1. Estimating the Value of a Parameter
Chapter 9
Estimating the Value of a Parameter
9.1
9.2
9.3

2. Estimating a Population Proportion
Section
9.1
Estimating a Population Proportion

3. Objectives Obtain a point estimate for the population proportion
Construct and interpret a confidence interval for the population proportion
Determine the sample size necessary for estimating the population proportion within a specified margin of error

4. Objective 1 Obtain a Point Estimate for the Population Proportion
A point estimate is the value of a statistic that estimates the value of a parameter.
For example, the point estimate for the population proportion is , where x is the number of individuals in the sample with a specified characteristic and n is the sample size.

5. Parallel Example 1: Calculating a Point Estimate for the
Parallel Example 1: Calculating a Point Estimate for the Population Proportion
In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder were in favor.
Obtain a point estimate for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

6. Objective 2
Construct and Interpret a Confidence Interval for the Population Proportion
A confidence interval for an unknown parameter consists of an interval of numbers based on a point estimate.
The level of confidence represents the expected proportion of intervals that will contain the parameter if a large number of different samples is obtained. The level of confidence is denoted (1 – α)·100%.

7. Interpretation of a Confidence Interval
For example, a 95% level of confidence (α = 0.05) implies that if 100 different confidence intervals are constructed, each based on a different sample from the same population, we will expect 95 of the intervals to contain the parameter and 5 not to include the parameter.

8. Point estimate ± margin of error.
Confidence interval estimates for the population proportion are of the form
Point estimate ± margin of error.
The margin of error, E, in a (1 – α) 100% confidence interval for a population proportion is given by

9. The margin of error depends on three factors:
Level of confidence: As the level of confidence increases, the margin of error also increases.
Sample size: As the size of the random sample increases, the margin of error decreases.
Standard deviation of the population: The more spread there is in the population, the wider our interval will be for a given level of confidence.

10. Constructing a (1 – α)·100% Confidence
Interval for a Population Proportion
Suppose that a simple random sample of size n is taken from a population. A (1 – α)·100% confidence interval for p is given by the following quantities
Lower bound:
Upper bound:
Note: It must be the case that and
n ≤ 0.05N to construct this interval.

11. Ex 2: Constructing a Confidence Interval for a Population Proportion
In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder were in favor.
Obtain a 90% confidence interval for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.
n = x = p hat =
1783
1123
1123/1783 = 0.63
sample size is definitely less than 5% of the population size
α = so zα/2 = z0.05 = 1.645 +
We are 90% confident that the proportion of registered voters who are in favor of the death penalty for those convicted of murder is between 0.61and 0.65.

12. Objective 3
Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error

13. Sample size needed for a specified margin of error, E, and level of confidence (1 – α):
Problem: The formula uses which depends on n, the quantity we are trying to determine!

14. Two possible solutions:
Use an estimate of based on a pilot study or an earlier study.
Let = 0.5 which gives the largest possible value of
n for a given level of confidence and a given margin of error.

15. Sample Size Needed for Estimating the Population Proportion p
The sample size required to obtain a (1 – α)·100% confidence interval for p with a margin of error E is given by
is a prior estimate of p.
If a prior estimate of p is unavailable
Always round up to the next integer

16. Parallel Example 4: Determining Sample Size
A sociologist wanted to determine the percentage of residents of America that only speak English at home. What size sample should be obtained if she wishes her estimate to be within 3 percentage points with 90% confidence assuming she uses the 2000 estimate obtained from the Census 2000 Supplementary Survey of 82.4%?
E = 0.03
We round this value up to The sociologist must survey 437 randomly selected American residents.

17. 9.1 Summary Obtain a point estimate for the population proportion
Construct and interpret a confidence interval for the population proportion
Determine the sample size necessary for estimating the population proportion within a specified margin of error

18. Estimating a Population Mean
Section
9.2
Estimating a Population Mean

19. Objectives Obtain a point estimate for the population mean
Determine t-values
Construct and interpret a confidence interval for a population mean
Find the sample size needed to estimate the population mean within a given margin or error

20. Objective 1
A point estimate is the value of a statistic that estimates the value of a parameter.
For example, the sample mean, , is a point estimate of the population mean μ.

21. The point estimate of μ is 2.464 grams.
Parallel Example 1: Computing a Point Estimate
Pennies minted after 1982 are made from 97.5% zinc and 2.5% copper. The following data represent the weights (in grams) of 17 randomly selected pennies minted after 1982.
Treat the data as a simple random sample. Estimate the population mean weight of pennies minted after 1982.
The point estimate of μ is grams.

22. Objective 2
Determine t-Values

24. Parallel Example 2: Finding t-values
Find the t-value such that the area under the t-distribution to the right of the t-value is 0.2 assuming 10 degrees of freedom. That is, find t0.20 with 10 degrees of freedom.
The unknown value of t is labeled, and the area under the curve to the right of t is shaded. The value of t0.20 with 10 degrees of freedom is

25. Objective 3
Construct and Interpret a Confidence Interval for the Population Mean

26. Constructing a (1 – α)100% Confidence Interval for μ
Provided
Sample data come from a simple random sample or randomized experiment
Sample size is small relative to the population size (n ≤ 0.05N)
The data come from a population that is normally distributed, or the sample size is large (n > 30)
Confidence Interval for the μ at a given value for α is found by the following:
Lower Upper
bound: bound:
where is the critical value with n – 1 df.

