1. Chapter 8 Large-Sample Estimation
General Objectives:
This chapter presents a method for estimating population parameters and illustrates the concept with practical examples. The Central Limit Theorem and the sampling distributions presented in Chapter 7 play a key role in evaluating the reliability of the estimates.
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2. Specific Topics 1. Types of estimators
2. Picking the best point estimator
3. Point estimation for a population mean or proportion
4. Interval estimation
5. Large-sample confidence intervals for a population mean or proportion
6. Estimating the difference between two population means
7. Estimating the difference between two binomial proportions
8. One-sided confidence bounds
9. Choosing the sample size
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3. 8.1 Where We’ve Been Descriptive Statistics
Probability and Probability Distributions
Binomial and Normal Distributions
Link Between Probability and Statistical Inference
The Central Limit Theorem
Populations
Statistics
Inferential Statistics
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4. 8.2 Where We’re Going— Statistical Inference
Applications of inference:
- The government needs to predict short- and long-term interest rates.
- A broker wants to forecast the behavior of the stock market
- A metallurgist wants to decide whether or new type of steel is more resistant to high temperatures than the old type was.
- A consumer wants to estimate the selling price of her house before putting it on the market.
Statistical inference is concerned with making decisions or predictions about parameters.
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5. - Estimation: Estimating or predicting the value of the parameter
Methods for making inferences about population parameters fall into one of two categories:
- Estimation: Estimating or predicting the value of the parameter
- Hypothesis testing: Making a decision about the value of a parameter based on some preconceived idea about what its value might be
Examples 8.1 and 8.2 give illustrations of estimation and hypothesis testing problems respectively.
Statistical procedures are important because they provide two types of information:
- Methods for making the inference
- A numerical measure of the goodness or reliability of the inference.
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6. Example 8.1
The circuits in computers and other electronics equipment consist of one or more printed circuit boards (PCB), and computers are often repaired by simply replacing one or more defective PCBs. In an attempt to find the proper setting of a plating process applied to one side of a PCB, a production supervisor might estimate the average thickness of copper plating on PCBs using samples from several days of operation. Since he has no knowledge of the average thickness m before observing the production process, his is an estimation problem.
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7. 8.3 Types of Estimators
Definition: An estimator is a rule, usually expressed as a formula, that tells us how to calculate an estimate based on information in the sample.
Estimators are used in two different ways:
- Point estimation: Based on sample data, a single number is calculated to estimate the population parameter.
The rule or formula that describes this calculation is called the point estimator, and the resulting number is called a point estimate.
- Interval estimation: Based on sample data, two numbers are calculated to form an interval within which the parameter is expected to lie.
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8. Example 8.3 discusses a possible estimator based on sample data.
The rule or formula that describes this calculation is called the interval estimator, and the resulting pair of numbers is called an interval estimate or confidence interval.
Example 8.3 discusses a possible estimator based on sample data.
Example 8.3
A veterinarian wants to estimate the average weight gain per month of 4-month-old golden retriever pups that have been placed on a lamb and rice diet. The population consists of the weight gains per month of all 4-month-old golden retriever pups that are given this particular diet. Hence, it is a hypothetical population where m is the average monthly weight gain for all 4-month-old golden retriever pups on this diet. This is the unknown parameter that the veterinarian wants to estimate. One possible estimator based on sample data is the sample mean,
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9. It could be used in the form of a single number or point estimate—for instance, 3.8 pounds—or you could use an interval estimate and estimate that the weight gain will fall into an interval such as 2.7 to 4.9 pounds.
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10. 8.4 Point Estimation
Point estimation is used in estimating population means and proportions.
To choose the best statistic or estimator, observe their behavior in repeated sampling, described by their sampling distributions.
Two characteristics are valuable in a point estimator:
1. The sampling distribution of the point estimator should be centered over the true value of the parameter to be estimated.
2. The spread (as measured by the variance) of the sampling distribution should be as small as possible.
Figure 8.1 illustrates differences in centering and spread for sampling distributions.
