Introduction to Estimation
POINT & INTERVAL ESTIMATION Basic definitions and concepts AND The assignment of value(s) to a population parameter (INTRODUCTION TO TESTING) based on a value of the corresponding sample statistic is called estimation.
The value(s) assigned to a population parameter based on the value of a sample statistic is called an estimate.
The sample statistic used to estimate a population parameter is called an estimator.
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Estimation steps Point Estimators
The estimation procedure involves the following steps: A Point Estimation 1. Select a sample. The value of a sample statistic that is used to estimate a population parameter is called a point estimate. Usually, 2. Collect the required information from the members of whenever we use point estimation, we calculate the margin of the sample. error associated with that point estimation, which is s 3. Calculate the value of the sample statistic. calculated as follows: Margin of error 1.96 x or 1.96 x 4. Assigg()pgppn value(s) to the corresponding population Point estimate is based on just one sample, we can not parameter. expect it to be equal to the corresponding population parameter. Indeed, each sample will have a different, non of them is equal to . But they are all unbiased estimates of . (Recall that unbiased means their expected value is equal to .)
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Statistics Point Estimates Statistics Parameters Z A statistic is a property of a sample from the Z In statistical inference, the term parameter is used to denote population. a quantity, say , that is a property of an unknown probability Z A statistic is defined to be any function of random distribution. variables. So, it is also a random variable. For Z For exampp,le, the mean, variance, or a p articular quantile of example, the sample mean, sample variance, or a the probability distribution particular sample quantile. Z Parameters are unknown, and one of the goals of statistical Z The observed value of the statistic can be calculated calculated inference is to estimate them . from the observed data values of random variables. Estimation Examples of statistics: Z A procedure of “guessing” properties of the population from which data are collected . X XX sample mean X 12 n Z A point estimate of an unknown parameter is a statistic that n represents a “guess” of the parameter of interest. n 2 Z B ()XXi There may be more than one sensible point estimate of a sample variance S 2 i1 n parameter. 5 1 6 The relationship between an unknown Properties of Estimators that We Desire parameter and its point estimator UbiUnbiased ness: E( ˆ ) In other words we would wish that the expected value of the estimator is the same as its true value. We define bias of an estimator as the difference between the expected value of the estimator and the true value in the pppopulation: Efficiency: we wish to minimize the mean square error around the true value. The efficiency tells us how well the estimator performs in predicting. Among unbiased estimators therefore, we want the one with the smallest variance. Consistency. As sample size increases, variation of the estimator from the true population value decreases.
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Unbiasedness Efficiency
P(X) Sampling Unbiased Biased P(X) Distribution of Median Sampling Distribution of Mean
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Consistency Interval estimation: Larger General approach P(X) sample size B Smaller sample size A
11 12 Interval Estimation Confidence Interval Estimation
Definition Outline: In interval estimation, an interval is constructed around Procedure: the point estimate, and it is stated that this interval is likely 1. Sample point estimator ( X or p ) to contain the corresponding population parameter. 2. Confidence level and Table Zortn-1 3. Formulas compute UCL and LCL: point estimator margin of error
x x $1370
13 $1130 $1610
Interval Estimation of the Population Mean Each interval is constructed with regard to a given confidence level Interval estimation of a population and is called a confidence interval. The confidence level associated with a confidence interval states how mean: The case of known much confidence we have that this interval contains the true population parameter. The confidence level is denoted by (1 – ))%100%. The (1 – )100% confidence interval for (population mean) is:
