Parameter Estimation Statistics PDF PDF

 Estimation of Parameters Estimation of Parameters is alo refering to the sampling distribution of a sample statistic. It is important to learn estimation of parameters to enables us to draw conclusions about the corresponding population parameter based on a random sample. The valu...

 Estimation of Parameters Estimation of Parameters is alo refering to the sampling distribution of a sample statistic. It is important to learn estimation of parameters to enables us to draw conclusions about the corresponding population parameter based on a random sample. The value sample mean based on the sample at hand is an estimate of the population mean. Estimation - the process of making inferences about a population based on information obtanined from a sample. Point Estimate - the sample mean x of the population or mean µ. It is the numerical valuewhich gives an estimate of a parameter. Interval Estimate - a range of values used to estimate the parameter. It can be calculated using two numbers or values which may or may not contain the value pf the parameter being estimated. Illustrative Example: Using the same data that we used previously, 86, 89, 92, 95, 98 Randomsample( n 3) Samplemean( x ) Sample median 86, 89, 92 89 89 86, 89, 95 90 89 86, 89, 98 91 89 86, 92, 95 91 92 86, 92, 98 92 92 86, 95, 98 93 95 89, 92, 95 92 92 89, 92, 98 93 92 89, 95, 98 94 95 92, 95, 98 95 95 Looking at the second column, obseve that the sample mean, 89, 90, and 95 appeared only once. Thus, their probabilities are all 0.1. SInce the mean grades 91, 92, and 93 appeared twice, their probabilities are 0.02. Hence, we obtain from the first table the following values: SampleMean x  P (x ) 89 0.1 90 0.1 91 0.2 92 0.2 93 0.2 94 0.1 95 0.1 Constucting a Probability Histogram of the Sample Mean: P (x ) 0.2 0.1 0.0 89 90 91 92 93 94 95 Illustrative Example: Estimate the mean consumption of 6 families in one month if their expenses are ₱14,200, ₱15,500, ₱17,500, ₱20,000, and ₱27,000. Solution:   X  14,200  15,500  16,800  17,500  20,000  27,000 N 6 111,000  6 x   ₱18,000 is the point estimator Interval Estimation - gives us a range of values which is likely to contain the population parameter. It can be dtermined by two values. Biased and Unbiased Estimators Unbiased Estimator - is an estimator in which the mean of a sampling distribution is equal to the population parameter, or x . x  Biased Estimator - is an estimator in which the mean of the mean of the distribution is not equal to the population, or x . x  Error of Estimate - is the distance between the estimate and the true value of a parameter. Confidence level - expressed as percent, it sets a portion of the sample to be included within a known range of the true population. If   is the allowable sampling error, the confidence level is 1 -. Confidence interval - the interval defined within the true population were members of the sample are expected to be found. Illustrative Example: The graph shows that 95% of the population distribution is contained (95% shaded region) in the confidence interval. A con- 2.5% 2.5% fidence level of 1 - α, when α = 5% means that there are 95% probabi- From the z-table of normal curve, the value of z  at A = 0.475 is ± 1.96. 2 The Confidence Interval of the population mean with a given confidence level of (1 - α)100% when the population varis unknown is:  x    x   x  x z    or  x  z   , x  z    2 n  2 n 2 n  Where: x = sample mean  x = sample standard deviation oro the square root of the sample variance;  x = the standard error of the mean; and n   x  = margin of error z   2 n illustrative Example 1: A marketing officer wishes to collect 300 female receptionists from a group of employees. The selected group has an average height of 170 cm and a sample standard deviation of 25 cm. What is the 95% confidence interval of all the employees' heights? Solution: Given: sample x size n = 300 sample  xmean = 170 sample std. deviation = 25 Confidence Interval gives 95%; 95/2 = 47.5% = z 1.960 0.4750 2 and from z-table:   x   x  Substitute  x  zthe values  , on x the z formula:      n  n   2   2    Sustituting values:  25   25  170  1.960  , 170  1.960   300   300   25   25  170  1.960  , 170  1.960   17.32   17.32  167.171, 172.829 (167, 173) Thus, the marketing officer in 95% confident that the employeeshave a mean height of 167 to 173 