1. 8 Chapter Estimation © 2012 Pearson Education, Inc.
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2. Chapter Outline 8.1 Estimating µ when σ is Known
8.2 Estimating µ when σ is Unknown
8.3 Confidence Intervals for Population Proportions
8.4 Estimating 𝜇 1 - 𝜇 2 and 𝑝 1 - 𝑝 2
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3. Estimating µ when σ is Known
Section 8.1
Estimating µ when σ is Known
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4. Section 8.1 Objectives Find a point estimate and a margin of error
Construct and interpret confidence intervals for the population mean
Determine the minimum sample size required when estimating μ
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5. Point Estimate for Population μ
A single value estimate for a population parameter
Most unbiased point estimate of the population mean μ is the sample mean
Estimate Population Parameter…
with Sample Statistic
Mean: μ
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6. Exercise 1: Point Estimate for Population μ
A social networking website allows its users to add friends, send messages, and update their personal profiles. The following represents a random sample of the number of friends for 40 users of the website. Find a point estimate of the population mean, µ. (Source: Facebook)
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7. Solution: Point Estimate for Population μ
The sample mean of the data is
Your point estimate for the mean number of friends for all users of the website is friends.
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8. Interval Estimate Interval estimate
An interval, or range of values, used to estimate a population parameter.
( )
Interval estimate
Right endpoint
Left endpoint
115
120
125
130
135
140
150
145
Point estimate
Point estimate
115
120
125
130
135
140
145
150
How confident do we want to be that the interval estimate contains the population mean μ?
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9. Confidence Levels
A confidence level, c, is any value between 0 and 1 that corresponds to the area under the standard normal curve between −𝑧 𝑐 and +𝑧 𝑐 .

10. Level of Confidence Level of confidence c
The probability that the interval estimate contains the population parameter.
c is the area under the standard normal curve between the critical values. c z
z = 0
–zc zc
½(1 – c)
½(1 – c)
The values of –zc and zc define the acceptable margin of error. The blue shaded area represents the probability that the absolute value of the sampling error does not exceed the margin of error.
Use the Standard Normal Table to find the corresponding z-scores.
Critical values
The remaining area in the tails is 1 – c .
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11. Common Confidence Levels
Appendix II, Table 5: Page A23

12. Level of Confidence
If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ. z
z = 0 zc
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
–zc = –1.645
zc = 1.645
The corresponding z-scores are ±1.645.
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13. Sampling Error Sampling error
The difference between the point estimate and the actual population parameter value.
For μ:
the sampling error is the difference – μ
μ is generally unknown
varies from sample to sample
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14. Margin of Error
The greatest possible distance between the point estimate and the value of the parameter it is estimating for a given level of confidence, c.
Denoted by E
𝐸=| 𝑥 − 𝜇| [Magnitude of sampling error]
𝐸= 𝑧 𝑐 𝜎 𝑥 = 𝑧 𝑐 𝜎 𝑛
Sometimes called the maximum error of estimate or error tolerance.
When n ≥ 30, the sample standard deviation, s, can be used for σ
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15. Distribution of Sample Means
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16. Distribution of Sample Means
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17. Distribution of Sample Means
STATEMENT
REASON
𝑃( 𝑥 −𝐸< 𝜇< 𝑥 +𝐸 )
Probability that the generated confidence interval contains 𝜇
= 𝑃 [ 𝑥 −𝐸< 𝜇 𝐴𝑁𝐷 𝜇< 𝑥 +𝐸 ]
Equivalent inequality
= 𝑃 [ 𝑥 < 𝜇+𝐸 𝐴𝑁𝐷 𝑥 > 𝜇−𝐸 𝑥 > 𝜇−𝐸 ]
= 𝑃 [ 𝜇−𝐸< 𝑥 < 𝜇+𝐸 ]
= 𝑐
Shaded area in diagram
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18. Exercise 2: Finding the Margin of Error
Use the social networking website data and a 95% confidence level to find the margin of error for the mean number of friends for all users of the website. Assume that the population standard deviation is about 53.0.
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19. Solution: Finding the Margin of Error
Step 1: Find the critical values [Table 5 – Page A23] z zc
z = 0
0.95
0.025
0.025
–zc = –1.96
zc
zc = 1.96
95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n = 40 ≥ 30.)