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11. Figure 8.1
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12. Definition: An estimator is said to be unbiased if the mean of its distribution is equal to the true value of the parameter. Otherwise, the estimator is said to be biased.
Figure 8.2 shows distributions for biased and unbiased estimators. Figure 8.3 compares estimator variability.
Definition: The distance between an estimate and the estimated parameter is called the error of estimation.
Figure 8.2
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13. Point Estimation of a Population Parameter:
- Point estimator: A statistic calculated using sample measurements
- Margin of error: 1.96 ´ Standard error of the estimator.
Figure 8.4 shows the sampling distribution of an unbiased estimator.
Figure 8.4
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14. Estimating a population mean or proportion:
To estimate the population mean m for a quantitative population, the point estimator is unbiased with standard error given as
- The margin of error is calculated as
- If s is unknown and n is 30 or larger, the sample SD s can be used to approximate s .
To estimate the population proportion p for a binomial population, the point estimator is unbiased with standard error given as
- The margin or error is calculated as
and estimated as
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15. Assumptions: np > 5 and nq > 5. Since p and q are unknown, use
Table 8.1 displays some calculated values of
Examples 8.4 and 8.5 estimate the population mean and the population proportion, respectively.
In calculating the standard error for these two point estimates, you need to estimate s with s, p with and q with
Table 8.1
p pq p pq
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16. Example 8.4
An investigator is interested in the possibility of merging the capabilities of television and the Internet. A random sample of n = 50 Internet users who were polled about the time they spend watching television produced an average of 11.5 hours per week with a standard deviation of 3.5 hours. Use this informa-tion to estimate the population mean time Internet users spend watching television.
Solution
The random variable measured is the time spent watching television per week. This is a quantitative random variable best described by its mean m. The point estimate of m , the average time Internet users spend watching television, is hours. The margin of error is
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17. Although s is unknown, the sample size is large, and you can approximate the value of s by using s. Therefore, the margin of error is approximately
You can feel fairly confident that the sample estimate of hours of television watching for Internet users is within ±1 hour of the population mean.
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18. Figure 8.5 Plot of treatment means and their standard errors
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19. 8.5 Interval Estimation
An interval estimator is a rule for calculating two numbers—say, a and b—that create an interval that you are fairly certain contains the parameters of interest, e.g., a confidence interval.
The concept of “fairly certain” can be quantified using a statistical concept called the confidence coefficient and designated by (1- a).
Definition: The probability that a confidence interval will contain the estimated parameter is called the confidence coefficient.
Constructing a Confidence Interval:
- For the standard normal random variable z, 95% of all values lie between and 1.96.
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20. Figure 8.6 shows the 95% confidence limits for a population parameter.
Figure 8.7 displays the locations of za / 2. Table 8.2 displays values of z commonly used.
Figure % confidence limits for a population parameter
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21. Figure 8.7 Location of za / 2
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22. A (1- a )100% Large-Sample Confidence Interval:
- For an unbiased point estimator with a normal sampling distribution, 95% of all point estimates lie with 1.96 standard errors of the parameter of interest.
- Consider constructing the interval as: point estimate ± 1.96SE. As long as the point estimate is within 1.96SE of the parameter of interest, the interval centered at this estimate will contain the parameter of interest. This will happen 95% of the time.
You may want to change the confidence coefficient from (1- a ) = .95 to another confidence level (1- a ). To do this, you will need to adjust the z-value.
A (1- a )100% Large-Sample Confidence Interval:
where za / 2 is the z-value with an area of a / 2 in the right tail of a standard normal distribution. This formula generates two values; the lower confidence limit (LCL) and the upper confidence limit (UCL).
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23. s = Standard deviation of the sampled population.
A (1- a )100% Large-Sample Confidence Interval for a Population Mean m :
where za / 2 is the value corresponding to an area a / 2 in the upper tail of a standard normal z distribution, and n = Sample size
s = Standard deviation of the sampled population.