cm. Illustrative Example 2: A nutritionist is interested in monitoring the calorie intake of women. It was found that in a selected random of 50 females adults, the average daily intake of meat products is 2,200 calories per day with std. deviation of 300 calories. Construct a 99% confidence interval for the mean intake of calories from meat products of female adults. Solution: Given: n = 50; x = 2,200; and  x = 300 Confidence interval of 99% = 99/2 = 49.5% = 0.495; z  = 2.57 2 Substituting values:  300   300  2,200  2.57  , 2,200  2.57   50   50  = = ; =2,200.005 = 2,091; 2,200 Thus there is 95% confidence interval between 2,091 and 2,200 Confidence Interval for the Difference Between Two Population Means There are several cases or problems in real life in which two means are mentioned. Such as: The result of final exam in Math based on the mean scores of the students from College of Business Adminsitration compared to scores of the students from College of Arts and Sciences. The average weight of children using two different vitamins. The mean output of workers executing two different strategies of a manager. The probability interval for the difference between two population means, we use formula as: 2 2 s1 s2 x1  x2 z   2 n1 n2 Illustrative Example 1: Two popular inventors created different mechatronics machines. Their efficiency was tested based on average number of product finished by each of machine per day. Samples of both machines were taken and their resulting mean output and standard deviations are as follows: Machine Mean output Standard Number of Deviation machines DP Machine x1 3,550 items s1 130 items 50 DF Machine x2 3,500 items s2 120 items 40 1. Construct a 95% confidence interval of the difference of mean outputs of the two types of mechatronics mchines. 2. Is it possible to conclude that there is no significant difference between the tyo types of machines based on their mean outputs. Solution: At 95% confidence interval. 95/2 = 47.5% = 0.4750, then from the z-table, z 1.96. 2 2 2 s1 s2 (130) 2 (120) 2 1. Using the: x1  x2 z   (3,550  3,500) 1.96  n1 n2 50 40 2 50 1.96 338  360 50 1.96 698 50 1.96(26.42) 50 51.78 ( 1.78, 101.78) Thus, the 95% confidence interval of the diffrence of the mean outputs of the two mechatronics machine is (-1.78, 101.78) items a day. ACTIVITY: Compute for the 90% confidence interval of the difference between two means when: x1 28, x2 25, s1 4, s2 3, and n1 n2 50 Estimating Sample Size with unknown Population Size The sample size n can be estimated based on the margin of error ME using the following formulas:    ME  z    2 n To be able to get the formula for n for estimating sample size n, we have to derive the above formula by multiplying each side by n ME Cont'n of derivation: n    n     ME  z    n z   ME 2 n  ME 2  ME  Square both sides of the equation: 2  2    n  z     2  ME   Thus: 2     n  z     2  ME   Illustrative Example: A office is conducting an investigation in a crime scene. He wanted to have a 95% confidence interval with a margin of error of 2 and a standard deviation of 8. How many respondents must he interview to achieve the desired result? Solution: Given: ME = 2; σ = 8; α = 95%; 95/2 = 47.5 = 0.475 From z-table: z  1.96 Unknown: n 2 2    Using formula: n  z      ME   2 Substituting values: 2   8  n 1.96   7.84 61.47 2   2  n 61 or 62 CHAPTER ACTIVITY: 1. An interval that gives an estimated range of values within which the population parameter x is expected ta fall is called____________. 2. Which of the expression n for? a. standard error of the mean c. the level of confidence b. the level of confidence minus one d. the level of confidence plus one 3.-5. A random sample of 150 lemon fruits was drawn from 3,000 delivered lemons in the market. The average weight of lemon is 110 g with a z standard deviation of 10 g. What is the 99% confidence interval of the 2 2 population mean? a. (11.8, 12.9) c. (13.6, 14.9) b. (14.16, 115.82) d. (9.75, 12.0)  Cont'n of CHAPTER ACTIVITY.. 8.-10. The set of data below shows the comparison between the average sales (in millions) of the two leading fast food chains in the country from their randml selected 30 branches. Fast Food Chain Sample Space Mean sales Sandard Deviation (in millions) (in millions) FFCA n1 = 30 µ1 = 3.5 σ1 = 1.2 FFCB n2 = 30 µ2 = 3.8 σ2 = 1.4 What is the confidence interval of the difference between the two means?