1. 6 Chapter Confidence Intervals © 2012 Pearson Education, Inc.
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2. Chapter Outline 6.1 Confidence Intervals for the Mean (Large Samples)
6.2 Confidence Intervals for the Mean (Small Samples)
6.3 Confidence Intervals for Population Proportions
6.4 Confidence Intervals for Variance and Standard Deviation
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3. Confidence Intervals for the Mean (Large Samples)
Section 6.1
Confidence Intervals for the Mean (Large Samples)
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4. Section 6.1 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. Example: 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)
Enter this in list 1. We will use this for future examples in this section
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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.
The problem with a Point Estimate is that the real world probability of hitting that point is virtually zero. -In other words, who has friends?
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8. Interval Estimate Therefore, we estimate µ using an 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. 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)
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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10. 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.
Previously- we have calculated this as InvNorm(.05)
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11. Levels of Confidence
There are 5 standard levels of confidence in statistics:
80%  z = (InvNorm(.10)
90%  z = (InvNorm(.05)
95%  z = (InvNorm(.025)
98%  z = (InvNorm(.01)
99%  z = (InvNorm(.005)
KNOW THESE
Write them down
It will save you time
NOTE: To have a 90% confidence level, we must have an area under the curve = .90
Therefore, our left limit is , and our right limit is 1.645
When using this Z value as a Z critical value (and not as a left limit/right limit) we only use the POSITIVE Z value

12. 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
Therefore the sampling error is hard to calculate –if at all
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13. Margin of Error Margin of error
The greatest possible distance between the point estimate (sample mean) and the value of the parameter (Population mean) it is estimating for a given level of confidence, c.
Denoted by E.
Sometimes called the maximum error of estimate or error tolerance.
When n ≥ 30, the sample standard deviation, s, can be used for σ.
Critical Z Score
NOTE: If we had entered the long list of numbers into list 1, we could then calculate S using STAT: 1 –VAR Stats
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14. Example: 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 the sample standard deviation is about 53.0.
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15. Solution: Finding the Margin of Error
First find the critical values
This slide is showing how they got a Z-score
We have this on our cheat sheet of scores 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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16. Solution: Finding the Margin of Error
You don’t know σ, but since n ≥ 30, you can use s in place of σ.
On a Calculator:
Stat TestZinterval
Choose Stats
Enter ơ
Enter “0” for x-bar (we don’t know it)
Enter n
Enter c-Level
Calculate
You are 95% confident that the margin of error for the population mean is about 16.4 friends.
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17. On the Calculator
The problem with calculating “The maximum error of Estimate” E is calculating the sample Standard Deviation S (they may not give you a population standard deviation to use)
This is easily overcome, however, if you make a list, then run 1-var Stats
You need this to see S the sample deviation
If you make a list, you can then go to Stats TestsZinterval
Here, it asks for input
You can either type in the mean, SD etc. or choose “Data” if you made a list.
You then choose the confidence level  for example .95
It will then tell you the bottom and top value for your interval estimate. It will also give you the sample means (which is also the point estimate) and the sample standard deviation S
Nice…