If s is unknown, it can be approximated by the sample standard deviation s when the sample size is large (n > 30) and the approximate confidence interval is
Example 8.6 constructs a 95% confidence interval for the mean daily intake of dairy products for men.
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24. Example 8.6
A scientist interested in monitoring chemical contaminates in food, and thereby the accumulation of contaminates in human diets, selected a random sample of n = 50 male adults. It was found that the average daily intake of dairy products was grams per day with a standard deviation of s = 35 grams per day. Use this sample information to construct a 95% confidence interval for the mean daily intake of dairy products for men.
Solution
Since the sample size of n = 50 is large, the distribution of the sample mean is approximately normally distributed with mean m and standard error Since the population standard
deviation is unknown, you can use the sample standard deviation s as its best estimate.
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25. The approximate 95% confidence interval is
Hence, the 95% confidence interval for m is from to grams per day.
Interpreting the Confidence Interval:
95% of the confidence intervals constructed using different
sample information would contain within their upper and lower bounds.
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26. Figure 8.8 shows 20 confidence intervals for the mean for Example 8.6.
Example 8.7 constructs a 99% confidence interval for the mean daily intake of dairy products for adult men in Example 8.6.
Figure 8.10 shows the sampling distributions for based on random samples for a normal distribution with n = 5, 20, and 80.
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27. Figure 8.8 Twenty confidence intervals for the mean for Example 8.6
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28. a / 2 = .005 to its right, is found from Table 8.2 to be z = 2.58.
Example 8.7
Construct a 99% confidence interval for the mean daily intake of dairy products for adult men in Example 8.6.
Solution
To change the confidence level to .99, you must find the appropriate value of the standard normal z that puts area (1 - a ) = .99 in the center of the curve. This value, with tail area
a / 2 = .005 to its right, is found from Table 8.2 to be z = 2.58.
The 99% confidence interval is then
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29. Large-Sample Confidence Interval for a Population Proportion p
- When the sample size is large, the sample proportion,
- The sampling distribution is approximately normal, with mean p and standard error
A (1- a )100% Large-Sample Confidence Interval for a Population Proportion p
where za / 2 is the z-value corresponding to an area a / 2 in the right tail of a standard normal z distribution. Estimate p and q with Sample size is considered large when np > 5 and nq > 5.
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30. 8.6 Estimating the Difference Between Two Population Means
Table 8.3 shows how the estimates of the population parameters are calculated from the sample data.
Properties of the Sampling Distribution of , the Difference Between Two Sample Means:
When independent random sample of n1 and n 2 have been selected from populations with means m 1 and m 2 and variances, respectively, the sampling distribution of the differences has the following properties:
1. The mean and the standard error of are
and
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31. Since m 1 - m 2 is the mean of the sampling distribution,
2. If the sampled populations are normally distributed, then the sampling distribution of is exactly normally distributed, regardless of the sample size.
3. If the sampled populations are not normally distributed, then the sampling distribution of is approximately normally distributed when n1 and n 2 are large, due to the CLT.
Since m 1 - m 2 is the mean of the sampling distribution,
is an unbiased estimator of (m 1 - m 2 ) with an approximately normal distribution.
The statistic
has an approximately standard normal z distribution.
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32. Point Estimation of (m 1 - m 2 ) : Point estimator: Margin of error:
If are unknown, but both n1 and n 2 are 30 or more, you can use the sample variances to estimate
A (1-a )100% Confidence Interval for (m 1 - m 2 ) :
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33. Example 8.9 illustrates the calculation of confidence intervals.
If are unknown, they can be approximated by the sample variances and the approximate confidence interval is
Example 8.9 illustrates the calculation of confidence intervals.
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34. The point estimate of ( m 1 - m 2 ) is and the standard error of is
Example 8.9
The wearing qualities of two types of automobile tires were compared by road-testing samples of n1 = n 2 = 100 tires for each type. The number of miles until wearout was defined as a specific amount of tire wear. The test results are given in Table 8.4. Estimate ( m 1 - m 2 ), the difference in mean miles to wearout, using a 99% confidence interval. Is there a difference in the average wearing quality for the two types of tires?