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20. Solution: Finding the Margin of Error
Step 2: Compute E
𝐸= 𝑧 𝑐 𝜎 𝑥 = 𝑧 𝑐 𝜎 𝑛 =1.96 ⋅ ≈16.4
We are 95% confident that the margin of error for the population mean is about 16.4 friends. Our estimate might be too big, or too small.
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21. Confidence Intervals for the Population Mean
A c-confidence interval for the population mean μ
The probability that the confidence interval contains μ is c.
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22. Constructing Confidence Intervals for μ
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23. Exercise 3: Construct a Confidence Interval
Construct a 95% confidence interval for the mean number of friends for all users of the website.
Solution: Recall and E ≈ 16.4
Left Endpoint:
Right Endpoint:
114.4 < μ < 147.2
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24. Solution: Constructing a Confidence Interval
114.4 < μ < 147.2 •
With 95% confidence, you can say that the population mean number of friends is between and
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25. Exercise 4: Construct Confidence Interval: Case σ is Known
A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. Construct a 90% confidence interval of the population mean age.
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26. Solution: Constructing a Confidence Interval σ Known
First find the critical values z
z = 0 zc
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
–zc = –1.645
zc
zc = 1.645
zc = 1.645
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27. Solution: Constructing a Confidence Interval σ Known
Margin of error:
Confidence interval:
Left Endpoint:
Right Endpoint:
22.3 < μ < 23.5
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28. Solution: Constructing a Confidence Interval σ Known
22.3 < μ < 23.5
Point estimate
22.3
22.9
23.5
( ) •
With 90% confidence, we can say that the mean age of all the students is between 22.3 and 23.5 years.
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29. Interpreting the Results
μ is a fixed number. Once the interval is fixed, it either contains or does not contain μ .
Incorrect: “There is a 90% probability that the actual mean μ is in the interval (22.3, 23.5).”
Correct: “If a large number of samples is collected and a confidence interval is created for each sample, approximately 90% of these intervals will contain μ.
Correct: “We are 90% confident that the interval (22.3,23.5) is one which contains the actual population parameter μ.”
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30. Interpreting the Results
The horizontal segments represent 90% confidence intervals for different samples of the same size.
In the long run, 9 of every 10 such intervals will contain μ. μ
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31. Interpreting the Results
Since 𝑥 is a random variable, so are the endpoints 𝑥 −𝐸 and 𝑥 +𝐸.
After the confidence interval is numerically fixed for a specific sample, it either does or does not contain µ. μ
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32. TI-84: Z-Interval Function
The ZInterval (STAT / TESTS / 7: ZInterval) function computes a confidence interval for a population mean µ when the population standard deviation is known.
Input
σ, 𝑥 , n, 𝑐−𝑙𝑒𝑣𝑒𝑙
Output
𝑥 −𝐸 𝑡𝑜 𝑥 +𝐸
Margin of Error
𝐸= 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ
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33. Exercise 5: Construct Confidence Interval via TI-84: Case σ is Known
The ZInterval (STAT / TESTS / 7: ZInterval) function computes a confidence interval for a population mean µ when the population standard deviation is known.
Inputs
σ = 1.5, 𝑥 =22.9, 𝑛=20, 𝑐=0.90
Output
(22.3, 23.5)
Margin of Error
𝐸= −22.3 =0.6
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34. Minimum Sample Size
Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean µ is given by
Round up using CELING
𝑛= 𝑧 𝑐 𝜎 𝐸 2
If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30
Result should be rounded up by using the CELING function
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35. Exercise 6: Minimum Sample Size
You want to estimate the mean number of friends for all users of the website. How many users must be included in the sample if you want to be 95% confident that the sample mean is within seven friends of the population mean? Assume the population standard deviation is about 53.0.
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36. Solution: Minimum Sample Size
First find the critical values z
z = 0 zc
0.95
0.025
0.025
–zc = –1.96
zc
zc = 1.96
zc = 1.96
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37. Solution: Minimum Sample Size
zc = σ ≈ s ≈ E = 7
When necessary, round up to obtain a whole number.
Interpretive Statement:
You should include at least 221 users in your sample in order to be 95% confident that the sample mean will be within 7 friends of the population mean.
