1. Estimation of Parameters
Umar khayam

2. Introduction
The process of drawing inferences about a population on the basis information contained in a sample taken from the population is called statistical inference. Statistical inference is divided into two major areas: estimation of parameters and testing of hypothesis.

3. Major Areas of Inferential Statistics
Estimation of Parameters
Testing
of Hypothesis

4. Estimation
Estimation is a procedure by which we obtain an estimate of the true but unknown value of a population parameter by using the sample observations from the population. For example we may estimate the mean and the variance of population by computing the mean and the variance of a sample drawn from the population.

5. Testing of hypothesis
Testing of hypothesis is a procedure which enable
us to decide on the basis of information obtained by
sampling whether to accept or reject any specified
statement or hypothesis regarding the value of the
parameter in a statistical problem.
We shall discuss estimation in this chapter and we shall deal with
testing of hypothesis in the next chapter

6. Estimate and Estimators
An estimate is a numerical value of the unknown parameter obtained by applying a rule or a formula , called an estimator, to a sample of size n, taken from the population.

7. Categories of Estimation
Point Estimation
Interval Estimation

8. Point estimate and Interval Estimate
When an estimate for the unknown population parameter is expressed by a single value, it is called a point estimate. An estimate expressed by a range of values within which the true value of the parameter is believed to lie, is referred to as an interval estimate.

9. Example A random sample of n = 3 has the elements
1, 3, and 5. Compute a point
estimates of
The population mean
Population standard deviation

10. Solution
(i) The sample mean is X = = = 3 Thus the point estimate of the population mean µ is 3. (ii) S = - = - = – 9 S = 1.63
∑ X
1+3+5 n 3 2 2
∑ X
∑ X 2 X 1 3 9 5 25 35 n n 35 9 2 3 3

11. A random sample of n = 6 has the elements
1, 3, 6, 9, 12 and 5. Compute a point
estimates of
The population mean
Population standard deviation

12. Estimation by confidence interval
A confidence interval estimate of the unknown parameter Ө is an interval computed from a random sample of n values with a statement of how confident (90%, 95% or 99%) we are that the interval contains the unknown parameter Ө. A confidence interval estimate is in the form (L< Ө< U) , where L is the lower confidence limit for Ө and U is the upper confidence limit for Ө.

13. List of Formulae Note: When σ is unknown, replace σ by s.
(1) 95 % Confidence interval for µ with σ known.
(1) 90 % Confidence interval for µ with σ known.
(1) 99 % Confidence interval for µ with σ known.
Note: When σ is unknown, replace σ by s.

14. Exercise No 4 Page 263.
A soft drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with a standard deviation equal to 1.5 deciliters. Find a 90% confidence interval for the mean of all drinks dispensed by the machine if random sample of 36 drinks had an average content of 22.5 deciliters.

15. Solution of Exercise No 4 Page 263.
Here n =36 x =22.5 σ = % confidence interval for population mean µ or or 22.5 – 0.41 , or (22.09, 22.91)

16. Exercise No 5 Page 263
The heights of a random sample of 50 college students showed a mean of cm and a standard deviation of 6.9 cm. Construct a 95% confidence interval for the mean height of all college students.

17. Solution of Exercise No 5 Page 263.
Here n = 50 X = S = 6.9 The 95% confidence interval for µ or or – 1.91 , or ( , )

18. Exercise 6 page 263
A random sample of 100 automobile owners shows that an automobile is driven on the average miles per year , in the state of Virginia, with a standard deviation of 3900 miles. Construct a 99% confidence interval for the average number of miles an automobile is driven annually in Virginia.

19. Solution of Exercise No 6 Page 263.
Here n = 100 , X = 23500, S = 3900 Therefore 99% C.I. for µ is or or , or ( , )

20. The Kryptonite Corporation personnel director wishes to estimate the mean scores for a proposed aptitude test that may be use in screening applicants for clerical positions. The population standard deviation is assumed to be б=15 For a sample of 100 applicants the sample mean scores is Construct a 95% confidence interval estimate of the true mean

21. Confidence Interval for the difference between the means of two Populations (i.e. 1 – 2):

22. Example 4 page 254
A standardized chemistry test was given to 75 boys and 50 girls. The girls made an average grade of 76 with a standard deviation of 6, while the boys made an average grade of 82 with a standard deviation of 8. find a 95% confidence interval for the difference , where is the mean score of all boys and is the mean score of all girls who might take this test.

