1. Chapter Eight
Estimation

2. Estimating  When  is Known
Assumptions:
We have a simple random sample of size n.
If the distribution is normal, methods work for any sample size.
If the distribution is unknown, a sample size of at least 30 (sometimes even more) is required.

3. Point Estimate
an estimate of a population parameter given by a single number

4. Examples of Point Estimates
x is used as a point estimate for m .
s is used as a point estimate for s.

5. Margin of Error
the magnitude of the difference between the point estimate and the true parameter value

6. The margin of error using x as a point estimate for m is

7. Confidence Level
A confidence level, c, is a measure of the degree of assurance we have in our results.
The value of c may be any number between zero and one.
Typical values for c include 0.90, 0.95, and 0.99.

8. Critical Value for a Confidence Level, c
the value zc such that the area under the standard normal curve falling between – zc and zc is equal to c.

9. Confidence Level

10. Find z0.90 such that 90% of the area under the normal curve lies between z0.90 and z0.90
P(– z0.90 < z < z0.90 ) = 0.90

11. Find z0.90 such that 90% of the area under the normal curve lies between –z0.90 and z0.90
P(z < z0.90 ) = (1 – 0.90)/2 = .05

12. Find z0.90 such that 90% of the area under the normal curve lies between –z0.90 and z0.90
According to Appendix Table 3, lies exactly halfway between two values in the table (.0505 and .0495).
Averaging the z values associated with areas gives z0.90 = 1.645.
z0.90 =

13. Common Levels of Confidence and Their Corresponding Critical Values

14. Confidence Interval
A c confidence interval for  is an interval computed from sample data.
In a c confidence interval for , c is the probability of generating an interval containing the actual value of .

15. To Find a Confidence Interval for  When  is Known:
Let x represent the appropriate random variable.
Obtain a simple random sample (of size n) of x values
Compute the sample mean,
If you cannot assume x has a normal distribution, use a sample size of 30 or more.

16. Confidence Interval for  When  is Known:

17. Create a 95% confidence interval for the mean driving time between Philadelphia and Boston. Assume that the mean driving time of 64 trips was 6.4 hours and that the standard deviation is 0.9 hours.

18. Creating a 95% confidence interval
x = 6.4 hours  = 0.9 hours c = 95%, so zc = n = 64

19. x = 6.4 hours  = 0.9 hours 95% confidence interval will be from

20. x = 6.4 hours  = 0.9 hours c = 95%, so zc = 1.96 n = 64

21. 95% Confidence Interval:
6.4 – < m <
< m <
We are 95% sure that the true time is between 6.18 and 6.62 hours.

22. We may get different confidence intervals for different samples.

23. We may get different confidence intervals for different samples.
For each sample the c confidence interval goes from
If we select many samples of the same size and find the corresponding confidence intervals, then the proportion of these intervals that actually contain  is c.

24. When estimating the mean, how large a sample must be used in order to assure a given level of confidence?
Use the formula:

25. How do we determine the value of the population standard deviation, s?
Use the standard deviation, s, of a preliminary sample of size 30 or larger to estimate s.

26. Determine the sample size necessary to determine (with 99% confidence) the mean time it takes to drive from Philadelphia to Boston. We wish to be within 15 minutes of the true time. Assume that a preliminary sample of 45 trips had a standard deviation of 0.8 hours.

27. ... determine with 99% confidence...
z0.99 = 2.58

28. ... We wish to be within 15 minutes of the true time. ...
E = 15 minutes = 0.25 hours

29. ...a preliminary sample of 45 trips had a standard deviation of 0.8 hours.
Since the preliminary sample is large enough, we can assume that the population standard deviation is approximately equal to 0.8 hours.

30. Minimum Sample Size =

31. Rounding Sample Size
Any fractional value of n is always rounded to the next higher whole number.

32. Minimum Sample Size n » 68.16 Round to the next higher whole number.
To be 99% confident in our results, the minimum sample size = 69.

33. Estimating  When  is Unknown
Apply the Student’s t distribution.

34. Student’s t Variable

35. The shape of the t distribution depends only on the sample size, n, if the basic variable x has a normal distribution.
When using the t distribution, we will assume that the x distribution is normal.

36. Appendix Table 4 (Page A8) gives values of the variable t corresponding to the number of degrees of freedom (d.f.)

37. Degrees of Freedom
d.f. = n – 1
where n = sample size

38. The t Distribution has a Shape Similar to that of the the Normal Distribution

39. Properties of a Student’s t Distribution
Symmetric about the mean 0.
Depends on the degrees of freedom.
Bell-shaped with thicker tails than the normal distribution.
As the degrees of freedom increase, the t distribution approaches the standard normal distribution

40. Appendix Table 4
Gives various t values for different degrees of freedom

41. Using Table 4 to Find Critical Values tc for a c Confidence Level

42. If the required d.f. are not in the table:
Use the closest d.f. that is smaller than the needed d.f.
This results in a larger critical value tc.
The resulting confidence interval will be longer and have a probability slightly higher than c.