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38. Exercise 7: Salmon Sample Size
A sample of 50 salmon is caught and weighed. The sample standard deviation of the 50 weights is 2.15 lb. How large of a sample should be taken to be 97% confident that the sample mean is within 0.20 lb of the mean weight of the population? Summarize your results in a complete sentence relevant to this application.
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39. Solution: Salmon Sample Size
Critical Values
𝑧 𝑐 =−𝑖𝑛𝑣𝑁𝑜𝑟𝑚 1− =2.17
n= z c σ E 2 =
2.17 ⋅ =
Apply Ceiling Function
545 salmon
Interpretive Statement
A sample size of 545 salmon will be large enough to ensure that we are 97% confident that the sample mean is within 0.20 lb. of the population mean.
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40. Section 8.1 Summary Found a point estimate and a margin of error
Constructed and interpreted confidence intervals for the population mean
Determined the minimum sample size required when estimating μ
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41. Estimating µ when σ is Unknown
Section 8.2
Estimating µ when σ is Unknown
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42. Section 8.2 Objectives
Interpret the t-distribution and use a t-distribution table
Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown
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43. The t-Distribution
When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a t-distribution.
Critical values of t are denoted by tc.
Distribution was discovered in 1908 by Louis Gosset while he worked for the Guinness Brewing Company
𝑡= 𝑥− 𝜇 𝑠 𝑛
When we use 𝑠 𝑛 as an estimate of the standard deviation of the mean, the shape of the sampling model changes from the normal model to a t-distribution.
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44. The t-Distribution
𝑡= 𝑥− 𝜇 𝑠 𝑛
If many samples of size n are drawn, then we get many t-scores using the equation above
For each value of n, the t-scores can be organized into a frequency table and converted into a histogram
They resulting graphs are known collectively as the Student’s t distribution
When we use 𝑠 𝑛 as an estimate of the standard deviation of the mean, the shape of the sampling model changes from the normal model to a t-distribution.
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45. Properties of the t-Distribution
The t-distribution is bell shaped and symmetric about the mean.
The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size.
d.f. = n – Degrees of freedom
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46. Properties of the t-Distribution
The total area under a t-curve is 1 or 100%.
The mean, median, and mode of the t-distribution are equal to zero.
As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t-distribution is very close to the standard normal z-distribution.
d.f. = 5
The tails in the t-distribution are “thicker” than those in the standard normal distribution.
d.f. = 2 t
Standard normal curve
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47. Exercise 1: Critical Values of t Appendix II, Table 6– p. A24
Find the critical value tc for a 95% confidence level when the sample size is 15.
Solution: d.f. = n – 1 = 15 – 1 = 14
Table 5: t-Distribution
tc = 2.145
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48. Example: Critical Values of t Appendix II, Table 5(a) – p. A23
Rounding Convention: t-scores should be rounded to three decimal places
Larson/Farber 5th ed

49. Solution: Critical Values of t
95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = ±2.145. t
–tc = –2.145
tc = 2.145
c = 0.95
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50. Confidence Intervals for the Population Mean
A c-confidence interval for the population mean μ
The probability that the confidence interval contains μ is c.
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51. Confidence Intervals and t-Distributions
In Words In Symbols
Identify the sample statistics n, , and s.
Identify the degrees of freedom, the level of confidence c, and the critical value tc.
Find the margin of error E.
d.f. = n – 1
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52. Confidence Intervals and t-Distributions
In Words In Symbols
Find the left and right endpoints and form the confidence interval.
Left endpoint: Right endpoint:
Interval:
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53. Exercise 2: Construct a Confidence Interval
You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed.
Solution:
Use the t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed).
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54. Solution: Construct a Confidence Interval
n =16, x = s = c = 0.95
df = n – 1 = 16 – 1 = 15
Critical Value
Table 5: t-Distribution
tc = 2.131
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55. Solution: Construct a Confidence Interval
Margin of error:
Confidence interval:
Left Endpoint:
Right Endpoint:
156.7 < μ < 167.3
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56. Solution: Construct a Confidence Interval
156.7 < μ < 167.3
Point estimate
156.7
162.0
167.3
( ) •
With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF.
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57. Normal or t-Distribution?
Yes
Is n ≥ 30?
Use the normal distribution with
If σ is unknown, use s instead. No
Is the population normally, or approximately normally, distributed? No
Cannot use the normal distribution or the t-distribution.
