1. INFERENTIAL STATISTICS

2. By using the sample statistics, researchers are usually interested in making inferences about the population. INFERENTIAL STATISICS. This branch of statistics is referred to as INFERENTIAL STATISICS. INFERENTIAL STATISICS i.Estimation ii.Hypothesis testing

3. The distribution of values for that sample statistic obtained from all possible samples of a population. The samples must all be the same size, and the sample statistic could be any descriptive sample statistic. Sampling Distribution The distribution is used for judging the quality (precision) of that estimate.

4. Example As a hypothetical example, suppose that in a population of 5 hospitalized patients, the length of hospitalization (in days) are given x 1 =5 x 2 =7 x 3 =4 x 4 =5 x 5 =3 Assume that a sample of size 2 will be selected. The list of all possible samples of size2, and the estimated avarege length of hospitalization per subject will be as follows

5. x 4,x 5 x 3,x 5 x 3,x 4 x 2,x 5 x 2,x 4 x 2,x 3 x 1,x 5 x 1,x 4 x 1,x 3 x 1,x 2 Estimated Average Possible Samples The distribution of sample means is called the Sampling Distribution of mean. This distribution is a Normal Distribution.

6. x 4,x 5 x 3,x 5 x 3,x 4 x 2,x 5 x 2,x 4 x 2,x 3 x 1,x 5 x 1,x 4 x 1,x 3 x 1,x 2 Estimated AveragePossible Samples If say x 4 were 15 instead of 5. What would change?

7. x 3,x 4,x 5 x 2,x 4,x 5 x 2,x 3,x 5 x 2,x 3,x 4 x 1,x 4,x 5 x 1,x 3,x 5 x 1,x 3,x 4 x 1,x 2,x 5 x 1,x 2, x 4 x 1,x 2,x 3 Estimated AveragePossible Samples Now, suppose that a sample of size 3 (n=3) is going to be selected from this population. List of all posible samples of size 3, and estimated average length of hospitalization will be:

8. x 3,x 4,x 5 x 2,x 4,x 5 x 2,x 3,x 5 x 2,x 3,x 4 x 1,x 4,x 5 x 1,x 3,x 5 x 1,x 3,x 4 x 1,x 2,x 5 x 1,x 2, x 4 x 1,x 2,x 3 Possible Samples Estimated Average Again assuming that x 4 =15, sample means are

9. A Normal Distribution is characterized by its mean and its standard deviation (or its variance). The mean of the sampling distribution of the sample mean is always equal to the population mean.

10. The population mean,  (which is unknown) in our hypothetical example is When n=2 mean of the sample means  =(5+7+4+5+3)/5=4.8 days  =(5+7+4+15+3)/5=6.8 days =(6+4.5+5+...+4)/10=48/10=4.8 =(6+4.5+10+...+9)/10=68/10=6.8 When n=3 mean of the sample means =(6+4.5+5+...+4)/10=48/10=4.8 =(6+4.5+10+...+9)/10=68/10=6.8

11. The scatter of all possible sample means, around the unknown population mean is measured by the variance (or standard deviation) of the sampling distribution of the sample means. The variance of all possible sample means is directly proportional with the population variance (  2 ) and inversly proportional with the sample size (n). Thus the variance of the sample means: the standard error of the mean: The standard deviation of sample means is also known as

12. When  (the population standard deviation) is unknown, it can be estimated from the sample. Then, the estimated standard error

13. The distribution of sample mean will possess the following properties: 1.The distribution of will be normal. 2.The mean of the distribution of will be equal to the mean of the population from which the samples were drawn. 3.The variance of the distribution of will be equal to the variance of the population divided by the sample size, Its square root, the standard deviation is referred to as the standard error

14. n=4n=3

15. How we can use these properties: Remember x i ~ N( ,  )  P(X  x 0 ) can be found by transforming x to z, where Instead of individual x i values, if our concern is the probability that the mean of a sample of size n, is greater than (or smaler than) x 0, that is P(  x 0 ) can be calculated similar to the previous transformation:

16. Individual x i ’sSample Mean

17. Example (Regarding X i ’s) If the mean and standard deviation of serum iron values for healty men are 120 and 15  gr/100 ml., respectively, what is the probability that a random selected healty man will have a serum iron value less than 115  gr/100 ml.? P(X i  115)=? P(Z i  -0.33)=0.37

18. (Regarding means, ‘s) If the mean and standard deviation of serum iron values for healty men are 120 and 15  gr/100 ml., respectively, what is the probability that a random sample of 50 men will yield a mean less than 115  gr/100 ml.? P(  115)=? P(Z i  -2.36)=0.009

19. The above properties can also be used to make INTERVAL ESTIMATION. A point estimation is a single numerical value used to estimate the corresponding population parameter  estimates An interval estimate consists of two numerical values defining an interval which, with a specified degree of confidence, we feel includes the parameter being estimated. This interval is also named as “The Confidence Interval” for a parameter.

