Download review - Mr. Kretsch`s Probability and Statistics Class

Basic Premise
• A sample will represent a population
• But a single mean, proportion, or standard deviation
derived from the sample might not be accurate.
• The confidence interval says if, instead of a sample,
we survey the entire population (called a _________),
there is a specific probability (called the
___________________) that the population’s mean,
proportion, or standard deviation will fall within the
confidence interval.
Concepts and Definitions
(Match them!!)
Point estimate
Interval estimate
Degree of confidence
Degrees of freedom
Margin of error
Critical value
Standard normal (Z) distribution
Student (T) distribution
Chi-squared distribution
Distribution for standard deviation
The sample size minus 1
A single estimate of the population
calculated from the sample
The boundary between what's in and
what's our of the interval estimate
Distribution for small samples
The difference between the mean and
the interval boundaries
The probability that the interval estimate
contains the population parameter
A range that probably contains the
population parameter
Distribution for large samples
Question to Ponder
• If an mean interval estimate of (34 to 47) has a 95%
degree of confidence, it would be ___________ for
the population mean to be 32.
A. Impossible
B. Unusual
C. Likely
D. Weird
Questions to Ponder
• The _________ the sample
size, the ________ the
interval.
A. larger, larger
B. smaller, smaller
C. larger, smaller
D. smaller, larger
• The _________ the degree
of confidence, the ________
the interval.
A. larger, larger
B. smaller, smaller
C. larger, smaller
D. smaller, larger
Skills and Procedures
• Finding confidence interval for
– Mean
– Proportion
– Standard deviation (and variance)
• Finding sample size needed to achieve a margin of
error and degree of confidence.
– Mean
– Proportion
Mean
X
s
n
Degree of
confidence
550
15
25
99%
4.5
0.25
55
95%
$12,000 $850
30
90%
Point
estimate
Interval estimate
Proportion
Sample
size
Successes
64
24
120
25
10
p
q
DoC
99%
0.85
22
95%
99%
0.1
90%
Point
Interval estimate
Sample size For Mean
Degree of
confidence
Margin of error
Standard deviation
95%
5
10
90%
1.2
2.0
99%
100
350
95%
1500
2700
n
Sample size for Proportion
Degree of
confidence
Margin of error
p
99%
10%
0.65
95%
3%
Unknown
90%
5%
0.05
95%
7.5%
Unknown
n
Standard Deviation
Degree of
Confidence
s
n
99%
18
45
95%
0.45
32
90%
145
75
90%
2.78
81
95%
1500
23
95%
15.4
12
Point
Interval estimate
Case study #1
• An insurance company that is considering opening an
office in town would like to know the average age of
its adults. Our class sampled their parents and found the
mean age is 52, the standard deviation is 7.
• Identify the point estimate:____________________
• Decide on a degree of confidence (________) and
calculate the confidence interval: _________________
• How many people would we have to sample if we want
to reduce the margin of error by one-half?
Case Study #2
• The insurance company would also like to know the
percent of cars that have antilock brakes. Of the 120
people we polled, 96 said they had ABS
• Calculate the point estimate:____________________
• Decide on a degree of confidence (________) and
calculate the confidence interval:
_________________
Case study #3
• The insurance company
would like to know the
percent of residents that own
hybrid cars, and they would
like a margin of error of 7.5%
• Select a degree of confidence
(_________) and calculate the
sample size needed to meet
the company’s needs.
________
• The insurance company
would like to know the
percent of residents that have
been in accidents. They
would like a margin of error
of 7.5%. A study done in a
nearby town suggests 8%
• Select a degree of confidence
(_______) and calculate the
sample size needed to meet
the company’s needs:
_______
Case study #4
• An insurance company would also like to know the
average accident repair bill. A sample of 41 accident
records showed an average bill of $1250. The standard
deviation was $525.
• Identify the point estimate:____________________
• Decide on a degree of confidence (________) and
calculate the confidence interval: _________________
Case Study #5
• An insurance company would also like to know the
percent of households that have teenage drivers. Of
the 75 people we asked at the local supermarket, 16
said they had teenage drivers
• Calculate the point estimate:____________________
• Decide on a degree of confidence (________) and
calculate the confidence interval:
_________________
Answers
• Mean
– (541.6, 558.4)
– (4.43, 4.57)
– (11,745, 12255)
• Proportion
–
–
–
–
p = 0.375, q = 0.625, point = 0.375, (0.219, 0.531)
x = 102, q = 0.15, point = 0.85, (0.786, 0.914)
p = 0.88, q = 0.12, point = 0.88, (0.70, 1.06)
x = 1, q = 0.9, point = 0.1, (-0.07, 0.27)
More answers
• Mean sample size
–
–
–
–
16
8
82
13
• Proportion Sample Size
–
–
–
–
151
1068
52
169
More Answers
• Standard Deviation
–
–
–
–
–
–
Point = 18, Interval = (14.6, 26.2)
Point = 0.45, interval = (0.366, 0.611)
Point = 145, interval = (131.1, 173.4)
Point = 2.78, interval = (2.463, 3.200)
Point = 1500, interval = (1160.1, 2123.1)
Point = 15.4, interval = (10.91, 26.15)
More answers
• Case study #1
– Point = 52
– Assume n = 25 and Degree of Confidence = 95%, then
Confidence interval = (49.1, 54.8)
• Case Study #2
– Point estimate = 0.8,
– If degree of confidence is 90%, then confidence interval =
(0.740, 0.860)
– If degree of confidence is 95%, then confidence interval =
(0.728, 0.872)
– If degree of confidence is 99%, then confidence interval =
(0.706, 0.894)
More answers
• Case Study #3 (hybrid cars)
– If degree of confidence is 90%, n = 121
– If degree of confidence is 95%, n = 171
– If degree of confidence is 99%, n = 295
• Case study #3 (accidents)
– If degree of confidence is 90%, n = 36
– If degree of confidence is 95%, n = 51
– If degree of confidence is 99%, n = 88
More answers
• Case study #4
–
–
–
–
Point estimate = 1250
If DoC = 90%, interval estimate = (1115.10, 1384.90)
If DoC = 95%, interval estimate = (1089.30, 1410.70)
If DoC = 99%, interval estimate = (1038.80, 1461.20)
• Case study #5
–
–
–
–
Point = 0.21
If DoC = 90%, interval estimate = (0.136, 0.291)
If DoC = 95%, interval estimate = (0.121, 0.306)
If DoC = 99%, interval estimate = (0.091, 0.335)