Download QBM117 - Business Statistics

QBM117 - Business Statistics
Estimating the population mean , when the
population variance 2, is unknown
Estimating the population mean  when the
population variance 2 is unknown

In reality, if we do not know the population mean , it is
unlikely that we will know the population standard
deviation .

Therefore we use the sample standard deviation, s, to
estimate the population standard deviation  and hence the
standard error s to estimate 
x

However
x
x   is not normally distributed.
s/ n
W.S Gosset showed that
x   has a particular
s/ n
distribution called the student t distribution or simply the
t distribution when the population from which the
sample is drawn is normal.
x
t
is called the t statistic
s/ n
What if the population from which we are
sampling is not normal?

The t distribution is said to be.robust. This means that the t
distribution also provides an adequate approximate
sampling distribution of the t statistic for moderately nonnormal populations.

In actual practice, we should draw the histogram of any
random variable that you are assuming is normal, to ensure
that the assumption is not badly violated.

If the assumption is not satisfied at all, due to extreme
skewness, we have two options:
• transform the data (perhaps with logarithms) to bring about a
normal distribution, or
• use non parametric methods (studied in QBM217)
What do we know about the t distribution?







It looks very much like the standard normal probability
density function, but with fatter tails and slightly more
rounded peaks.
It is more widely dispersed than the normal probability
density function.
The graph of the t probability density function changes for
different sample sizes.
The t statistic has n - 1 degrees of freedom.
The similarity between the t pdf and the standard normal
pdf increases rapidly, as the degrees of freedom for the t
pdf increases.
The two distributions are virtually indistinguishable when
the degrees of freedom exceed 30.
The values for t / 2, are identical to the corresponding Z / 2
Estimating the population mean  when the
population variance 2 is unknown
The (1-α)100% confidence interval for µ is given by
x  t / 2.n 1
s
n
where
x
is the sample mean
t / 2.n 1
is the value of t for the given level of
confidence (S&S Table 4 in appendix)
s
n
is the standard deviation of the sample mean,
known as the standard error
Example 1 – Exercise 8.14 p264



Here we want to estimate the population mean .
The sample mean x is the best estimator of .
x
We have sampled from a normal population therefore,
s/ n
will follow the t distribution.
x  27.3 n  75 s  7.8
1    0.99   0.01 t0.005,74  t0.005,70  2.648
Therefore the confidence interval is given by
x  t / 2,n 1
s
7.8
 27.3  (2.648)(
)
n
75
 27.3  2.38
 99% CI ( )  24.92 to 29.68
Two confidence interval estimators of 
We now have two different interval estimators of the
population mean. The basis for determining which
interval estimator to use 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 the degrees of freedom exceed 200, we approximate
the required t statistic by the t / 2, value.
Example 2
A foreman in a manufacturing plant wishes to estimate the
average amount of time it takes a worker to assemble a
certain device. He randomly selects 81 workers and
discovers that they take an average of 29 minutes with a
standard deviation of 4.5 minutes. Assuming the times are
normally distributed, find a 90% confidence interval estimate
for the average amount of time it takes the workers in this
plant to assemble the device. What can you report to the
foreman?
Example 1
A foreman in a manufacturing plant wishes to estimate the
average amount of time it takes a worker to assemble a
certain device. He randomly selects 81 workers and
discovers that they take an average of 29 minutes with a
standard deviation of 4.5 minutes. Assuming the times are
normally distributed, find a 90% confidence interval estimate
for the average amount of time it takes the workers in this
plant to assemble the device. What can you report to the
foreman?
Since  is unknown and the population is normally
distributed, the confidence interval estimator of the
population mean  is
x  t / 2,n 1
s
n
X  29 s  4.5 n  81
1    0.9   0.1 t / 2,n1  t.05,80  1.664
Therefore the confidence interval is given by
x  t / 2,n 1
s
4.5
 29  (1.664)(
)
n
81
 29  0.83
We are 90% confident that the mean assembly time
lies between 28.17 and 29.83 minutes.
Example 3

A random sample of 26 airline passengers at the local
airport showed that the mean time spent waiting in line to
check in at the ticket counter was 21 minutes with a
standard deviation of 5 minutes.

Construct a 99% confidence interval for the mean time
spent waiting in line by all passengers at this airport.

Assume the waiting times for all passengers are normally
distributed.
Reading for next lecture

S&S Chapter 8 Sections 8.5 - 8.7
Exercises to be completed before next lecture

S&S 8.21
8.23