x z x if is known and x zsx if is not known,
where x / n and sx s/ n
The value of z used here can be found from the standard normal distribution table, for the given confidence level. The maximum error of estimate for , denoted by E, is the quantity that is subtracted from and added to the value of x to obtain a confidence interval for . Thus, E z x or zsx 16 17
Interval Estimation of the Population Mean Interval Estimation of the Population Mean when is when is known: Example known: Answers to the Example x A publishing company has just published a new college textbook. Here we take advantage of our knowledge on distribution of to Before the comppyany decides the p rice at which to sell this develop a confidence interval for . 4.50 textbook, it wants to know the average price of all such a) n = 36, x = $70.50, and = $4.5, thus: $.75 x n textbooks in the market. The research department at the Point estimate of = x = $70.50 36 company took a sample of 36 comparable textbooks and Margin of error = 1.96 x 1.96(.75) $1.47 collected information on their prices. This information produces b) Confidence level is 90% or .90; and z = 1.65. a mean price of $70.50 for this sample. It is known that the x z standddiidard deviation of fh the pri ces of fllh all such textb oo ki$450ks is $4.50. x 70.50 1.65(.75) 70.50 1.24 (a) What is the point estimate of the mean price of all such (70.50 -1.24) to (70.50 1.24) textbooks? What is the margin of error for the estimate? $69.26 to $71.74 (b) Construct a 90% confidence interval for the mean price of all Based on our results, we can say that we are 90% confident that such college textbooks. the mean price of all such college textbooks is between $69.26 and $71.74. 18 19 Example 1: 1. Interval Estimation for Population Mean Answer: Example 1: ( known case) • n = 49 In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant. Data were collected X $24.8 for a sample of 49 customers over a three-week period. Assume a population standard deviation of $5. =$5 a. At the 95% confidence, what is the margin error? • Z: (1- )/2 = 0.95/2 = 0.475 Table 1: Z = 1.96 5 •1. Z x (1.96) 1.4 b. If the sample mean is $24.80, What is the 95% / 2 n 49 confidence interval for the population mean? 2. UCL X Z x 24 .8 1.4 26 .2 / 2 n LCL X Z x 24 .8 1.4 23 .4 / 2 n : [23.4, 26.2]
Interval Estimation of the Population Mean when is unknown
Interval estimation of a population Instead of population standard deviation we have mean: The case of unknown sample standard deviation s.
Instead of normal distribution, we have t distribution The t distribution is used to construct a confidence interval about if: 1. The popul ati on f rom whi c h the samp le is drawn is (approximately) normally distributed; 2. The sampp(,);le size is small (that is, n < 30); 3. The population standard deviation, , is not known.
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t The Distribution The t Distribution: Example The t distribution is a specific type of bell shaped distribution with a lower height and a wider spread than the standard normal Find the value of t for 16 degrees of freedom and .05 area in distribution. As the samppg,le size becomes larger, the t t distribution approaches the standard normal distribution. A the r ig ht ta il o f a dis tr ibu tion curve. specific t distribution depends on only one parameter, called the Area in the Right Tail Under the t Distribution Curve degrees of freedom (df). The mean of the t distribution is equal df .10 .05 .025 ….001 df df to 0 and its standard deviation is found by . The /( 2) 1 3.078 6.314 12.706 … 318.309 graph below depicts the case of df=3. 2 1.886 2.920 4.303 … 22.327 3 1.638 2.353 3.182 … 10.215 The standard deviation of the The standard deviation of the t . … … … … … standard normal distribution is 1.0 distribution is 9 /(9 2) 1.134 16 1.337 1.746 2.120 … 3.686 . … … … … …
The required value of t for 16 df and .05 area in the right tail. 24 25 = 0 Confidence Interval for Population mean Using t (continued) The Distribution the t Distribution
The t distribution with 16 degrees of freedom, areas under the right and the left tails . The (1 – )100% confidence interval for is s x ts where s x x n
The value of t is obtained from the t distribution table for n – 1 dffddhfdlldegrees of freedom and the given confidence level. .05 1.746
