Transcript 9.3 Confidence Interval for a Population Mean

9.3 Confidence Interval for a
Population Mean
Wednesday, April 5, 2017
Conditions
1. Random sample
2. Sample size n is large n  30 if n<30, pop.
dist. must be approx normal (if not told
and have data graph it (boxplot) to show
symmetry and no outliers)
3.  , pop std dev known (drawback)
Confidence Interval for Pop Mean
 I
F
x  z *G J
Hn K
When

is unknown!
• We use s to estimate 
• This results in a new standardized variable
t
x
t
s
n
• s introduces extra variability (t dist. is more
spread out than z)
• ex. IQ tests are normed (normally distributed)
with a m = 100 and s = 15. A random sample of
100 students from a high school has a mean IQ
score of 112 and standard deviation of 16.
Calculate a 95% CI for IQ scores at this school.
•
if you have a known population SD, use
it—use a z-interval!
• 1. Describe variable
• 2. requirements: randomness/large
sample/independence
• 3. formula/substitution:
•
• 4. compute interval:
•
• 5. answer in context of the problem:
• 1. Describe variable of interest
– x is the estimate of the average IQ score at
this school ( x is used to estimate μ)
• 2. requirements: randomness/large
sample/independence
– told have a random sample
– large enough sample n=100≥30
– reasonable to assume more than 1000
students so n≤10%N
– given the population standard deviation σ
• OK to use a z-interval
3. formula/substitution:

95% CI : x  z *
n
15
112  196
.
100
112  196
. 15
.
bg
4. compute interval:
– 95% CI: 112 ± 2.94
5. answer in context of the problem:
– I am 95% confident that the true mean IQ
score of students at this school is contained in
the interval (109.06, 114.94)
t distribution:
• Characterized by degrees of freedom (df) like
normal curve characterized by its mean and
standard deviation
• Bell-shaped and centered at zero
• More spread out than z curve
• As df increases, spread of t curve decreases
• As df increases, t curve approached z curve
• See figure 9.6 pg. 498
t distribution:

s I
F
x  t *G J
Hn K
• Interval for
becomes:
• 3rd condition becomes :
•
is unknown, use s to estimate
and t dist with ______ df.
• Ex 9.8 Executive Salaries
• Ex 9.9 Walking a Straight Line
• Ex 9.10 Housework


t distribution:
Choosing the Sample Size
• Set margin of error value given equal to
margin of error part of CI formula and
solve for n
• Ex 9.11 Cost of textbooks
• (range is reasonable estimate of  and z*
for t*)
• ex. Questionnaires were sent to a SRS of 160
major U.S. hotel chain managers and 114
responses were returned. The average
reported time that the managers had spent with
their current company was 11.78 years with
standard deviation of 3.2 years. Give a
99%
confidence interval for the time spent by all U.S.
hotel chain managers.
• not given a population SD—use a t-interval!
• 1. describe variable
• 2. requirements: randomness/large
sample/independence
• 3. formula/substitution:
• 4. compute confidence interval:
• 5. answer in context of the problem:
•
1. Describe variable:
–
x is the average time managers have spent
with their current company (
estimate μ)
x is used to
2. requirements: randomness/large
sample/independence
– told have a SRS
– large enough sample n = 114 which is ≥ 30
– reasonable to assume there are more than
1140 hotel managers so n ≤ 10%N
• OK to use a t-interval
3. formula/substitution: df = n–1 = 114–1=113
use df =100 on table B
s
99% CI : x  tn 1
n
3.2
1178
.  2.626
114
1178
.  2.626 .2997
b g
4. compute confidence interval:
– 11.78 ± 0.787
5. answer in context of the problem:
– I am 99% confident that the interval 10.993 to
12.567 contains the true mean number of
years spent by U.S. hotel chain managers with
their current companies.