Lecture Probability & statistics: Chapter 8

Lecture 8. ESTIMATION Lecture 8. ESTIMATION

 [1] Chapter 8, Chapter 11

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 Concept of Estimate  Point Estimate  Maximum Likelihood Estimate  Interval Estimate  For Mean  For Proportion  For Variance (pp. 484 – 488)

8.1. Concept of Estimate 8.1. Concept of Estimate

 Estimation is determining the approximate value of a

unknown parameter on given data.

 Two types of estimate:  Point estimate: single value  Interval estimate: an interval that parameter falls into

it with a determined probability level

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Ex. “The average height of VN’s male is 163 cm” “The average height of VN’s male is from 160 to 165 cm”

8.2. Point Estimate 8.2. Point Estimate

 Point estimate of parameter is denoted by  Estimator: a statistic calculated on random sample, an approximation to unknown parameter. Estimator is a random variable

 Estimate: specific value calculated on observed

sample. Estimate is number.

Ex. Formula ̅ = is an estimator

∑

Value ̅ = = 5 is an estimate

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Point Estimate Point Estimate

 Population parameters ,, are unknown  Estimate by statistics from sample

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 Sample mean ̅ is point estimate for  Sample variance is point estimate for  Sample proportion ̅ is point estimate for

Properties of Point Estimate Properties of Point Estimate

 Estimate for by  Unbiased: is unbiased estimator of

=  Efficient: and are unbiased, is more efficient

than :

<  If is smallest in every unbiased estimator: most

efficient, best estimator

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 BUE (Best Unbiased Estimator)

Example Example

Example.8.1. Population mean , sample (,,… ,) (a) Which of the followings are unbiased estimator? (b) Which of the unbiased estimator is more efficient?

= + = +

= + = + +

= + = + + ⋯ +

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Example Example

Example.8.2. Population mean of , there are two sample. The first sample has size of 4, and mean of ̅; the second sample has size of 8 and sample mean ̅. Which are the unbiased estimator, and which is more efficient in the followings:

= ̅ + ̅ = ̅ + ̅

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8.3. Maximum Likelihood (ML) Estimate 8.3. Maximum Likelihood (ML) Estimate

Ex. Probability that a student pass an exam is = 0.6.  Which of the following sample is most likely to occurs = (,,) = (,,) = (,,) = (,,)

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Likelihood function is Probability  = 0.6 ∗ 0.4 ∗ 0.4 =  = 0.4 ∗ 0.6 ∗ 0.4 =  = 0.4 ∗ 0.4 ∗ 0.4 =  = 0.6 ∗ 0.6 ∗ 0.6 =

Maximum Likelihood (ML) Estimate Maximum Likelihood (ML) Estimate

Ex. The sample is: = (,,,)  Which of the following value of is most likely?

= 0.2; = 0.4; = 0.6; = 0.8

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Likelihood function  = 0.2  = 0.2 ∗ 0.8 ∗ 0.2 ∗ 0.2 =  = 0.4  = 0.4 ∗ 0.6 ∗ 0.4 ∗ 0.4 =  = 0.6  = 0.6 ∗ 0.4 ∗ 0.6 ∗ 0.6 =  = 0.8  = 0.8 ∗ 0.2 ∗ 0.8 ∗ 0.8 = Find the maximum likelihood estimate?

8.4. Interval Estimate for Mean 8.4. Interval Estimate for Mean

 The interval (,)is interval estimate of , then

( < < ) is equal to (1 − )

 (,)is confidence interval  (1 − )is confidence level  = − is the width of confidence interval  : Lower Limit (LL)  : Upper Limit (UL)  The shorter (narrower) the is, the more accurate

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the estimate is.

Confidence Interval for Mean Confidence Interval for Mean

< < − ()/ − ()/

 Population is unknown, estimate for the mean  Sample size , mean of ̅, variance of  Confidence level (1 − ), confidence interval

 or ± ()/

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 Value in the Table 2 (p. 976); ≈

Confidence Interval for Confidence Interval for

 Marignal Error = ()/

 Confidence interval: ±

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= /  Narrower the Confidence Interval  Reduce standard deviation  Reduce confidence level  Increase sample size

Example Example

Example 8.3. From the normality population, collect two samples  Sample 1: 15, 17, 16, 20, 17  Sample 2: 18, 13, 14, 20, 15, 13

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With confidence level of 95%, find confidence interval of mean by using data in Sample 1, Sample 2, and Pooled sample

Example Example

 Example 8.3 (cont.)  Sample 1:

= ̅ =

???

̅ ̅ ̅ ̅ ̅ ???

= = 5; 1 − = 0.95

= 8.3

= 6.164

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 Sample 2: ̅ = 15.5  Pooled sample: ̅ = 16.182

Example Example

Example 8.4. To estimate the average score of students in Maths, 40 students’ score are collected, and the sample mean is 74.5, and sample variance is 64. Assumed that score is normal distributed (a) Find the 95% confidence interval of average score (b) Find the 90% confidence interval of average score (c) To reduce interval width to less than 4:

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(c1) With level 95%, how many observations should be surveyed (c2) With above sample, how much confidence level should be

8.5. Interval Estimate for Proportion 8.5. Interval Estimate for Proportion

 Population proportion is unknown  Sample has size of , sample proportion ̅  Confidence interval level (1 − )

− / < < + /

( − ) ( − )

()

 Or ± with = /

() /

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 Sample size: =

Example Example

Example 8.5. In 200 observed visitors, 50 of them buying goods, and 40 using services. With confidence level of 95% (a) Estimate of buyer proportion in the visitors (b) To have interval that narrower than 10%, how many

observations should be surveyed?

services

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(d) In 4000 visitors, estimate the total number of buyers

8.6. Interval Estimate for Variance 8.6. Interval Estimate for Variance

< <

 Population variance is unknown  Sample has size of , sample proportion  Confidence interval level (1 − ) ( − ) / ( − ) /

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Example 8.6. To estimate the variability of the time spent on producing, manager random observed 40 times and calculated the variance was 12.5 minutes2. Find the 95% interval estimate of the producing time’s variance, assumed that time is normal distributed.

Key Concepts Key Concepts

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 Point Estimate  Unbiased, Efficient Estimator  Confidence Interval  Confidence Level, Marginal Error  Confidence Interval for Mean  Confidence Interval for Proportion  Confidence Interval for Sample