Chapter 9: Introduction to Estimation - U of L Class Index
Chapter 10: Introduction to Estimation 10.1
Introduction
This chapter introduced estimation, a form of statistical inference. Make certain that you understand the sampling distribution of the mean before proceeding. At the completion of this chapter, you are expected to know the following: 1. 2. 3. 4.
10.2
Understand the fundamental concepts of estimation. How to produce interval estimates of the population mean. How to interpret interval estimates. How to determine the sample size to estimate a population mean.
Concepts of Estimation
This section dealt with the fundamental concepts of estimation. You are expected to know the meaning of the terms introduced in this chapter and understand why estimation is necessary in most practical situations.
EXERCISES 10.1
What is the difference between an estimator and an estimate?
10.2
Define each of the following terms: a) Unbiased estimator
b) Consistent estimator
10.3
Explain why x is an unbiased estimator of µ.
10.4
Explain why s2 is an unbiased estimator of σ2.
10.5
Explain why x is a consistent estimator of µ.
115
10.3
Estimating the Population Mean Value When the Population Standard Deviation Is Known
We presented the technique to produce a interval estimate of a population mean when the population variance is known. The interval estimator of µ is x ± zα / 2 σ / n
It is assumed that the population variance is known. The confidence level is equal to 1 – α, and from this value you need to be able to determine zα / 2 . Other than plugging the numbers into the formula, no other arithmetic skills are required. Question:
How can I tell when the population variance is known?
Answer:
The question must state explicitly that the population variance σ2 is known, and it must also specify the value.
Question:
How do I find zα / 2 ?
Answer:
The question will specify the confidence level 1 – α. From the confidence level, determine a and divided it by 2. For example, if the confidence level is .84, then since 1 – α = .84 it follows that α = .16 and α/2 = .08. To find zα / 2 = z.08 , we need to find the value of z such that the area to its right is .08.
Since Table 3 is designed to provide the areas between 0 and z, we need to subtract .08 from .5 to produce .42. (Notice that we can get .42 more directly by dividing the confidence level by 2.) Now in Table 3 find the area closest to .42 (.4200). As you should be able to see, the area between 0 and 1.40 is .4192 and the area between 0 and 1.41 is .4207. Since the latter area is closer to .4200, we find z.08 = 1.41.
116
Example 10.1 A statistician would like to estimate the mean number of cars making legal right turns on the red light at a particular busy intersection. She observes 25 red light sequences and counts the number of legal right turns. She finds the mean to be 7.3. If it is known that the population standard deviation is 1.8, find the 98% confidence interval estimate of the mean number of legal right turns made per sequence.
Solution The interval estimator of µ is x ± zα / 2 σ / n
The confidence level is specified as 98%. Thus, 1 – α = .98 which means that α = .02 and α/2 = .01 We need to find zα / 2 = z.01 . From Table 3 in Appendix B, we find z.01 = 2.33. The population standard deviation is σ = 1.8. Thus, the 98% confidence interval estimate of µ is x ± zα / 2 σ / n = 7.3 ± 2.33 (1.8) / 25 = 7.3 ± .84
The lower confidence limit (LCL) is 6.46, and the upper confidence level (UCL) is 8.14.
Question:
Does this result mean that there is a 98% probability that the population mean falls between 6.46 and 8.14?
Answer:
No. The interval estimate in this case is LCL = 6.46 and UCL = 8.14. If we applied this technique repeatedly, 98% of the samples would produce interval estimates that include the actual population mean and 2% would exclude it. For the sample of 25 in this application, we do not know if this estimate includes µ or excludes it.
117
EXERCISES 10.6
In a random sample of 50 observations drawn from a normal population whose standard deviation is 250, the sample mean was found to be 1,275. a) Find the 95% confidence interval estimate of µ.
b) Find the 92.5% confidence interval estimate of µ.
10.7 Given the following observations from a normal population whose variance is 25, find the 99% confidence interval estimate of the population mean. 8, 12, 15, 3, 8, 10, 12, 18
10.8 In a time and motion study, the amount of time required to paint a 10-foot by 10-foot room is measured for 15 painters. The sample mean is calculated and found to be x = 73. If we assume that the population standard deviation is 8 minutes, find the 95% confidence interval estimate of the mean time for all painters to paint a 10-foot by 10-foot room.