Yes
Yes
Is σ known?
Use the normal distribution with No
Use the t-distribution with
and n – 1 degrees of freedom.
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58. Example: Normal or t-Distribution?
You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost?
Solution:
Use the normal distribution (the population is normally distributed and the population standard deviation is known)
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59. Exercise 3: Construct Confidence Interval
An archeologist discovered a new, but extinct, species of miniature horse. The only seven known samples show shoulder heights (in cm) of 45.3, 47.1, 44.2, 46.8, 46.5, 45.5, and Find the 99% confidence interval for the mean height of the entire population of ancient horses and the margin of error E. Then summarize your results in a complete sentence relevant to this application.
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60. Exercise 3: Construct Confidence Interval𝑛= 7
𝑑𝑓=𝑛−1 6
𝑥 =
14.6 𝑠=
1.19
𝑡 𝑐 =
𝑡 𝑐 =3.707
𝐸 = 𝑡 𝑐 𝑠 𝑛
𝐸 = ⋅ ≈1.67
Left Endpoint: 𝑥 −𝐸
46.14 – 1.67 = 44.5
Right Endpoint: 𝑥 +𝐸
= 47.8
Confidence Interval
44.5 < 𝜇<47.8
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61. Section 8.2 Summary
Interpreted the t-distribution and used a t-distribution table
Constructed confidence intervals when n < 30, the population is normally distributed, and σ is unknown
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62. Confidence Intervals for Population Proportions
Section 8.3
Confidence Intervals for Population Proportions
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63. Section 8.3 Objectives
Find a point estimate for the population proportion
Construct a confidence interval for a population proportion
Determine the minimum sample size required when estimating a population proportion
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64. Point Estimate for Population p
Population Proportion
The probability of success in a single trial of a binomial experiment.
Denoted by p
Point Estimate for p
The proportion of successes in a sample.
𝑝 = 𝑟 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒
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65. Point Estimate for Population p
Estimate Population Parameter…
with Sample Statistic
Proportion: p
Point Estimate for q, the population proportion of failures
Denoted by
𝑞 = 1− 𝑝
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66. Example: Point Estimate for p
In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal while at work. Find a point estimate for the population proportion of U.S. adults who say it is acceptable to check personal while at work. (Adapted from Liberty Mutual)
Solution: n = and r = 662
𝑝 = 𝑟 𝑛 = =0.662=66.2%
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67. Confidence Intervals for p
A c-confidence interval for a population proportion p
The probability that the confidence interval contains p is c.
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68. Constructing Confidence Intervals for p
In Words In Symbols
Identify the sample statistics n and r.
Find the point estimate
Verify that the sampling distribution of can be approximated by a normal distribution.
Find the critical value zc that corresponds to the given level of confidence c.
𝑝 = 𝑟 𝑛
Use the Standard Normal Table or technology.
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69. Constructing Confidence Intervals for p
In Words In Symbols
Find the margin of error E.
Find the left and right endpoints and form the confidence interval.
Left endpoint:
Right endpoint: Interval:
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70. Exercise 1: Confidence Interval for p
In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal while at work. Construct a 95% confidence interval for the population proportion of U.S. adults who say that it is acceptable to check personal while at work.
Solution: Recall
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71. Solution: Confidence Interval for p
Verify the sampling distribution of can be approximated by the normal distribution
Margin of error:
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72. Solution: Confidence Interval for p
Left Endpoint:
Right Endpoint:
0.633 < p < 0.691
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73. Solution: Confidence Interval for p
Point estimate
With 95% confidence, you can say that the population proportion of U.S. adults who say that it is acceptable to check personal while at work is between 63.3% and 69.1%.
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74. Exercise 2: Confidence Interval for p
Suppose that 800 students were selected at random from a student body of 20,000 and given flu shots. All 800 students were exposed to the flu, and 600 of them did not get the flu. Let p represent the probability that the shot will be successful for any single student selected at random from the entire population. Find a 99% confidence interval for p.