20. Example The distribution of ‘s are normal. The shaded area covers 95% of the total area. In other words, 95% of the sample means, of the same size, fall within this interval 0.950.025

21. Example Suppose we wish to estimate the average score on the 2nd committee exam. To do this, a random sample of 10 students is selected with the following scores: 64, 37, 78, 73, 44, 59, 81, 51, 33, 46. From previous years experience, I can assume that the standad deviation of scores is approximately equal to 17. Point estimate of the population mean:

22. 95% CI for the population mean: 90% CI for the population mean:

23. STANDARD NORMAL DISTRIBUTION TABLE P(0<Z<z i ) 0.000.010.020.030.040.050.060.070.080.09 0.00.00000.00400.00800.01200.01600.01990.02390.02790.03190.0359 0.10.03980.04380.04780.05170.05570.05960.06360.06750.07140.0753 0.20.07930.08320.08710.09100.09480.09870.10260.10640.11030.1141 0.30.11790.12170.12550.12930.13310.13680.14060.14430.14800.1517 0.40.15540.15910.16280.16640.17000.17360.17720.18080.18440.1879 0.50.19150.19500.19850.20190.20540.20880.21230.21570.21900.2224 0.60.22570.22910.23240.23570.23890.24220.24540.24860.25170.2549 0.70.25800.26110.26420.26730.27040.27340.27640.27940.28230.2852 0.80.28810.29100.29390.29670.29950.30230.30510.30780.31060.3133 0.90.31590.31860.32120.32380.32640.32890.33150.33400.33650.3389 1.00.34130.34380.34610.34850.35080.35310.35540.35770.35990.3621 1.10.36430.36650.36860.37080.37290.37490.37700.37900.38100.3830 1.20.38490.38690.38880.39070.39250.39440.39620.39800.39970.4015 1.30.40320.40490.40660.40820.40990.41150.41310.41470.41620.4177 1.40.41920.42070.42220.42360.42510.42650.42790.42920.43060.4319 1.50.43320.43450.43570.43700.43820.43940.44060.44180.44290.4441 1.60.44520.44630.44740.44840.44950.45050.45150.45250.45350.4545 1.70.45540.45640.45730.45820.45910.45990.46080.46160.46250.4633 1.80.46410.46490.46560.46640.46710.46780.46860.46930.46990.4706 1.90.47130.47190.47260.47320.47380.47440.47500.47560.47610.4767 2.00.47720.47780.47830.47880.47930.47980.48030.48080.48120.4817 2.10.48210.48260.48300.48340.48380.48420.48460.48500.48540.4857 2.20.48610.48640.48680.48710.48750.48780.48810.48840.48870.4890 2.30.48930.48960.48980.49010.49040.49060.49090.49110.49130.4916 2.40.49180.49200.49220.49250.49270.49290.49310.49320.49340.4936 2.50.49380.49400.49410.49430.49450.49460.49480.49490.49510.4952 2.60.49530.49550.49560.49570.49590.49600.49610.49620.49630.4964 2.70.49650.49660.49670.49680.49690.49700.49710.49720.49730.4974 2.80.49740.49750.49760.4977 0.49780.4979 0.49800.4981 2.90.49810.4982 0.49830.4984 0.4985 0.4986 3.00.4987 0.4988 0.4989 0.4990

24. The t-distribution In the previous sections procedures were outlined for constructing CI for a population mean. These procedures require a knowledge about the population standard deviation, , from which the sample is drawn. This condition presents a problem when constructing confidence intervals and when finding probabilities related with the sample mean.

25. is normally distributed When  is unknown, use sample standard deviation to replace  The quantityfollows a t-distribution. Like the standard normal distribution, the t-distribution is also tabulated

26. When  is unknown, its estimate, the sample standard deviation is used to construct CI’s. The inequality becomes:

27. Example A researcher wishes to estimate the average age of the mother at first birth. He selects 10 mothers at random, and gathers the following data: Mother No12345678910 Age at first birth24202619202328221825

28. Point estimate of the population mean: Sample standard deviation: Estimated standard eror of the mean:

29. If the researcher wishes to be 95% confident in his estimate:

30. DISTRIBUTION OF THE SAMPLE PROPORTION From the population, take all possible samples of a given size and for each sample compute the sample proportion, p. The frequency distribution of p is normal. The mean of the distribution,  p, is equal to the population proportion, P. The standard deviation of the distribution (standad error of proportion) is equal to

31. How we can use this property: Example: To find some probabilities related to the sample proportion If x i ’s are normally distributed: Since ’s are normally distributed: Since p’s are normally distributed:

32. Example: Suppose we know that in a certain human population 8% are colorblind. If we randomly select 50 individuals from this population what is the probability that the proportion in the sample who are colorblind will be more than 10%?

33. CONFIDENCE INTERVAL FOR A POPULATION PROPORTION When P, population proportion is unknown, its estimate, the sample proportion, p can be used.

34. A researcher wishes to estimate, with 95% confidence, the proportion of woman who are at or below 20 years of age at first birth. Example

35. Point estimate of the population proportion: p=a/n=4/10=0.4 Estimated standard error of the mean:

36. In the above example if the sample size were 100 instead of 10, then the 95% confidence interval would be:

37. Among 250 students of Hacettepe University interwieved 185 responded that they reqularly read a daily newspaper. With 95% confidence, find an interval within which the proportion of students who regularly read a newspaper in Hacettepe University lie. Point estimate of the proportion of students who read a newspaper. The standard error of the estimate is 0.028.

38. In oder words, the standard deviation of the proportions that can be computed from all possible samples of size 250 is 0.028. The 95% Confidence Interval is: