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1 8.1 Introduction The field of statistical inference consist of those methods used to make decisions or to draw conclusions about a population. These methods utilize the information contained in a sample from the population in drawing conclusions. 2 Estimator • In statistics, the method used Estimate • The value that obtained from a sample I have a sample of 5 numbers and I take the average. The estimator is taking the average of the sample. The estimator of the mean. Let say, the average = 4 the estimate. Point estimate – an estimate of the parameter using a single number. ◦ E.g : x is the point estimate for In choosing the point estimators, we have to depends on the properties of the estimators. Unbiased Consistent Efficient Sufficient 2 x x x2 ... xn E(x ) E 1 n 1 E x1 E x2 ... E xn n 1 1 2 ... n n 1 n i n i 1 1 n n Example 8.2 If a random sample size n is taken from a population with mean, and variance 2, hence x (mean sample) is a consistent estimator for . x x ... xn V x V 1 2 n 1 2 V x1 x2 ... xn n 1 2 V x1 V x2 ... V xn n 1 2 2 2 ... 2 n 1 2 2 2 n n n lim var ˆ 0 n 2 lim 0 n n 6 . Example 8.3 Sample mean is an efficient estimator compares to sample median in estimating the population mean. Proof: 2 Variance for sample mean, var x n 2 if n is odd Variance for sample median, var x 2 if n is even 2 This gives var x var x if n2 Thus, significantly, mean is more efficient in estimating 7 Sufficient •If it used all the sample’s information 8 X 2 S 2 n 1 X I X n 1 i 1 P̂ 2 Definition 8.1: An Interval Estimate In interval estimation, an interval is constructed around the point estimate and it is stated that this interval is likely to contain the corresponding population parameter. Definition 8.2: Confidence Level and Confidence Interval Each interval is constructed with regard to a given confidence level and is called a confidence interval. The confidence level associated with a confidence interval states how much confidence we have that this interval contains the true population parameter. The confidence level is denoted by 1 100% . 10 The (1 )100% Confidence Interval of Population Mean, (i) x z 2 n if is known and normally distributed population or x z x z 2 2 n n s (ii) x z if is unknown, n large (n 30) 2 n s s or x z x z 2 2 n n 11 s (iii) x tn 1, if is unknown, normally distributed population 2 n and small sample size n 30 s s or x tn 1, x tn 1, 2 2 n n 12 If a random sample of size n 20 from a normal population with the variance 2 225 has the mean x 64.3, construct a 95% confidence interval for the population mean, . 13 It is known that, n 20, x 64.3 and 15 For 95% CI, 95% 100(1 – )% 1 – 0.95 0.05 2 0.025 z z0.025 1.96 2 14 Hence, 95% CI x z n 2 15 64.3 1.96 20 64.3 6.57 [57.73, 70.87] @ 57.73 70.87 Thus, we are 95% confident that the mean of random variable is between 57.73 and 70.87 15 Example 8.5 : A publishing company has just published a new textbook. Before the company decides the price at which to sell this textbook, it wants to know the average price of all such textbooks in the market. The research department at the company took a sample of 36 comparable textbooks and collected the information on their prices. This information produced a mean price RM 70.50 for this sample. It is known that the standard deviation of the prices of all such textbooks is RM4.50. Construct a 90% confidence interval for the mean price of all such college textbooks. 16 solution It is known that, n 36, x RM70.50 and RM 4.50 For 90% CI, 90% 100(1 – )% 1 – 0.90 0.1 2 0.05 z z0.05 1.65 2 17 Hence, 90% CI x z n 2 4.50 70.50 1.65 36 70.50 1.24 [ RM 69.26, RM 71.74] Thus, we are 90% confident that the mean price of all such college textbooks is between RM69.26 and RM71.74 18 The (1 )100% Confidence Interval for p for Large Samples ( n 30) pˆ z 2 pˆ 1 pˆ n or pˆ z pˆ 1 pˆ 2 n p pˆ z pˆ 1 pˆ 2 n 19 Example 8.6 According to the analysis of Women Magazine in June 2005, “Stress has become a common part of everyday life among working women in Malaysia. The demands of work, family and home place an increasing burden on average Malaysian women”. According to this poll, 40% of working women included in the survey indicated that they had a little amount of time to relax. The poll was based on a randomly selected of 1502 working women aged 30 and above. Construct a 95% confidence interval for the corresponding population proportion. 20 Solution Let p be the proportion of all working women age 30 and above, who have a limited amount of time to relax, and let pˆ be the corresponding sample proportion. From the given information, n 1502 , pˆ 0.40, qˆ 1 pˆ 1 – 0.40 0.60 ˆ Hence, 95% CI p z 2 ˆˆ pq n 0.40(0.60) 0.40 1.96 1502 0.40 0.02478 [0.375, 0.425] or 37.5% to 42.5% Thus, we can state with 95% confidence that the proportion of all working women aged 30 and above who have a limited amount of time to relax is between 37.5% and 42.5%. 21 Definition 8.3: If x is used as an estimate of , we can be 100(1- )% confident that the error | x | will not exceed a specified amount E when the sample size is z /2 n E 2 22 Example 8.7: A team of efficiency experts intends to use the mean of a random sample of size n=150 to estimate the average mechanical aptitude of assembly-line workers in a large industry (as measured by a certain standardized test). If, based on experience, the efficiency experts can assume that 6.2 for such data, what can they assert with probability 0.99 about the maximum error of their estimate? 23 Solutions Substituting n 150, 6.2, and z0.005 2.575 into the expression for the maximum error, we get E z / 2 n 2.575(6.2) 1.30 150 Thus, the efficiency experts can assert with probability 0.99 that their error will be less than 1.30. 24 x If pˆ is used as an estimate of p, we can assert with n (1- )100% confidence that the error is less than z /2 If we set E z /2 pˆ (1 pˆ ) . n pˆ (1 pˆ ) and solve for n, the appropriate n sample size is 2 z /2 n pˆ (1 pˆ ) E 25 Example 8.8: A study is made to determine the proportion of voters in a sizable community who favor the construction of a nuclear power plant. If 140 of 400 voters selected at random favor the 140 project and we use pˆ 400 0.35 as an estimate of the actual proportion of all voters in the community who favor the project, what can we say with 99% confidence about the maximum error? 26 Solution Substituting n 400, pˆ 0.35, and z0.005 2.575 into the formula, we get E z /2 pˆ (1 pˆ ) n (0.35)(0.65) 2.575 0.061 400 Thus, if we use pˆ 0.35 as an estimate of the actual proportion of voters in the community who favor the project, we can assert with 99% confidence that the error is less than 0.061. 27 Example 8.9: How large a sample required if we want to be 95% confident that the error in using p̂ to estimate p is less than 0.05? If pˆ 0.12, find the required sample size. 28 Solution 2 z0.025 n pˆ (1 pˆ ) E 2 1.96 0.12(0.88) 162 0.05 29