10.4
Selecting the Sample Size
In this section, we discussed how to determine the sample size necessary to produce a interval estimate of the population mean. In order to calculate n, we need to know the confidence level and the error bound. Once these are known, the calculation is quite simple. We use the formula ⎡z σ ⎤ n=⎢ α/2 ⎥ ⎣ W ⎦
2
118
Question: Answer:
Where do I get σ from? The population standard deviation may be known from previous experiments.
Question: Answer:
Where do I get W from? The exercise will state that we wish to estimate µ with a certain degree of confidence to within a specified amount. The specified amount is the value of W.
Example 10.2 A research scientist would like to estimate the tensile strength of a newly created synthetic fiber. She would like to estimate the mean strength to within 2 pounds with 90% confidence. In a preliminary study, she estimates the standard deviation to be 35. What sample size should the research scientist use?
Solution The confidence level is 90%. Thus, 1 – α = .90 and zα/2 = z.05 = 1.645 Because we wish to estimate µ to within 2, we have W=2 Finally, we're told that the standard deviation is estimated to be 35. Thus, 2
⎡z σ ⎤ ⎡1.645( 35 ) ⎤ n=⎢ α/2 ⎥ =⎢ ⎥ = 829 2 ⎦ ⎣ ⎣ W ⎦ 2
The research scientist should draw a random sample of 829.
EXERCISES 10.9 Determine the sample size necessary to estimate µ to within 10 with 99% confidence and σ = 200.
119
10.10 Repeat Exercise 10.9 with σ = 2,000.
10.11 A dietician would like to estimate the mean weight loss of a new type of diet. He would like to estimate the parameter to within 1/2 pound with 95% confidence. From past experience, he believes that the population standard deviation is about 10 pounds. How large a sample of dieters should he draw?
10.5
Simulation Experiments
Like the simulation experiments described in Chapter 9, these experiments allow you to discover important statistical concepts. Each experiment should be run following the textbook's instructions. Write a brief report describing your findings.
120
Introduction
This chapter introduced estimation, a form of statistical inference. Make certain that you understand the sampling distribution of the mean before proceeding. At the completion of this chapter, you are expected to know the following: 1. 2. 3. 4.
10.2
Understand the fundamental concepts of estimation. How to produce interval estimates of the population mean. How to interpret interval estimates. How to determine the sample size to estimate a population mean.
Concepts of Estimation
This section dealt with the fundamental concepts of estimation. You are expected to know the meaning of the terms introduced in this chapter and understand why estimation is necessary in most practical situations.
EXERCISES 10.1
What is the difference between an estimator and an estimate?
10.2
Define each of the following terms: a) Unbiased estimator
b) Consistent estimator
10.3
Explain why x is an unbiased estimator of µ.
10.4
Explain why s2 is an unbiased estimator of σ2.
10.5
Explain why x is a consistent estimator of µ.
115
10.3
Estimating the Population Mean Value When the Population Standard Deviation Is Known
We presented the technique to produce a interval estimate of a population mean when the population variance is known. The interval estimator of µ is x ± zα / 2 σ / n
It is assumed that the population variance is known. The confidence level is equal to 1 – α, and from this value you need to be able to determine zα / 2 . Other than plugging the numbers into the formula, no other arithmetic skills are required. Question:
How can I tell when the population variance is known?
Answer:
The question must state explicitly that the population variance σ2 is known, and it must also specify the value.
Question:
How do I find zα / 2 ?
Answer:
The question will specify the confidence level 1 – α. From the confidence level, determine a and divided it by 2. For example, if the confidence level is .84, then since 1 – α = .84 it follows that α = .16 and α/2 = .08. To find zα / 2 = z.08 , we need to find the value of z such that the area to its right is .08.
Since Table 3 is designed to provide the areas between 0 and z, we need to subtract .08 from .5 to produce .42. (Notice that we can get .42 more directly by dividing the confidence level by 2.) Now in Table 3 find the area closest to .42 (.4200). As you should be able to see, the area between 0 and 1.40 is .4192 and the area between 0 and 1.41 is .4207. Since the latter area is closer to .4200, we find z.08 = 1.41.
