Sampling theory

14,261 views

Published on

1 Comment
13 Likes
Statistics
Notes
No Downloads
Views
Total views
14,261
On SlideShare
0
From Embeds
0
Number of Embeds
15
Actions
Shares
0
Downloads
630
Comments
1
Likes
13
Embeds 0
No embeds

No notes for slide

Sampling theory

  1. 1. SAMPLING THEORY
  2. 2. Sampling TheoryTwo ways of collection of statistical data: 1.Complete Enumeration (or Census) 2. Sample Survey Population (or Universe): Totality of statistical data forming a subject of investigation. Sample : Portion of population which is examined with a view to estimating the characteristics of population.
  3. 3. Methods of Sampling1.Simple Random Samplinga) [Simple Random Sampling without replacement]b) [Simple Random Sampling with replacement]2. Systematic Sampling3.Stratified Sampling4.Cluster Sampling5.Quota Sampling6. Purposive Sampling ( or Judgment Sampling)
  4. 4. Some Important terms associated with samplingParameter : A characteristic of a population based on all the units of the population.Statistics: A statistical measure of sample observation and as such it is a function of sample observations.Statistical inferences are drawn about population values i. e. parameters based on the sample observations i.e. statistics. Usually the following notations are used: Measure Parameter Statistic Mean μ X Proportion P p Standard deviation σ s
  5. 5. Sampling Distribution:Starting with a population of N units, we can draw many samples of a fixed size n. In case of sampling with replacement, the total number of samples that can be drawn is Nn and when sampling is without replacement, the total number of samples that can be drawn is NCn.If it is possible to obtain the values of a statistic (t) from all possible samples of a fixed sample size along with corresponding probabilities, then we can arrange these values of statistics (treating them as random variables ) , in the form of probability distribution. Such a probability Distribution is called Sampling Distribution.
  6. 6. Basic Statistical Laws:1. Law of Statistical Regularity:- It states that a reasonably large number of items selected at random from a large group of items, will on the average represent the characteristics of the group.2. Law of Inertia of Large Number: It states that large groups of data show high degree of stability because there is a greater possibility that one side are compensated by the extremes on the other side.3. Central Limit Theorem : If x1, x2, x3, …….. xn is a random sample of size n drawn from any population (having mean µ and variance σ2), then the distribution sample mean (x) is normally distributed with mean µ and variance σ2/n, provided n is sufficiently large, i.e. n→∞, where µ and σ2 respectively are population mean and variance.
  7. 7. The Mean of the statistic is called ‘Expectation’ and standard deviation of statistic t is called Standard Error.Standard Errors (S.E.) of common statistics: Statistic Standard Error(S.E.)1.Single Mean (x) : σ/√n2.Differences of Means (x-y) : √[σ’ 2/n’ + σ”2/n” ]3. Single Proportion (p) : √[PQ/n]4. Differences of proportion (p’-p”): √[PQ(1/ n’ +1/ n”]The factor √[N-n / N-1] is known as finite population correction factor (fpc)This is ignored for large population. It is used when n/N is greater than 0.05
  8. 8. Examples :1. A simple random sample of size 36 is drawn from finite population consisting of 101 units. If the population S.D. is 12.6, find the standard error of sample mean when the sample is drawn(a) with replacement (b) without replacement . [Ans: a) 2.1 b) 1.69]2. A random sample of 500 oranges was taken from a large consignment and 65 were found to be defective. Show that the S.E. of the proportion of bad ones in a sample of this size is 0.015.
  9. 9. Theory of Estimation :Point Estimation ; When a single sample value (t) is used to estimate parameter (θ), is called point estimation.Interval Estimation: Instead of estimating parameter θ by a single value, an interval of values is defined. It specifies two values that contains unknown parameter.i.e. P ( t ’≤ θ ≤ t” ) = 1 – α. Then [ t’ , t” ] is called confidence interval. α is called level of significance e.g. 5% or 1% l.o.s. 1 – α is called confidence level e.g. 95% or 99% .Confidence LevelThe confidence level is the probability value associated with a confidence interval.It is often expressed as a percentage. For example, say , then the confidence level is equal to (1-0.05) = 0.95, i.e. a 95% confidence level.
  10. 10. Determination of sample size for Mean :The following factors must be known:i) The desired confidence level.ii) The permissible sampling error E = x - µ.iii) The standard deviation σ.The size of sample mean n is given by n = ( σ Z / E )2 .
  11. 11. Determination of sample size for Proportion:The following factors must be known:i) The desired confidence level.ii) The permissible sampling error E = P - p.iii) The estimated true proportion of success.The size of sample mean n is given by n = ( Z2pq / E 2 ). Where q = 1-p
  12. 12. Problems:1. It is known that the population standard deviation in waiting time for L.P.G. gas cylinder in Delhi is 15 days. How large a sample should be chosen to be 95% confident, the waiting time is within 7 days of true average. [18]2. A manufacturing concern wants to estimate the average amount of purchase of its product in a month by the customers whose standard error is Rs.10. Find the sample size if the maximum error is not to exceed Rs.3 with a probability of 0.99 [74]3. The business manager of a large company wants to check the inventory records against the physical inventories by a sample survey. He wants to almost assure that maximum sampling error should not be more than 5% or below the true proportion of accurate records. The proportion of accurate records is estimated as 35% from past experience. Determine the sample size. [819] ************
  13. 13. • Standard deviation and confidence intervals
  14. 14. If t is statistic then 95% confidence interval is given by [ t ± 1.96 S.E.of t] 99% confidence interval is given by [ t ± 2.58 S.E.of t]