Solution
The point estimate of ( m 1 - m 2 ) is
and the standard error of is
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35. where are used to estimate respectively.
The 99% confidence interval is calculated as
or < ( m 1 - m 2) < The difference in the average miles to wearout for the two types of tires is estimated to be between LCL = and UCL = miles of wear.
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36. Based on this confidence interval, can you conclude that there is a difference in the average miles to wearout for the two types of tires? If there were no difference in the two population means, then m 1 and m 2 would be equal and ( m 1 - m 2 ) = 0. If you look at the confidence interval you constructed, you will see that 0 is not one of the possible values for ( m 1 - m 2 ). Therefore, it is not likely that the means are the same; you can conclude that there is a difference in the average miles to wearout for the two types of tires. The confidence interval has allowed you to make a decision about the equality of the two population means.
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37. The sampling distribution of
has a standard normal distribution for all sample sizes when both sampled populations are normal and an approximate standard normal distribution when the sampled populations are not normal but the sample sizes are large (greater than or equal to 30).
When are not known and are estimated by the sample estimated by the sample estimates the resulting statistic will also have an approximate standard normal distribution when the sample sizes are large.
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38. 8.7 Estimating the Difference Between Two Binomial Proportions
Properties of the Sampling Distribution of the Difference Between Two Sample Proportions:
Assume that independent random samples of n1 and n 2 observations have been selected from binomial populations with parameters p 1 and p 2 , respectively. The sampling distribution of the difference between sample proportions
has these properties:
1. The mean and the standard error of are
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39. To use a normal distribution to approximate the distribution of
2. The sampling distribution of can be approximated by a normal distribution when n1 and n 2 are large, due to the CLT.
Although the range of a single proportion is from 0 to 1, the difference between two proportions ranges from -1 to +1.
To use a normal distribution to approximate the distribution of
, both p1 and p 2 should be approximately normal; that is
Point estimation of
Point estimator:
Margin of error :
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40. A (1-a ) 100% Large-Sample Confidence Interval for (p 1 - p 2):
Note: The estimates must be substituted for p 1, p 2, q 1, and q 2 to estimate the margin of error.
A (1-a ) 100% Large-Sample Confidence Interval for (p 1 - p 2):
Assumption: n 1 and n 2 must be sufficiently large so that the sampling distribution of can be approximated by a normal distribution—namely, if n 1p 1, p 1q 1, n 2 p 2, and n 2q 2 are all greater than 5.
Example 8.11 displays the calculation of a confidence interval and the margin of error.
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41. 8.8 One-Sided Confidence Bounds
The confidence intervals discussed in Section s 8.5–8.7are sometimes called two-sided confidence intervals because they produce both a UCL and an LCL for the parameter of interest.
Sometimes an experimenter needs only an upper or lower limit.
You can construct a one-sided confidence bound for the parameter of interest, such as m, p, m 1 - m 2., or p 1 - mp 2 .
A (1 - a)100% Lower Confidence Bound (LCB):
(Point estimator) - za ´ (Standard error of the estimator)
A (1 - a)100% Upper Confidence Bound (UCB):
(Point estimator) + za ´ (Standard error of the estimator)
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42. Figure 8.11 illustrates a z-value for a one-sided confidence bound.
The z-value used for a (1 - a)100% one-sided confidence bound, za , locates an area in a single tail of the normal distribution.
Figure 8.11 illustrates a z-value for a one-sided confidence bound.
Example 8.12 finds the 95% upper confidence bound for the mean interest rate that a corporation will have to pay for notes.
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43. 8.9 Choosing the Sample Size
The total amount of relevant information in a sample is controlled by two factors:
- The sampling plan or experimental design: the procedure for collecting the information
- The sample size n: the amount of information you collect
In a statistical estimation problem, the accuracy of the estimation is measured by the margin of error or the width of the confidence interval.