26 -1.746 0 27
Confidence Interval for Population mean Using Confidence Interval for Population mean Using the t Distribution: Example the t Distribution: Example Answered
Dr. Moore wanted to estimate the mean cholesterol level for Confidence level is 95% or .95, with df = n – 1 = 25 – 1 = 24 Area in each tail = .5 – (.95/2) = .5 .4750 = .025 all adult men living in Hartford. He took a sample of 25 adult s The value of t in the right tail is 2.064, and s 12 men from Hartford and found that the mean cholesterol x 2.40 n 25 level for this sample is 186 with a standard deviation of 12. Assume that th e c hlholestero lll leve lfls for a lldlll adult men in df Hartford are (approximately) normally distributed. Construct = 24 a 95% confidence interval for the population mean . .025 .025 .4750 .4750
x tsx 186 2.064(2.40) 186 4.95 181.05 to 190.95
Thus, we can state with 95% confidence that the mean cho les tero l l eve l f or all ad u lt men li v ing i n H arf or d li es b e tween 181.05 and 190.95. 28 29
Example 2: Example 2: ( known case) Given: n = 100,,,, X = 49, S = 8.5, 1- = .95 The mean flying time for pilots at Continental Think: What to estimate? Use Z or t? Airlines is 49 hours per month. This mean was Answer: based on a sample of 100 pilots and the sample • Sample info (given): n = 100,X = 49, S = 8.5 standard deviation was 8.5 hours. • t: 1- =0.95, so /2=0.025, d.f.=n-1=99 Table 2: dfd.f.=100, /2=0.025 t=1.984 a. At 95% confidence, what is the margin of error? d.f.=80, /2=0.025 t=1.990 100 99 b. What is the 95% confidence interval estimate of *Interpolation: t 1.984 (1.990 1.984) 1.9843 the population mean flying time? 100 80 S 8.5 c. The mean flying time for pilots at United Airlines a. m.o.e.: m.o.e. t 1.9843 1.69 / 2 n is 36 hours per month. Discuss difference 100 between the flying times at two airlines. b. UCL = 49+1.69 =50.69 LCL = 49 – 1.69 = 47.31 : [47315069][47.31, 50.69] c. 36 < LCL. The mean flying time is lower at United. Example Solution Twenty five randomly selected adults who buy books for general reading were asked how much they Confidence level is 99% or .99 s 300 usually spend on books per year. sx $60 n 25 df n The sample produced a mean of $1450 and a = – 1 = 25 – 1 = 24 standar d dev iat ion of $300 for suc h annua l expenses. Area in each tail = . 5 – (.99/2) = .5 -.4950 = . 005 The values of t are 2.797 and -2.797 Assume that such expenses for all adults who buy books for general reading have an approximate The 99% confidence interval for is normal distribution. x tsx $1450 2.797(60) Determine a 99% confidence interval for the corresppgpponding population mean. $1450 $167.82 $1282.18 to $1617.82 32 33
Interval estimation of a population proportion: The case of larggpe samples Interval estimation of a population Estimator of the Standard Deviation of pˆ The value of , s which gives a point estimate of , is proportion: The case of large samples pˆ pˆ calculated as pq s ˆ ˆ pˆ n
The (1 – )100% confidence interval for the population proportion, p, is p zs ˆ pˆ
The value of z used here is obtained from the standard normal distribution table for the given confidence level, and s pq/n pˆ ˆ ˆ 34 35
Interval estimation of a population proportion in Interval estimation of a population proportion in case of large samples: Example case of large samples: Example solved According to a 2002 survey by FindLaw.com, 20% of n = 1000, = .20, and, = .80 pˆ qˆ Americans needed legal advice during the past year to Note that npˆ and nqˆ are both greater than 5. resolve such thorny issues as family trusts and landlord pq disputes (CBS.MarketWach.com, August 6, 2002). Suppose s ˆ ˆ (.20)(.80) pˆ .01264911 a recent sample of 1000 adult Americans showed that 20% n 1000 of them needed legal advice during the past year to resolve a) Point estimate of p = = .20pˆ s such family related issues. pˆ a) What is the point estimate of the population MifMargin of error = ±1961.96 = ±1.96( . 