  15. 15. There are five ingredients to any statistical test :(a) Null Hypothesis (Ho)(b) Alternate Hypothesis(c) Test Statistic(d) Rejection/Critical Region or Acceptance of Ho(e) Conclusion
  16. 16. Null HypothesisH0: there is no significant difference between the two values (i. e. statistic and parameter or two sample values)Alternative hypothesisH1: The above difference is significant[the statement to be accepted if the null is rejected ]
  17. 17. Type I ErrorIn a hypothesis test, a type I error occurs whenthe null hypothesis is rejected when in fact it istrue; that is, Ho is wrongly rejected.P(type I error) = significance level = 1 – α.Type I error = ( Reject Ho / Ho is true)Type II ErrorIn a hypothesis test, a type II error occurs whenthe null hypothesis Ho, is not rejected when in fact it is false .Type II error = ( Accept Ho / Ho is not true)
  18. 18. Decision Reject Ho Accept HoTruth Ho Type I Error Right decision H1 Right decision Type II ErrorP(RejectHo/Ho is true) = Type I Error = Level of significance (Producer’s risk)P(AcceptHo/Ho is not true) = Type II Error (Consumer’s risk)A type I error is often considered to be more serious, and therefore more important to avoid, than a type II error.
  19. 19. • One tailed test : Here the alternate hypothesis HA is one- sided and we test whether the test statistic falls in the critical region on only one side of the distribution• Two tailed test : Here the alternate hypothesis HA is formulated to test for difference in either direction
  20. 20. Common test statistics Name Formula1.One-sample z-test2.Two-sample z-test3.One-proportion z-test4.Two-proportion z-test,
  21. 21. Critical Value(s)The critical value(s) for a hypothesis test is a threshold to which the value of the test statistic in a sample is compared to determine whether or not the null hypothesis is rejected.For Normal Tests:Critical value (Ztable) Level of Significance 1% 5%Two tailed test 2.58 1.96One tailed test 2.33 1.645
  22. 22. Decision:*If modulus of the computed value of Z is less than table value of Z, then Accept Null Hypothesis Ho. i.e. Calculated |z| < Table z then Accept Ho*If modulus of the computed value of Z is greater than table value of Z, then Reject Null Hypothesis Ho. i.e. Calculated |z| > Table z then Reject Ho
  23. 23. Steps in Hypothesis Testing1. Identify the null hypothesis Ho and the alternate hypothesis H A.2. Choose 1- α (level of significance). The value should be small, usually less than 10%. It is important to consider the consequences of both types of errors.3. Select the test statistic and determine its value from the sample data. This value is called the observed value of the test statistic.4. Compare the observed value of the statistic to the critical value obtained for the chosen l.o.s..5. Make a decision. : -If the test statistic falls in the critical region: Reject Ho in favour of H1. -If the test statistic does not fall in the critical region: Conclude that there is not enough evidence to reject Ho.
  24. 24. Chi Square Goodness of Fit(One Sample Test)This test allows us to compare a collection of categorical data with some theoretical expected distribution.Ho: There is no considerable difference between observed value and theoretical value.H1: The difference is significantChi Square Test of IndependenceFor a contingency table that has r rows and c columns, the chi square test can be thought of as a test of independence. In a test of independence the null and alternative hypotheses are:Ho: The two categorical variables are independent.H1: The two categorical variables are related.
  25. 25. Calculate the chi square statistic x2 by completing the following steps:1.For each observed number in the table subtract the corresponding expected number (O — E).2.Square the difference [ (O —E)2 ].3.Divide the squares obtained for each cell in the table by the expected number for that cell [ (O - E)2 / E ].4.Sum all the values for (O - E)2 / E. This is the chi square statistic .
  26. 26. Example . Incidence of three types of malaria in three tropical regions.  Asia Africa South America Totals  Malaria A 31 14 45 90  Malaria B 2 5 53 60  Malaria C 53 45 2 100 Totals  86 64 100 250 Solution: We now set up the following table  Observed Expected |O -E|  (O — E) 2  (O — E)2/E 31  30.96  0.04  0.0016  0.0000516  14  23.04  9.04 81.72 3.546  45  36.00  9.00 81.00 2.25  2  20.64  18.64 347.45 16.83  5  15.36  10.36 107.33 6.99  53  24.00  29.00 841.00 35.04  53  34.40  18.60 345.96 10.06 45  25.60  19.40 376.36 14.70  2  40.00  38.00  1444.00 36.10
  27. 27. Test Statistic:Chi Square = 125.516(Calculated value)Degrees of Freedom = (c - 1)(r - 1) = 2(2) = 4Reject Ho because 125.516 is greater than 9.488 (for alpha 5% l.o.s.)(Table value)
  28. 28. Oneway analysis of varianceIf the variances in the groups (treatments) are similar, we can divide the variation of the observations intothe variation of the groups (variation of the means) andthe variation in the groups. The variation is measured with the sum of the squares
  29. 29. Analysis of Variance (By Coding Method)Steps in Short Cut Method1.Set the null hypothesis Ho & Alternate hypothesis H12. Steps of computing test statistici] Find the sum of all the values of all the items of all the samples (T)ii] Compute the correction factor C = square of T / N N – the total number of observations of all the samples.iii] Find sum of squares of all the items of all the samples.iv] Find the total sum of squares SST [ Total in (iii) – C]v] Find sum of squares between the samples SSC. [Square the totals of the sample total ,divide by no. of elements in that samples & subtract C from it.]vi] Set up ANOVA table and calculate F, which is the test statistic.vii] If calculated F is less than table F , Accept Ho otherwise Reject Ho.
  30. 30. ANOVA Table Source of Sum of d.o.f. Mean Squares F variation squares Between SSC c-1 MSC=SSC/c-1 Samples Within SSE c(r-1) MSE=SSE/ c(r-1) MSC/MSSamples [orMSE/MSC] (As F ratio is greater than 1)Total SST cr-1 -

×