23. Solution of Example 4 Page 254
Here
Boys Girls
n1= n2= 50
X1 = 82 X2 = 76
σ1 = 8 σ2 = 6
σ1 = σ2 = 36 2 2

24. Solution of Example 4 Page 254 cont,
Therefore 95% confidence interval for or or , or ( , 8.458)

25. Exercise 14 page 264
A random sample of size 25 taken from a population with standard deviation 5 has a mean 80. A second sample of size 36 taken from a different population with a standard deviation 3 has a mean 75. Find a 90% confidence interval for

26. Solution of Exercise 14 Page 264
Here
Population I Population II
n1= n2= 36
X1 = 80 X2 = 75
σ1 = 5 σ2 = 3
σ1 = σ2 = 9 2 2

27. Solution of Exercise 14 Page 264
Therefore 90% confidence interval for or or ( , ) or (3.161, 6.839)

28. Exercise 15 page 264
Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms, while brand B had an average tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms. Construct a 99% confidence interval for the difference of the population means.

29. Solution of Exercise 15 Page 264
Here Brand B Brand A n1= 50 n2= 50 X1 = 87.2 X2 = 78.3 S1 = 6.3 S2 = 5.6 S1 = S2 = 31.36 2 2

30. Solution of Exercise 15 Page 264 cont;
Therefore 99% confidence interval for or or (8.9 – 3.08 , ) or ( 5.82 , 11.98)

31. A study was made to estimate the difference in salaries of collage professor in the private and state colleges of Virginia. A random sample of 100 professor in private collages showed an average 9month salary of with a standard deviation of A random sample of 200 professor in state collages showed an average salary of with a standard deviation of 1400 find a 98% for the difference between the average salaries

32. What is population proportion ‘’p’’
In many situation one may be interested in estimating certain characteristic of the population. Most common case is that of a binary characteristic. For example in opinion polls the answer is in the form of Yes or No the result of an examination may be pass or fail. Such a binary characteristic of the population is generally referred to as population proportion and is denoted by “p”

33. List of Formulae
(1) 95 % Confidence interval for population proportion P. (1) 90 % Confidence interval for population proportion P. (1) 99 % Confidence interval for population proportion P.

34. Example 8 page 268
In a random sample of 500 people eating lunch at a hospital cafeteria on various Fridays, it was found that x=160 preferred seafood. Find a 95% confidence interval for the actual proportion of people who eat seafood on Friday at this cafeteria. Solution: Here = X / n = 160 /500 = 0.32

35. Solution of Example 8 page 268 cont;
Therefore 95% confidence interval for P or
or ( 0.32 – , )
or ( , 0.36)

36. Exercise 2 page 273
A random sample of 400 cigarette smokers is selected and 86 are found to have preference for Brand A. Find the 90% confidence interval for the fraction of the population of cigarette smokers who prefer brand A. Solution: Here X = 86 and n=400 = X / n = 86 / 400 = 0.215

37. Solution of Exercise 2 page 273 cont;
Therefore the 90% confidence interval for P or or ( , ) or (0.181 , 0.249)

38. Exercise 4 page 273
A sample of 75 college students is selected and 16 are found to have cars on campus. Use a 99% confidence interval to estimate the fraction of students who have cars on campus. Solution: Here n=75 and X= 16 = X/n = 16/75 = 0.21

39. Solution of Exercise 4 page 273
Therefore 99% confidence interval for P or or (0.21 – 0.12 , ) or (0.09 , 0.33)

40. (1) 90% confidence Interval for P1 – P2
Confidence Interval for the difference between two Population proportions P1 – P2:
(1) 90% confidence Interval for P1 – P2
(2) 95% confidence Interval for P1 – P2
(3) 99% confidence Interval for P1 – P2

41. Example 11 page 272
A poll is taken among the residents of a city and the surrounding county to determine the feasibility of a proposal to construct a civic centre. If 2400 of 5000 city residents favor the proposal and 1200 of 2000 county residents favor it, find a 90% confidence interval for the true difference in the fractions favoring the proposal to construct the civic centre.

42. Solution of Example 11 page 272
Here = 244/500= and = 1200/200 = 0.60
Therefore 90% confidence interval for P1 – P2 or
or ( – , )
or ( , )

43. Exercise 10 page 274
In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that 52 of 100 urban residents favor the construction while only 34 of 125 suburban residents are in favor. Find a 95% confidence interval for the difference between the proportion of urban and suburban residents who favor construction of the nuclear plant.

44. Solution of Exercise 10 page 274
Here =52/100 = 0.52 = 34/125 =0.272 Therefore 95% confidence interval for P1 – P2 or or ( , ) or ( , )

45. Exercise 12 Page 274
A geneticist is interested in the proportion of males and females in the population that have a certain minor blood disorders. In a random sample of 100 males, 24 are found to be afflicted, where as 13 of 100 females tested appear to have the disorders. Compute a 99% confidence interval for the difference between the proportion of males and females that have this blood disorder.

46. Solution of Exercise 10 page 274
Here = 24/100 = = 13/100 = 0.13
Therefore 99% confidence interval for P1 – P2 or
or ( , )
or ( , )