Transcript Advanced Math Topics

Advanced Math Topics
9.2/9.3 Estimating the Population Mean
From a Large Sample
Remember this from last chapter?
How would you find the limits of the sample means that lie in the middle 95% of
the normal distribution?
.475
.475
xz

y
μx
Lower Limit
x    1.96
Upper Limit

y
x    1.96
If you did not know the mean of a population (μ),
but you knew a sample mean ( x ), it would help to have a formula for μ.
Rearrange the formulas above to get…
  x  1.96

y
  x  1.96

y

y
95% Confidence Interval for μ
If x is known for a large sample (y > 30), and σ is the sample standard deviation,
then there is a 95% chance that the population mean (μ), lies within the interval…
  x  1.96

y
and
  x  1.96
μ

y
You may want to know a different interval than 95%...
Lower Boundary
90% Confidence
Interval
99% Confidence
Interval
  x  1.645
  x  2.58

y

y
Upper Boundary
  x  1.645
  x  2.58

y

y
A sample survey of 81 movie theaters showed that the average length of the main
feature film was 90 minutes with a standard deviation of 20 minutes. Find a…
a) 90% confidence interval for the mean of the population.
  x  1.645

  x  1.645
y
20
  90  1.645
81

y
20
  90  1.645
81
The 90% confidence interval for μ is 86.34 to 93.66 minutes.
b) 95% confidence interval for the mean of the population.
  x  1.96

y
20
  90  1.96
81
  x  1.96
  90  1.96

y
20
81
The 95% confidence interval for μ is 85.64 to 94.36 minutes.
From the HW
P. 445
1) A sample of 49 banks found that the average charge for a money order was $2.75
for amounts up to $1000. The standard deviation was $0.25. Find a 90% confidence
interval for the average charge for a money order.
The 90% confidence interval is $2.69 to $2.81.
HW
P. 445 #1-8