Transcript Week 9, Lecture 1, Revising interval estimation

Business Statistics - QBM117
Revising interval estimation
Objective

To develop confidence in identifying the correct formula to
use when calculating an interval estimate.
Identifying the parameter to be estimated

In this subject we need only concern ourselves with
estimating the population mean , or the population
proportion p.

Generally it is clear from the question, which parameter we
are estimating, if we read it carefully enough.
Two confidence interval estimators of 
There are two different interval estimators of the
population mean, and the basis for determining which
method is appropriate is quite simple.

If  is known, the confidence interval estimator of the
population mean  is
x  Z / 2


n
If  is unknown and the population is normally distributed,
the confidence interval estimator of the population mean  is
x  t / 2,n 1
s
n
When d.f > 200 we approximate t by the t / 2 , value.
Only one confidence interval estimator of p

There are is only one confidence interval estimator of the
population proportion so there is no choice.
pˆ  Z / 2

pˆ qˆ
n
One condition however must be satisfied before it is
appropriate to use this formula
both npˆ  5 and nqˆ  5
Exercises relating to interval estimation
Example 1
a.
Assume that the time it takes to assemble the parts of an electric
lamp is normally distributed. Construct a 95% confidence interval
for the average assembly time if a random sample of 16 trials has
an average assembly time of 400 seconds with a standard
deviation of 10 seconds.
b. Is it reasonable to suppose that the average assembly time will not
exceed 410 seconds?
c. How large a sample would be required if we want to estimate the
average assembly time to within 2 seconds with 99% confidence?
Example 1
Does the question ask us to estimate a mean
or a proportion?
mean
Do we know the population standard
deviation,  or do we only have the sample
standard deviation s?
S
x  t / 2,n 1
s
n
s  10
n  16 x  400
1    0.95   0.05  / 2  0.025 t.025,15  2.131
95%CI (  )  x  t0.025,15
s
n
 10 
 400  ( 2.131)

 16 
 400  5.33
The average assembly time is between
394.67 secs and 405.33 secs
C.
s  10
B 2
 z / 2 s 
n

 B 
z0.005  2.575
2
 (2.575)(10) 


2


 165.77
2
 166
Therefore 166 times should be sampled.
Example 2
a.
b.
c.
The daily business at a local restaurant is known
to be normally distributed with a standard
deviation of $571. Find a 95% confidence
interval for the mean daily income, if a random
sample of 17 days' receipts shows an average
daily income of $9832.
Would it be reasonable for the restaurant owner
to claim to the taxation department that her
average daily income is $9400?
How large a sample would be required if we
want to estimate the mean daily income to within
$50 with 90% confidence?
Example 2
Does the question ask us to estimate a mean
or a proportion?
mean
Do we know the population standard
deviation,  or do we only have the sample
standard deviation s?

x  Z / 2

n
  571 n  17
x  $9832
1    0.95   0.05  / 2  0.025 Z ..025

025 1.96
95%CI (  )  x  Z 0.025

n
 571 
 9832  (1.96)

 17 
 9832  271.4
The average daily income is between
$9560.60 and $10 103.40
C.
  571
B  50
 z / 2 
n

 B 
z0.05  1.645
2
 (1.645)(571) 


50


 352.9
2
 353
Therefore 353 businesses should be sampled.
Example 3
a. An interstate savings institution claims that 94% of its depositors
have both a savings account and a checking account with the
institution. Suppose a random sample of 250 customers shows that
226 have both a saving account and a checking account, whereas
the other 24 have either only a savings account or a checking
account, not both. Use this sample to construct a 90% confidence
interval for the true proportion of depositors who have both types
of accounts with this institution.
b. What size sample is required to estimate the proportion of
depositors with both types of accounts to within 0.1 with 90%
confidence?
Example 3
Does the question ask us to estimate a
mean or a proportion?
proportion
226
24
pˆ 
 0.904 qˆ 
 0.096
n  250
250
250
1    0.90   0.1  / 2  0.05 Z .05  1.645
npˆ  226  5
nqˆ  24  5
90%CI ( p )  pˆ  Z 0.05
pˆ qˆ
n
 (0.904)(0.096) 

 0.904  (1.645)

250


 0.966  0.031
The proportion of depositors with both types of
accounts is between 93.5% and 99.7%
C.
pˆ  0.904 qˆ  0.096
 z / 2 pˆ qˆ 

n
 B 


B  0.1
z0.05  1.645
2
 (1.645) (0.904)(0.096) 




0
.
1


 551.5
 552
2
Therefore 552 customers should be sampled.
Reading for next lecture
Chapter 10 Sections 10.1 - 10.2