18. 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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19. Constructing Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean (n ≥ 30 or σ known with a normally distributed population)
In Words In Symbols
Find the sample statistics n and .
Specify σ, if known. Otherwise, if n ≥ 30, find the sample standard deviation s and use it as an estimate for σ.
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20. Constructing Confidence Intervals for μ
In Words In Symbols
Find the critical value zc that corresponds to the given level of confidence.
Find the margin of error E.
Find the left and right endpoints and form the confidence interval.
Use the Standard Normal Table or technology.
Left endpoint:
Right endpoint: Interval:
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21. Example: Constructing 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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22. 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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23. Example: Constructing a Confidence Interval σ 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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24. 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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25. Solution: Constructing a Confidence Interval σ Known
Margin of error:
Confidence interval:
Sometimes this is easier than using Z interval…
Left Endpoint:
Right Endpoint:
22.3 < μ < 23.5
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26. Solution: Constructing a Confidence Interval σ Known
22.3 < μ < 23.5
Point estimate
22.3
22.9
23.5
( ) •
With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years.
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27. Interpreting the Results
μ is a fixed number. It is either in the confidence interval or not.
Incorrect: “There is a 90% probability that the actual mean 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.
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28. 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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29. 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
If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30 members.
You will need to do this manually. Make a note of this equation.
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30. Example: 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 sample standard deviation is about In this example, we are setting our margin of error –how close do we want to be- and still maintaining our confidence level
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31. Solution: Sample Size First find the critical values zc = 1.96 z z = 0
0.95
0.025
0.025
–zc = –1.96
zc
zc = 1.96
zc = 1.96
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32. Solution: Sample Size zc = 1.96 σ ≈ s ≈ 53.0 E = 7
When necessary, round up to obtain a whole number. Always round up.
You should include at least 221 users in your sample.
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33. Section 6.1 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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34. Assignment Page 311 5-32 skip 17-20 Page 312 35-50 skip 47,48
Larson/Farber 5th ed

35. Chapter 6 Quiz 1 ( 5 points each)
A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 25 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.
What is the point estimate of the mean age?
What is the Z critical value for 95%?
What is the Margin of Error (E) using this data (at 95%)
Construct a 95% confidence interval of the population mean age.
22.9 years
1.96
.59
22.3,23.5
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36. Confidence Intervals for the Mean (Small Samples)
Section 6.2
Confidence Intervals for the Mean (Small Samples)
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37. Section 6.2 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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38. 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.
How would we figure out if the variable x is approximately normally distributed?
Question: When do we use the T-Distribution instead of the Normal Distribution? Answer: When the population σ is unknown and the sample is less than 30. KNOW THIS
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39. 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
the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary.
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40. Degrees of Freedom
In statistics, the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary
Remember, this is calculated as n-1
For every n-value, there is the possibility of variance, so we account for that
Why n-1 then?
What if you had 25 chairs, and 25 students. As each student walked in the door, they would have a choice (freedom) of where to sit. Except for the last student. His choice is made. So you have 25-1, or 24, choices

41. 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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42. T-Distribution Curve
The tails in a t-distribution are “thicker” than those in the standard normal distribution

43. T-Distribution Curves
William S. Gosset developed this curve
He published it under the pseudo-name “Student” thus –if you Google a T-curve, you may find “Student’s T-Distribution”

44. Example: Critical Values of t
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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45. 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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46. Example (using a Calculator)
Finding the Critical Values of T
2ndVarsinvT(area,D.F.)
Area is almost the same as the confidence level –but you have to add one tail to get (in essence) the right limit, or critical T-value.
For example, find the critical value of T for a 95% confidence when the sample size is 15
Area = one tail (1+.95)/2 = .975
D.F. = 15-1 = 14
InvT(.975,14) =  which is the T-Critical value
Therefore, 95% of the area under the t-distribution curve with 14 degrees of freedom is between and 2.145

47. Chapter 6 Quiz 2 Provide the answers on a piece of paper Instructions
Using data from problem 64 on page 315, do the following:
Enter data into list 1
Calculate the sample standard deviation (s), the population standard deviation (Ơ) and sample mean (xbar) calc/1-Var Stats
Calculate a 95% confidence interval test/Zinterval (be sure to use Ơ since you have it, and input data, not “stats”) n=
xbar=
Ơ= S=_______
Z critical score=
Margin of Error (E) =
Confidence interval=
__________< µ < ________
Each question worth 5 points
N=34
Xbar=19.96
Ơ=2.36
Zcritical = 1.96
Margin of Error: .795ish
Confidence Interval = 19.15ish and 20.77ish
Larson/Farber 5th ed