43. Using Table 4 to Find Critical Values of tc
Find the column in the table with the given c heading
Compute the number of degrees of freedom:
d.f. = n  1
Read down the column under the appropriate c value until we reach the row headed by the appropriate d.f.

44. To find the critical value tc for a 95% confidence interval if n = 8.
Find the column in the table with c heading 0.950

45. To find the critical value tc for a 95% confidence interval if n = 8.
Compute the number of degrees of freedom:
d.f. = n  1 = 8  1 = 7

46. To find the critical value tc for a 95% confidence interval if n = 8.
Read down the column under the appropriate c value until we reach the row headed by d.f. = 7

47. Find the critical value tc for a 95% confidence interval if n = 8.

48. Finding Confidence Intervals for  When  is Unknown
c = confidence level (0 < c < 1)
tc = critical value for confidence level c, and degrees of freedom = n  1

49. The mean weight of eight fish caught in a local lake is 15
The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of 2.3 ounces. Construct a 90% confidence interval for the mean weight of the population of fish in the lake.

50. Mean = 15.7 ounces Standard deviation = 2.3 ounces.
n = 8, so d.f. = n – 1 = 7
For c = 0.90, Appendix Table 4
gives t0.90 =

51. Mean = 15.7 ounces Standard deviation = 2.3 ounces.
The 90% confidence interval is:
<  <
14.16 <  < 17.24

52. The 90% Confidence Interval: 14.16 < m < 17.24
We are 90% sure that the true mean weight of the fish in the lake is between and ounces.

53. Review of the Binomial Distribution
Completely determined by the number of trials (n) and the probability of success (p) in a single trial.
q = 1 – p
If np and nq are both > 5, the binomial distribution can be approximated by the normal distribution.

54. A Point Estimate for p, the Population Proportion of Successes

55. Point Estimate for q (Population Proportion of Failures)

56. For a sample of 500 airplane departures, 370 departed on time
For a sample of 500 airplane departures, 370 departed on time. Use this information to estimate the probability that an airplane from the entire population departs on time.
We estimate that there is a 74% chance that any given flight will depart on time.

57. Margin of Error for p as a Point Estimate for p

58. Maximal Margin of Error
the maximal error of estimate E for a confidence interval

59. Confidence Interval for p for Large Samples (np and nq > 5)
c = confidence level
zc = critical value for confidence level c taken from a normal distribution

60. For a sample of 500 airplane departures, 370 departed on time
For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time.
Is the use of the normal distribution justified?

61. Can we use the normal distribution?
For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time.

62. For a sample of 500 airplane departures, 370 departed on time
For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time.
so the use of the normal distribution is justified.

63. Out of 500 departures, 370 departed on time
Out of 500 departures, 370 departed on time. Find a 99% confidence interval.

64. Confidence interval is:
99% confidence interval for the proportion of airplanes that depart on time:
E =
Confidence interval is:

65. 99% confidence interval for the proportion of airplanes that depart on time
Confidence interval is
We are 99% confident that between 69% and 79% of the planes depart on time.

66. The point estimate and the confidence interval do not depend on the size of the population. The sample size, however, does affect the accuracy of the statistical estimate.

67. Interpretation of Poll Results
The proportion responding in a certain way is:
the sample estimate of the population proportion.

68. Interpret the following poll results:
“ A recent survey of 400 households indicated that 84% of the households surveyed preferred a new breakfast cereal to their previous brand. Chances are 19 out of 20 that if all households had been surveyed, the results would differ by no more than 3.5 percentage points in either direction.”

69. A 95% confidence interval for population proportion p is:

70. “Chances are 19 out of 20 …” 19/20 = 0.95
A 95% confidence interval is being used.

71. “... 84% of the households surveyed preferred …”
84% represents the percentage of households who preferred the new cereal.

72. “. the results would differ by no more than 3
“... the results would differ by no more than 3.5 percentage points in either direction.”
3.5% represents the margin of error, E.
The confidence interval is:
84% - 3.5% < p < 84% + 3.5%
80.5% < p < 87.5%

73. The poll indicates ( with 95% confidence):
between 80.5% and 87.5% of the population prefer the new cereal.

74. Sample Size for Estimating p for the Binomial Distribution

75. If p is an estimate of the true population proportion,
Formula for Minimum Sample Size for Estimating p for the Binomial Distribution
If p is an estimate of the true population proportion,

76. Formula for Minimum Sample Size for Estimating p for the Binomial Distribution
If we have no preliminary estimate for p, the probability is at least c that the point estimate r/n for p will be in error by less than the quantity E if n is at least:

77. The manager of a furniture store wishes to estimate the proportion of orders delivered by the manufacturer in less than three weeks. She wishes to be 95% sure that her point estimate is in error either way by less than Assume no preliminary study is done to estimate p.

78. She wishes to be 95% sure ...
z0.95 = 1.96

79. ... that her point estimate is in error either way by less than 0.05.

80. ... no preliminary study is done to estimate p.
The minimum required sample size (if no preliminary study is done to estimate p) is 385.

81. If a preliminary estimate estimate of p indicated that p was approximately equal to 0.75:
The minimum required sample size (if this preliminary study is done to estimate p) is 289.