate the P(x=7) using the Binomial distribution function and as well as the Normal distribution function. These values are given in these two columns. So, in the first case the P(x=7) = 0.177, whereas using the Normal distribution function this probability is 0.1826. Now in the next case it gives the P(5 ≤ x<9) this probability is 0.698 when we use Binomial distribution function and this is 0.6918 when we use the Normal distribution function. The last row of this table gives the P(x<7). This probability is 0.62 when we use Binomial distribution function and the same is 0.6026 when we use Normal approximation. So, we can see that these probabilities are very close to each other and we can see that how fair the Normal approximation works in case of the Binomial distribution when ‘n’ is large and ‘p’ is neither close to 1 nor close to 0. So in general the Binomial distribution can be approximated to normal random variable, as 1 1 ( x ) np ( x ) np 2 2 b(x; n, p) P( Z ) npq npq If n is large and p is not close to 0 or 1. Then Normal approximation to Binomial distribution is satisfactory. Also if p is close to ½, the approximation is rather close, even for n as low as 15. We are saying that if n is large and p is not close to 0 or 1 then this approximation is satisfactory. But even if n is small and p=1/2 then this approximation still works very well. So the Working Rule: If both np and nq are greater than 5, Normal approximation to Binomial distribution is satisfactory. Now we shall discuss the Normal approximation to Poisson distribution. As we know that the Poisson distribution function is given by e x f ( x) x! where, x 0,1,... And λ is a positive constant. Now this Poisson distribution can be approximated with Normal distribution for large value of λ. That is, as λ→∞ the Poisson distribution → Normal distribution. Now having discussed the relationship between Binomial, Poisson and Normal distribution we come to the point of estimation of the parameters of these distributions. And especially today we shall discuss about the estimation of population mean µ. Now in order to estimate the population parameters we select a random sample from the given population and by taking the observations on the selected sample, we calculate certain statistics to estimate the parameters of the distribution. 4. INTERVAL ESTIMATION Now we shall discuss the estimation of population mean µ. Now there are two types of estimation: Point estimation Interval estimation Now, if on the basis of a random sample of cars we say that the average of a car is 20 km/liter. Then we say that estimated value of µ =20. So this is Point estimation of µ. But if we make a statement on the basis of the sample observations that the average of a car is between 18 and 22 km/liter. Then it provides me an interval estimate of µ. That is, the µ lies between 18 & 22. Now in order to find the interval estimate of µ, if x1, x2,. . ., xn the random sample is selected from the normal population then, the P(1.96 x 1.96) 0.95......(1) / n Now, for any value of n if sample is selected from a normal population then this relation is true, that we know. But even if sample is not selected from the Normal population but n is large, then using the approximate relationship between Binomial and Normal distribution the relation (1) is true. Now the relation (1), which is the P(1.96 gives me x 1.96) 0.95......(1) / n P( x 1.96 / n x 1.96 / n ) 0.95 and this gives me 95% confidence interval estimate of µ as: ( x 1.96 / n , x 1.96 / n ) So this gives me the lower and upper limits of the confidence interval. Now in general P( z / 2 x n z / 2 ) 1 this is clear from this graph. 1 /2 z / 2 /2 1.96 z / 2 And this in turn gives me the ( 1 ) X 100% confidence interval estimate of µ as: ( x z / 2 n , x z / 2 n ) 5. INTERVAL ESTIMATION : AN EXAMPLE Now here we consider an example to see how a 90% Confidence Interval can be constructed for µ. Suppose we wish to estimate (µ) the average daily yield of a chemical manufactured in a chemical plant. In order to find an estimate of µ, a random sample of 50 days is selected. The daily yield recorded for these n=50 days. It gives (say) x 871tons and the value of σ is 21 tons. Then the 90% Confidence Interval estimate of µ is given by ( x 1.645 / n ) . And that in turns gives me the interval estimate of µ as the interval (866.11, 875.89). That is, the lower limit of this Confidence Interval is 866.11 and the upper limit of this Confidence Interval estimate is 875.89. Therefore, estimated average daily yield of µ is between 866.11 and 875.89 tons. The 90% confidence interval implies that in repeated sampling, if we calculate confidence interval for each sample, then 90% of the confidence intervals would enclose µ. Here, (1 ) 0.90 or in terms of percentage it is (1 ) 100 90% is called confidence coefficient. Now here in this table we give the confidence intervals for the µ i.e. different values of (1- α). So, the first column of this table gives the values of the confidence coefficient. The second column gives me the lower confidence limit and the third column gives the upper confidence limit. so, when (1- α) = 0.90 then the LCL= ( x 1.645 / n ) and UCL= ( x 1.645 / n ) . If I come to the last row of this table then (1- α) = 0.99 then the confidence limits will be ( x 2.58 / n ) and ( x 2.58 / n ) . These values 1.645 or 1.96 or 2.58 all these values comes from the normal table. And these values depend: what is my confidence coefficient? That depends the value of α or (1-α). And we know that the α is the level of significance and (1-α) we are calling the confidence coefficient. So, In general ( x z / 2 / n ) gives an interval estimate of µ with confidence coefficient (1-α). So, using this formula one can construct the different confidence intervals by taking different values of (1-α) not necessarily 0. 90 or 0.95 or 0.99. 6. SUMMARY In today’s lecture we have discussed two important aspects of the three basic distributions namely the Binomial, Poisson & Normal. First we have discussed the relationship between these three distributions. That is, when Poisson distribution can be taken as an approximation to Binomial distribution, or when the Normal distribution can be taken as an approximation of the Binomial distribution, or when we can take Normal distribution as an approximation to Poisson distribution. These approximate relationships help us to calculate the probabilities in a more simplified manner. Then in the second part of this lecture we have discussed the interval estimation of population parameter especially, the population mean µ. This estimation of µ we have discussed in both the cases when the sample is selected from the normal population and even when the sample is not selected from the normal population. So, with this lecture we have come to an end of our discussions on these three basic distributions of statistics namely: the Binomial, Poisson & the Normal. Thank You!