48. 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.
This should look familiar, it is the SAME EQUATION we used to calculate the margin of Error in a normal distribution, except instead of a Zcritical score, we have a Tcritical score and we use the sample standard deviation
 Same equation except use T instead of Z, and S instead of Ợ
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49. On a Calculator On a Calculator: Stat TestTinterval Choose Stats
Enter sx
Enter “0” for x-bar (we don’t know it)
Enter n
Enter c-Level
Calculate
Chose the positive answer
Larson/Farber 5th ed

50. 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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51. 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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52. Example: Constructing 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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53. Solution: Constructing a Confidence Interval
n =16, x = s = c = 0.95
df = n – 1 = 16 – 1 = 15
Critical Value
InvT(.975,15)
Table 5: t-Distribution
tc = 2.131
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54. Solution: Constructing a Confidence Interval
Margin of error:
Confidence interval:
Left Endpoint:
Right Endpoint:
156.7 < μ < 167.3
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55. Solution: Constructing 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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56. Example
You randomly select 16 restaurants and measure the temperature of the coffee
The sample mean is 162 degrees (F)
The sample Standard Deviation (S) = 10 deg.
Find the 95% confidence interval for the mean temperature. Assume they are normally distributed
Go to: StatTestTinterval. Choose “Stats” and enter information
You should get: , with a mean of 162

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. Section 6.2 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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60. Assignment
Page Skip 13-16
Larson/Farber 5th ed

61. Chapter 6 Quiz 3
Using data from problem 24 on page 324, and assuming a 98% confidence level, answer the following: n= s=
D.F.=
xbar=
Tcritical=
Confidence interval (based on 98% confidence level)
Hint: Use InvT for question 5
Hint: Use Tinterval for question 6
5 points each question
1) n=14
2) s=
3) D.F.= 13
4) xbar=
5) Tcritical= InvT(.99,13) = 2.65
6) Confidence Interval = (66258,70853)
Larson/Farber 5th ed

62. Chapter 6 Quiz 4 Solve question #35 on page 325
Calculate the confidence interval indicated in the problem
Is the company making acceptable tennis balls?
Why or why not?
For a Confidence Level we calculate the Tinterval for xbar = 56, sx = .25 and n = 25 with a confidence level of .99 = (55.86,56.14)
They want the average bounce height to be 55.5 inches.
We want this: < mean < 56.14, but the mean is not between these two numbers, therefore these are not acceptable tennis balls.
Larson/Farber 5th ed

63. Confidence Intervals for Population Proportions
Section 6.3
Confidence Intervals for Population Proportions
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64. Section 6.3 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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65. 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.
Denoted by
read as “p hat”
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66. Point Estimate for Population p
Estimate Population Parameter…
with Sample Statistic
Proportion: p
Point Estimate for q, the population proportion of failures
Denoted by
Read as “q hat”
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67. 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 x = 662
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68. 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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69. Constructing Confidence Intervals for p
In Words In Symbols
Identify the sample statistics n and x.
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.
Use the Standard Normal Table or technology.
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70. 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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71. Example: 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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72. Solution: Confidence Interval for p
Verify the sampling distribution of can be approximated by the normal distribution
Margin of error:
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73. Solution: Confidence Interval for p
Left Endpoint:
Right Endpoint:
0.633 < p < 0.691
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74. 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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75. On a Calculator
You can calculate a p confidence interval on a calculator
Stats Tests1PropZInt
Enter X (the number of successes in the sample)
Enter n (the size of the sample)
Enter the confidence level
This will give you the confidence level, and the p-hat
It will not give you the Margin of Error (E) should you need that, you gotta go old school 
Larson/Farber 5th ed

76. 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
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77. Example: 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 is available.
Solution:
Because you do not have a preliminary estimate for use and
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78. Solution: Sample Size c = 0.95 zc = 1.96 E = 0.03
Round up to the nearest whole number.
With no preliminary estimate, the minimum sample size should be at least 1068 voters.
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79. Example: 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
a preliminary estimate gives
Solution:
Use the preliminary estimate
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80. 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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81. Section 6.3 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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82. Assignment
Page
Larson/Farber 5th ed