01264911) = ± .025 or ±25%2.5% proportion? What is the margin of error of this b) The confidence level is 99%, or .99. estimate? b) Find, with a 99% confidence level, the percentage of all The z value for .4950 is approximately 2.58. adult Americans who needed legal advice during the pˆ zs .20 2.58( .0126 4911) .20 .033 past year to resolve such family related issues. pˆ 36 37 .167 to .233 or 16.7% to 23.3% Determining the sample for estimation of the Determining the sample size for estimating the mean ppproportion Given the confidence level and the values of p and q, the Given the confidence level and the standard deviation of the sample size that will produce a predetermined maximum population, the sample size that will produce a error E of the confidence interval estimate of p is: z 2 pq pq predetermined maximum error E of the confidence interval n , with E z p z estimate of is: z 2 2 E 2 ˆ n n E z z. E 2 , with x n p q Example: In case the values of and are not known: Take the most conservative estimate of the sample size n An alumni association wants to estimate the mean debt of this year’s by using p = .5 and q = .5. For a given E, these values of p and q will give the largest sample size in comparison to college graduates. It is known that the population standard any other pair of values of p = .5 and q = .5 since their deviation of the debts of this year ’s college graduates is $11 , 800. product is greater than the product of any other pair. How large a sample should be selected so that the estimate with a 99% confidence level is within $800 of the population mean? Take a preliminary sample of arbitrarily determined size and calculate p and q from this sample. Then use them z 2 2 (2.58) 2 (11,800) 2 n ˆ ˆ n 1448 .18 1449 to find . 38 E 2 (800) 2 39
Determining the sample size for estimating the Determining the sample size for estimating the proportion: Example ppproportion: Examp le (()continued)
Lombard Electronics Company has just installed a new machine that Consider the previous Example again. Suppose a preliminary makes a part that is used in clocks. The company wants to estimate the sample of 200 parts produced by this machine showed that proportion of these parts produced by this machine that are defective. The company manager wants this estimate to be within .02 of the 7% of them are defective. How large a sample should the population proportion for a 95% confidence level. What is the most comppyany select so that the 95% confidence interval for p is conservative estimate of the sample size that will limit the maximum within .02 of the population proportion? error to within .02 of the population proportion? Answer: Answer: The value of z for a 95% confidence level is 1.96; p = .5 and q = .5 2 2 p q , z pq (1.96) (.50)(.50) ˆ = .07 and = .93ˆ thus: n 2401 E 2 2 z 2 pq 2 (.02) n ˆ ˆ (1.96) (.07)(.93) E 2 2 Thus, if the company takes a sample of 2401 parts, there is 95% chance (.02) p (3.8416)(.07)(.93) that the estimate of will be within .02 of the population proportion. 625.22 626 40 41 .0004
INTERVAL ESTIMATION OF A POPULATION MEAN: An Interval Estimation cont. LARGE SAMPLES
Definition Confidence Interval for for Large Samples The (1 – )100% confidence interval for is Each interval is constructed with regggard to a given confidence level and is called a confidence interval. The x z x if is known confidence level associated with a confidence interval x zs states how much confidence we have that this interval x if is not known contains the true population parameter. The confidence where x / n and sx s / n level is denoted by (1 – )100%. The value of z used here is read from the standard normal distribution table for the given confidence level.
42 43 Repeat, again: Finding z for a 95% confidence level. Area in the tails – what this means?
Total shaded area is .9500 or 95%
2 2 (1 – ) .4750 .4750 x -z 0 z z -1.96 0 1.96 z 44 45
Interval Estimation for Population Proportion Example 3: Answer: x 152 El3Example 3: a. Point estimate of p: p .4393 A survey asked 346 job seekers. The answer selected most n 346 (152 times) was “higher compensation.” b. Confidence interval: • Z(1Z: (1- )/2= 0. 475, Ta ble 1 Z196Z=1.96 p(1 p) (0.4393)(1 0.4393) a. What is the point estimate of the proportion of job seekers • Margin of error = Z ( 1 . 96 ) = 0.0523 who would select “higher compensation ” as the reason of n 346 changing jobs? • Confidence interval UCL p m o e a. What is the 95% confidence interv al estimate of the . . . .4393 .0523 .4916 population proportion? LCL p m.o.e. .4393 .0523 .3870
p: [.3870 , . 4916]
Sample Size and Precision Quality of estimation: Interval estimation: • Confidence level: 1 - Sample size and precision • Precision: margin of error Confidence level: 1 - is guaranteed by procedure. sampling distribution for sample mean.