116
Example 10.1 A statistician would like to estimate the mean number of cars making legal right turns on the red light at a particular busy intersection. She observes 25 red light sequences and counts the number of legal right turns. She finds the mean to be 7.3. If it is known that the population standard deviation is 1.8, find the 98% confidence interval estimate of the mean number of legal right turns made per sequence.
Solution The interval estimator of µ is x ± zα / 2 σ / n
The confidence level is specified as 98%. Thus, 1 – α = .98 which means that α = .02 and α/2 = .01 We need to find zα / 2 = z.01 . From Table 3 in Appendix B, we find z.01 = 2.33. The population standard deviation is σ = 1.8. Thus, the 98% confidence interval estimate of µ is x ± zα / 2 σ / n = 7.3 ± 2.33 (1.8) / 25 = 7.3 ± .84
The lower confidence limit (LCL) is 6.46, and the upper confidence level (UCL) is 8.14.
Question:
Does this result mean that there is a 98% probability that the population mean falls between 6.46 and 8.14?
Answer:
No. The interval estimate in this case is LCL = 6.46 and UCL = 8.14. If we applied this technique repeatedly, 98% of the samples would produce interval estimates that include the actual population mean and 2% would exclude it. For the sample of 25 in this application, we do not know if this estimate includes µ or excludes it.
117
EXERCISES 10.6
In a random sample of 50 observations drawn from a normal population whose standard deviation is 250, the sample mean was found to be 1,275. a) Find the 95% confidence interval estimate of µ.
b) Find the 92.5% confidence interval estimate of µ.
10.7 Given the following observations from a normal population whose variance is 25, find the 99% confidence interval estimate of the population mean. 8, 12, 15, 3, 8, 10, 12, 18
10.8 In a time and motion study, the amount of time required to paint a 10-foot by 10-foot room is measured for 15 painters. The sample mean is calculated and found to be x = 73. If we assume that the population standard deviation is 8 minutes, find the 95% confidence interval estimate of the mean time for all painters to paint a 10-foot by 10-foot room.
10.4
Selecting the Sample Size
In this section, we discussed how to determine the sample size necessary to produce a interval estimate of the population mean. In order to calculate n, we need to know the confidence level and the error bound. Once these are known, the calculation is quite simple. We use the formula ⎡z σ ⎤ n=⎢ α/2 ⎥ ⎣ W ⎦
2
118
Question: Answer:
Where do I get σ from? The population standard deviation may be known from previous experiments.
Question: Answer:
Where do I get W from? The exercise will state that we wish to estimate µ with a certain degree of confidence to within a specified amount. The specified amount is the value of W.
Example 10.2 A research scientist would like to estimate the tensile strength of a newly created synthetic fiber. She would like to estimate the mean strength to within 2 pounds with 90% confidence. In a preliminary study, she estimates the standard deviation to be 35. What sample size should the research scientist use?
Solution The confidence level is 90%. Thus, 1 – α = .90 and zα/2 = z.05 = 1.645 Because we wish to estimate µ to within 2, we have W=2 Finally, we're told that the standard deviation is estimated to be 35. Thus, 2
⎡z σ ⎤ ⎡1.645( 35 ) ⎤ n=⎢ α/2 ⎥ =⎢ ⎥ = 829 2 ⎦ ⎣ ⎣ W ⎦ 2
The research scientist should draw a random sample of 829.
EXERCISES 10.9 Determine the sample size necessary to estimate µ to within 10 with 99% confidence and σ = 200.
119
10.10 Repeat Exercise 10.9 with σ = 2,000.
10.11 A dietician would like to estimate the mean weight loss of a new type of diet. He would like to estimate the parameter to within 1/2 pound with 95% confidence. From past experience, he believes that the population standard deviation is about 10 pounds. How large a sample of dieters should he draw?
10.5
Simulation Experiments
Like the simulation experiments described in Chapter 9, these experiments allow you to discover important statistical concepts. Each experiment should be run following the textbook's instructions. Write a brief report describing your findings.
120