83. Chapter 6 Quiz 5
You are running a political campaign and wish to estimate, with 98% 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. (hint: solve for n) You do not have a preliminary point estimate to the population proportion. Show equation 5 points for equation, 5 points for solution
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84. Solution: Sample Size c = 0.98 zc = 2.33 E = 0.03 2.33 1508.02 .50 .50
Round up to the nearest whole number.
Using the standard value of .50 for p-hat, and .50 for q-hat, the minimum sample size should be at least 1509 voters.
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85. Confidence Intervals for Variance and Standard Deviation
Section 6.4
Confidence Intervals for Variance and Standard Deviation
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86. Section 6.4 Objectives
Interpret the chi-square distribution and use a chi-square distribution table
Use the chi-square distribution to construct a confidence interval for the variance and standard deviation
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87. The Chi-Square Distribution
The point estimate for σ2 is s2
The point estimate for σ is s
s2 is the most unbiased estimate for σ2
Estimate Population Parameter…
with Sample Statistic
Variance: σ2 s2
Standard deviation: σ s
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88. The Chi-Square Distribution
You can use the chi-square distribution to construct a confidence interval for the variance and standard deviation.
If the random variable x has a normal distribution, then the distribution of
forms a chi-square distribution for samples of any size n > 1.
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89. Properties of The Chi-Square Distribution
All chi-square values χ2 are greater than or equal to zero.
The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ2, use the χ2-distribution with degrees of freedom equal to one less than the sample size.
d.f. = n – Degrees of freedom
The area under each curve of the chi-square distribution equals one.
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90. Properties of The Chi-Square Distribution
Chi-square distributions are positively skewed.
Chi-square Distributions
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91. Critical Values for χ2
There are two critical values for each level of confidence.
The value χ2R represents the right-tail critical value
The value χ2L represents the left-tail critical value. χ2 c
The area between the left and right critical values is c.
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92. Example: Finding Critical Values for χ2
Find the critical values and for a 95% confidence interval when the sample size is 18.
Solution:
d.f. = n – 1 = 18 – 1 = 17 d.f.
Use Table 6 on A-19
Each area in the table represents the region under the chi-square curve to the right of the critical value.
Area to the right of χ2R =
Area to the right of χ2L =
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93. Solution: Finding Critical Values for χ2
Table 6: χ2-Distribution
7.564
30.191
95% of the area under the curve lies between and
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94. Confidence Intervals for σ2 and σ
Confidence Interval for σ2:
Confidence Interval for σ:
The probability that the confidence intervals contain σ2 or σ is c.
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95. Confidence Intervals for σ2 and σ
In Words In Symbols
Verify that the population has a normal distribution.
Identify the sample statistic n and the degrees of freedom.
Find the point estimate s2.
Find the critical values χ2R and χ2L that correspond to the given level of confidence c.
d.f. = n – 1
Use Table 6 in Appendix B.
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96. Confidence Intervals for 2 and 
In Words In Symbols
Find the left and right endpoints and form the confidence interval for the population variance.
Find the confidence interval for the population standard deviation by taking the square root of each endpoint.
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97. Example: Constructing a Confidence Interval
You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation.
Solution:
d.f. = n – 1 = 30 – 1 = 29 d.f.
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98. Solution: Constructing a Confidence Interval
Area to the right of χ2R =
Area to the right of χ2L =
The critical values are χ2R = and χ2L =
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99. Solution: Constructing a Confidence Interval
Confidence Interval for σ2:
Left endpoint:
Right endpoint:
0.80 < σ2 < 3.18
With 99% confidence, you can say that the population variance is between 0.80 and 3.18.
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100. Solution: Constructing a Confidence Interval
Confidence Interval for σ :
0.89 < σ < 1.78
With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams.
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101. Section 6.4 Summary
Interpreted the chi-square distribution and used a chi-square distribution table
Used the chi-square distribution to construct a confidence interval for the variance and standard deviation
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102. Assignment
Page
Larson/Farber 5th ed