Probability that X falls between Z x and Z x / 2 n / 2 n is 1 - . In general, any sample mean that is within this range will provide an interval that contains the population mean . Margin of error: Given n, then (1 - ) margin of error . Given 1 - , then n margin of error . 48 Determine sample size to meet requirements for both Example 4: confidence level and margin of error: Bride’s magazine reported that the mean cost of a wedding is 1. Determine sample size for estimation of $19,000. Assume that the population standard deviation is $9,400. Use 95% confidence, F Z 2 2 V n G / 2 W a.What is t he reco mme nded sam pl e siz e if th e desir ed m ar gin HG E 2 XW of error is $1,000? b. What is the recommended sample size if the desired margin E: desired margin of error of error is $500? [ ]: round up (“Be conservative”) Answer: Z 1 Z 2. Determine samppple size for estimation of p a. / 2 : .475 / 2 1.96 2 F Z 2 p p V 2 2 / 2 (1 ) F (1.96 ) (9400 ) V n G W n G W 339 .44 340 H E 2 X H 1000 2 X b. p = .5 (“Be conservative”) F(1.96)2 (9400)2 V E: desir ed m ar gin of err oor n G W 1357.78 1358 H 500 2 X [ ]: round up (“Be conservative”)
Example 6: Example The League of American Theatres and Producers uses an According to the analysis of a CNN–USA TODAY–Gallup ongoing audience tracking survey that provides up-to-date poll conducted in October 2002, “Stress has become a in formati on ab ou t B road way thea ter audi ences. Every wee k, the common part of everyday life in the United States. The demands of work, family, and home place an increasing League distributes a one-page survey on random theater seats burden on the average American.” at a rotation roster of Broadway shows. a. How large a sample should be taken if the desired margin of According to this poll, 40% of Americans included in the error on any proportion is 0.04? Use 95% confidence. survey indicated that they had a limited amount of time to Answer: relax (Gallup. com, November 8, 2002). Z 1 Z The poll was based on a randomly selected national / 2 : .475 , Table 1 / 2 1 .96 sample of 1502 adults aged 18 and older. 2 F (1 .96 ) 2 .5(1 .5) V Construct a 95% confidence interval for the corresponding n G W 600 .25 601 popultilation propor tion. H (. 04 ) 2 X 53
Solution Example Confidence level = 95% or .95 According to a report by the Consumer Federation of The value of z for .95 / 2 = .4750 is 1.96. America,,, National Credit Union Foundation, and the Credit Union National Association, households with negative pˆqˆ (.40)(.60) assets carried s an average of $15,528 in debt in 2002 pˆ .01264069 n 1502 (CBS.MarketWatch.com, May 14, 2002).
Assume that this mean was based on a random sample pˆ zs .40 1.96(.01264069) of 400 households and that the standard deviation of pˆ debts for households in this sample was $4200. .40 .025 Make a 99% confidence interval for the 2002 mean debt .375 to .425 or 37.5% to 42.5% for all such households.
54 55 Solution Example Confidence level 99% or .99 Lombard Electronics Company has just installed a new s 4200 machine that makes a part that is used in clocks. The s $210 x n company wants to estimate the proportion of these parts 400 produced by this machine that are defective. The sample is large (n > 30) z Z Therefore, we use the normal distribution = 282.58 The company manager wants this estimate to be within .02 of the population proportion for a 95% confidence x zsx 15,528 2.58(210) 15,528 541.80 level. $14,986.20 to $16,069.80 What is the most conservative estimate of the sample size that will limit the maximum error to within . 02 of the Thus, we can state with 99% confidence that the 2002 population proportion? mean debt for all households with negative assets was between $14 ,986 . 20 and $16, 069. 80.
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Solution Example The value of z for a 95% confidence level is 1.96. Consider previous example again. p = .50 and q = .50 z 2 pq (1.96) 2 (.50)(.50) Suppose a preliminary sample of 200 parts produced by n 2401 this machine showed that 7% of them are defective. E 2 (.02) 2
Thus, if the company takes a sample of 2401 parts, there is How large a sample should the company select so that the 95% chance that the estimate of p will be within .02 of the 95% confidence interval for p is within .02 of the pppopulation population proportion. proportion?
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Solution
pˆ = .07 and q ˆ = .93
z 2 pq 2 n ˆ ˆ (1.96) (.07)(.93) E 2 (.02) 2 (3.8416)(.07)(.93) 625.22 626 .0004
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