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75. Solution: Confidence Interval for p
Critical Values
𝑧 𝑐 =2.58 𝑛=
800
𝑝 = r/n
600/800 = 0.75
𝑞 = 1− 𝑝 =
1 – 0.75 = 0.25
𝑛 𝑝 =
800 (0.75) = [exceeds 5]
𝑛 𝑞 =
800 (0.25) = [exceeds 5]
𝐸 = 𝑧 𝑐 𝑝 𝑞 𝑛 =
2.58 ⋅ (0.25) ≈0.0395
Left Endpoint: 𝑝 −𝐸
0.750−0.0395=0.710
Right Endpoint: 𝑝 + 𝐸
=0.790
Confidence Interval
0.710<𝑝<0.790
Interpretive Statement
With 99% confidence, we can say that the probability that a flu shot will be effective for a student selected at random will be between 71% and 79%.
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76. Exercise 3: Confidence Interval for p via TI-4
Suppose that 800 students were selected at random from a student body of 20,000 and given flu shots. All 800 students were exposed to the flu, and 600 of them did not get the flu. Let p represent the probability that the shot will be successful for any single student selected at random from the entire population. Find a 99% confidence interval for p using the 1-PropZinterval function on the TI-84.
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77. Exercise 3: Confidence Interval for p via TI-84 1-PropZInterval Function
Inputs
𝑥=600, 𝑛=800, 𝑐=0.99
Output
(0.711,0.789)
Margin of Error
𝐸= −0.711 =0.039
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78. Minimum Sample Size
Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is
This formula assumes you have an estimate for and .
If not, use and
Round up to nearest integer using the CEILING function
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79. Minimum Sample Size
Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is
If you have no preliminary estimate for p, use
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80. Exercise 4: Minimum Sample Size
You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if
No preliminary estimate for p is available
A preliminary estimate for p is 0.31
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81. Solution: Minimum Sample Size: No Preliminary Estimate for p is available
Critical Values
𝑧 𝑐 =1.96 𝐸=
0.03
𝑛= 𝑧 𝑐 𝐸 2 = ≈
Apply Ceiling Function
1068 voters
Interpretive Statement
The sample size should be at least 1068 voters in order to be 95% confident that our estimate is within 3% of the true population proportion.
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82. Solution: Minimum Sample Size: Case: Preliminary Estimate for p is Available
You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if a preliminary estimate gives
Solution:
Use the preliminary estimate
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83. Solution: Sample Size c = 0.95 zc = 1.96 E = 0.03
Round up to the nearest whole number.
With a preliminary estimate of , the minimum sample size should be at least 914 voters.
Need a larger sample size if no preliminary estimate is available.
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84. Exercise 5: Minimum Sample Size
You wish to estimate, with 90% confidence and within 2% of the true population, the proportion of adults age 18 to 29 who have high blood pressure. Find the minimum sample size needed if
No preliminary estimate for p is available
A previous survey found that 4% of adults in this age group had high blood pressure.
Solution:
Because you do not have a preliminary estimate for use and
© 2012 Pearson Education, Inc. All rights reserved.
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85. Solution: Minimum Sample Size When Preliminary Estimate for p is Unavailable
Critical Values
𝑧 𝑐 =1.645 𝐸=
0.02
𝑛= 𝑧 𝑐 𝐸 2 = ≈
Apply Ceiling Function
1692 adults
Interpretive Statement
With no preliminary estimate for p, the sample size should be at least 1692 adults in order to be 90% confident that our estimate is within 2% of the true population proportion.
© 2012 Pearson Education, Inc. All rights reserved.
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86. Solution: Minimum Sample Size When Preliminary Estimate for p is Available
Critical Values
𝑧 𝑐 =1.645 𝐸=
0.02
𝑛=𝑝𝑞 𝑧 𝑐 𝐸 2 =
≈259.78
Apply Ceiling Function
260 adults
Interpretive Statement
With a preliminary estimate for p of 4%, the sample size should be at least 260 adults in order to be 90% confident that our estimate is within 2% of the true population proportion
© 2012 Pearson Education, Inc. All rights reserved.
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87. Solution: Minimum Sample Size When Preliminary Estimate for p is known
c) Determine how many more adults are needed when no preliminary estimate is available compared to when a preliminary estimate for p is 4%.
Solution:
1692 −260=1432
© 2012 Pearson Education, Inc. All rights reserved.
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88. Section 8.3 Summary
Found a point estimate for the population proportion
Constructed a confidence interval for a population proportion
Determined the minimum sample size required when estimating a population proportion
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