27. Parallel Example 2: Using Simulation to Demonstrate the
Parallel Example 2: Using Simulation to Demonstrate the Idea of a Confidence Interval
We will use Minitab to simulate obtaining 30 simple random samples of size n = 8 from a population that is normally distributed with μ = 50 and σ = 10. Construct a 95% confidence interval for each sample. How many of the samples result in intervals that contain μ = 50 ?

28. Sample Mean % CI
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29. SAMPLE MEAN 95% CI
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30. Note that 28 out of 30, or 93%, of the confidence intervals contain the population mean μ = 50.
In general, for a 95% confidence interval, any sample mean that lies within 1.96 standard errors of the population mean will result in a confidence interval that contains μ.
Whether a confidence interval contains μ depends solely on the sample mean, .

31. Parallel Example 3: Constructing a Confidence Interval
Construct a 99% confidence interval about the population mean weight (in grams) of pennies minted after 1982.
Find sample mean and sample standard deviation… And show the data comes from a normal population

32. Weight (in grams) of Pennies
Sample:
n = 17
mean = 2.464
stand dev = 0.02
Weight (in grams) of Pennies

33. We are 99% confident that the mean weight of pennies
Construct a 99% confidence interval
Sample:
n = 17
mean =
stand dev =
2.921
Lower bound:
= – 2.921
= – = 2.452
Upper bound: =
= = 2.476
We are 99% confident that the mean weight of pennies
minted after 1982 is between and grams.
0.0172
0.0172

34. ($3171.50, $3828.50) 1.971 ± Construct a 95% confidence interval
In a random sample of 225 audited estate tax returns, it was determined that the mean amount of additional tax owed was $3500 with a standard deviation of $ Construct a 95% confidence interval for the mean additional tax owed for estate tax returns.
Construct a 95% confidence interval
Sample:
n = 225
mean = 3500
stand dev = 2500
1.971
($ , $ )

35. Objective 4
Find the Sample Size Needed to Estimate the Population Mean within a Given Margin of Error

36. Determining the Sample Size n
The sample size required to estimate the population mean, µ, with a level of confidence (1– α)·100% with a specified margin of error, E, is given by
where n is rounded up to the nearest whole number.

37. Parallel Example 7: Determining the Sample Size
Back to the pennies. How large a sample would be required to estimate the mean weight of a penny manufactured after 1982 within grams with 99% confidence? Assume  = 0.02. 6
s = E = 0.005 6
106.17
Rounding up, we find n = 107.

38. Which to use, t or z? Find CI for p Find CI for μ Zα/2 Know σ Know s
tα/2, df=n–1

39. 9.2 Summary Obtain a point estimate for the population mean
Determine t-values
Construct and interpret a confidence interval for a population mean
Find the sample size needed to estimate the population mean within a given margin or error

40. Putting It Together: Which Procedure Do I Use?
Section
9.3
Putting It Together: Which Procedure Do I Use?

41. Which to use, t or z? Find CI for p Find CI for μ Zα/2 Know σ Know s
tα/2, df=n–1
Zα/2

42. From 9.2 not used

43. From 9.2 not used Objective 2
State Properties of Student’s t-Distribution
From 9.2 not used

44. Student’s t-Distribution
Suppose that a simple random sample of size n is taken from a population. If the population from which the sample is drawn follows a normal distribution, the distribution of
follows Student’s t-distribution with n – 1 degrees of freedom where is the sample mean and s is the sample standard deviation.
From 9.2 not used

45. Parallel Example 1: Comparing the Standard Normal
Parallel Example 1: Comparing the Standard Normal Distribution to the t-Distribution Using Simulation
Obtain 1,000 simple random samples of size n = 5 from a normal population with μ = 50 and σ = 10.
Determine the sample mean and sample standard deviation for each of the samples.
Compute and for each sample.
Draw a histogram for both z and t.
From 9.2 not used

46. Histogram for z
Histogram for t
From 9.2 not used

47. From 9.2 not used CONCLUSION:
The histogram for z is symmetric and bell-shaped with the center of the distribution at 0 and virtually all the rectangles between –3 and 3. In other words, z follows a standard normal distribution.
From 9.2 not used

48. CONCLUSION (continued):
The histogram for t is also symmetric and bell-shaped with the center of the distribution at 0, but the distribution of t has longer tails (i.e., t is more dispersed), so it is unlikely that t follows a standard normal distribution. The additional spread in the distribution of t can be attributed to the fact that we use s to find t instead of σ. Because the sample standard deviation is itself a random variable (rather than a constant such as σ), we have more dispersion in the distribution of t.
From 9.2 not used

49. Properties of the t-Distribution
1. The t-distribution is different for different degrees of freedom.
2. The t-distribution is centered at 0 and is symmetric about 0.
3. The area under the curve is 1. The area under the curve to the right of 0 equals the area under the curve to the left of 0, which equals 1/2.
4. As t increases or decreases without bound, the graph approaches, but never equals, zero.
From 9.2 not used

50. Properties of the t-Distribution
5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution, because we are using s as an estimate of σ, thereby introducing further variability into the t- statistic.
6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because, as the sample size n increases, the values of s get closer to the values of σ, by the Law of Large Numbers.
From 9.2 not used

51. From 9.2 not used