Approximately 95% of the time in repeated sampling, the distance between the sample mean and the population mean m will be less than 1.96SE.
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44. If, for example, you want this quantity to be less than 4, then
Solving for n, you obtain
If you know s , the population standard deviation, you can substitute its value into the formula and solve for n.
If s is unknown — which is often the case — you can use the best approximation available:
- An estimate s obtained from a previous sample
- A range estimate based on knowledge of the largest and smallest possible measurements: s » Range ¤ 4
The bound B on the error of your estimate is
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45. Choosing the Sample Size:
Determine the parameter to be estimated and the standard error of its point estimator. Then proceed as follows:
1. Choose B, the bound on the error of your estimate, and a confidence coefficient (1- a).
2. For a one-sample problem, solve this equation for the sample size n: za ´ (Standard error of the estimator) = B
where za / 2 is the value of z having area a / 2 to its right.
3. For a two-sample problem, set n 1 = n 2 = n and solve the equation in step 2.
Note: For most estimators (all presented in this text), the standard error is a function of the sample size n.
Example 8.13 determines the sample size required for a given bound on error B and probability of error.
Example 8.14 determines the number of workers that must be in each training group for a given B and p.
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46. Example 8.13
Producers of polyvinyl plastic pipe want to have a supply of pipes sufficient to meet marketing needs. They wish to survey wholesalers who buy polyvinyl pipe in order to estimate the proportion who plan to increase their purchases next year. What sample size is required if they want their estimate to be within .04 of the actual proportion with probability equal to .90?
Solution
For this particular example, the bound B on the error of the estimate is .04. Since the confidence coefficient is (1- a ) = .90, a must equal .10 and a / 2 is .05. The z-value corresponding to an area equal to .05 in the upper tail of the z distribution is z.05 = You then require
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47. In order to solve this equation for n, you must substitute an approximate value of p into the equation. You could use the estimate based on the sample of n = 100; or if you want to be certain that the sample is large enough, you could use p = .5 (substituting p = .5 will yield the largest possible solution for n because the maximum value of pq occurs when p = q = .5). Substitute p =.5. Then or
Therefore the producers must include approximately 423 wholesalers in its survey if it wants to estimate the proportion p correct to within .04.
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48. Key Concepts and Formulas
I. Types of Estimators
1. Point estimator: a single number is calculated to estimate the population parameter.
2. Interval estimator: two numbers are calculated to form an interval that contains the parameter.
II. Properties of Good Estimators
1. Unbiased: the average value of the estimator equals the parameter to be estimated.
2. Minimum variance: of all the unbiased estimators, the best estimator has a sampling distribution with the smallest standard error.
3. The margin of error measures the maximum distance between the estimator and the true value of the parameter.
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49. III. Large-Sample Point Estimators
To estimate one of four population parameters when the sample sizes are large, use the following point estimators with the appropriate margins of error.
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50. IV. Large-Sample Interval Estimators
To estimate one of four population parameters when the sample sizes are large, use the following interval estimators.
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51. V. One-Sided Confidence Bounds
1. All values in the interval are possible values for the unknown population parameter.
2. Any values outside the interval are unlikely to be the value of the unknown parameter.
3. To compare two population means or proportions, look for the value 0 in the confidence interval. If 0 is in the interval, it is possible that the two population means or proportions are equal, and you should not declare a difference. If 0 is not in the interval, it is unlikely that the two means or proportions are equal, and you can confidently declare a difference.
V. One-Sided Confidence Bounds
Use either the upper (+) or lower (-) two-sided bound, with the critical value of z changed from za / 2 to za.
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52. VI. Choosing the Sample Size
1. Determine the size of the margin of error, B, that you are willing to tolerate.
2. Choose the sample size by solving for n or n = n 1 = n 2 in the inequality: 1.96 SE £ B, where SE is a function of the sample size n.
3. For quantitative populations, estimate the population standard deviation using a previously calculated value of s or the range approximation s » Range / 4.
4. For binomial populations, use the conservative approach and approximate p using the value p = .5.
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