Transcript Chapter 8 -- Confidence Intervals

Chapter 8 -- Confidence Intervals
m or p is unknown (another course: s)
 We want to estimate the parameter
 Confidence Intervals are accurate estimates
of the true value of the parameter

Decisions to make
How many to survey?
the entire population?
just one object or person?
 How confident do we want to be?
 What error can we live with?

Confidence
CL = 1 - a = confidence level
 CI = confidence interval

ex. Let’s build a 95% Confidence Interval
 1 - a = .95 (confidence level)
 a = “unconfidence” = 0.05
 a/2 = .025
z a/2 = z.025 = 1.96
Confidence Interval
Confidence Interval
CL = 1 - a/2
a/2
-6
-4
a/2
-2
0
2
4
6
Error Bounds
 EBP =
error bound for a proportion
 EBM = error bound for a mean
Error Bound for Proportions
P = X/n
 P = proportion of successes
 X ~ N(np, npq)
 P = X/n ~ N(np/n, npq/n2)
 P ~ N(p, pq/n)
 z = (p’ - p)/(pq/n)1/2
 EBP = p’ - p = z*(p’q’/n)1/2

Error Bounds for Means
X
~ N(m, s/n)
EBM  x  m
z
xm
s
n
EBM  z s
n
Error Bound For Means
 Student-t
Distribution
EBM  x  m
xm
t
s
n
s
EBM  t 
n
Confidence Interval (C. I.)

Means:
(sample mean - EBM,
sample mean + EBM)

Proportions:
(p’ - EBP, p’ + EBP)
C. I. For a Mean - Population
Standard Deviation Known

For her project, Maria bought 50 small bags
of jelly beans. The average number of jelly
beans in a bag was 25. It is known that the
standard deviation is 3 jelly beans. Maria
wants to construct a 90% confidence
interval for the true average number of jelly
beans in a small bag.
C. I. For a Mean - Population
Standard Deviation Known

C.L. = 0.90 implies
z = 1.645
EBM  z s
n
EBM  1645
. 3
EBM  0.70


50
C.I. =
(sample mean + EBM,
sample mean - EBM )
= (25 - 0.70, 25 + 0.70)
= (24.30 , 25.70)
We are 90% confident
that the true average
number of jelly beans is
in the range 24.30 to
25.70.
C. I. For a Mean - Population
Standard Deviation Unknown
Uses a new distribution called the Student-t
(invented by William Gossett)
 Notation: t degrees of freedom
degrees of freedom is abbreviated df.
 Use Student-t when the population standard
deviation is not known, the sample is
“small” (less than 30), and the population
from which the sample comes is normal.

C. I. For a Mean - Population
Standard Deviation Unknown

Seven fast food restaurants were surveyed
concerning the number of calories in four
ounces of french fries. The data is 296, 329,
306, 324, 292, 310, 350 calories. Construct
a 95% confidence interval for the true
average number of calories in a four ounce
serving of french fries. Assume the number
of calories in french fries
follows a normal distribution.
C. I. For a Mean - Population
Standard Deviation Unknown



Sample mean = 315.3
Sample stdev = 20.4
Use Student-t with 6
df. (df = n - 1 = 7 - 1)
EBM  t  s
n
EBM  2.45 20.4
EBM  18.9
7
t=2.45 comes from a
Student-t table where
df = 6 and C.I. = 0.95
 C. I. =
(sample mean - EBM,
sample mean + EBM)
= (315.3-18.9, 315.3+18.9)
=(296.4, 334.2)

C. I. For a Binomial Proportion

At a local cabana club, 102 of the 450
families who are members have children
who swam on the swim team in 1998.
Construct an 80% confidence interval for
the true proportion of families with children
who swim on the swim team in any year.
C. I. For a Binomial Proportion

C.L. = 0.80 implies
z = 1.28; n = 450

p’ = 102/450
q’ = 348/450
EBP  z p'*q'/ n
EBP  128
.  (102 / 450) *(348 / 450) / 450
EBP  0.0253
C. I. For a Binomial Proportion
C. I. = (p’ - EBP, p’ + EBP)
((102/450)-0.0253,
(102/450)+0.0253)
= (0.2014, 0.2520)
 We are 80% confident that the true
proportion of families that have children on
the swim team in any year is between 20%
and 25%.
