Note: if we use to estimate and = 10, then is the estimator whereas the estimate is 10.
When we choose an estimator to be used to make a point estimate of a particular population parameter, all we can do is compare the sampling distributions of the various estimators.
for example: sample mean, sample median, etc.
Such a comparison will show which estimators are likely to depart considerably from the population parameter.
We use 3 criteria:
thus, is an unbiased estimator because E() =
the variance of the sample mean is ^{2} / n
if the population is normal, the variance of the sampling distribution of the sample median is 1.57^{2}
for example, for > = / n^{1/2} > 0 as n >
Note: define P{X > 32} as probability that X is greater than 32. Now define P{a < X < b} as the probability that X lies between a and b. Thus, the probability that the value of the sample mean lies between
 1.96 / n^{1/2} and + 1.96 / n^{1/2}
is denote by
P{  1.96 / n^{1/2} < < + 1.96 / n^{1/2}}
thus from what we know about the sampling distribution of the sample mean implies:
P{  1.96 / n^{1/2} < < + 1.96 / n^{1/2}} = 0.95
because the probability that any normal random variable will lie within 1.96 standard deviations of its mean is 0.95.
To construct an interval estimate for the population mean :
P{ 1.96 / n^{1/2} <  < 1.96 / n^{1/2}} = 0.95
P{  1.96 / n^{1/2} < < + 1.96 / n^{1/2}} = 0.95
This means that before the sample is drawn, when is unknown, there is a 0.95 probability that the interval will include the population mean .
but
Example 1
= 10 it would be incorrect to say P{2.00  1.96 * 10 / n^{1/2} < < 2.00 + 1.96 * 10 / n^{1/2}} = 0.95 P{0.04 < < 3.96} = 0.95 Since is a constant, it is either in this interval or it is not.

All one can say is that if intervals of this sort are calculated repeatedly, they will include the population mean in about 95% of the cases.
Thus, what we have is a confidence interval with confidence coefficient (1  ). In the above example (1  ) = 0.95
or
= 0.05
Where Z_{a/2} is the value of the standard normal variable that is exceeded with a probability of / 2.
Example 2
Suppose we wanted a 90% C.I. instead of 95%. (1  ) = 0.90 so / 2 = 0.05 from the Standard Normal Table > Z_{0.05} = 1.64 so 2.00 1.64 (10 / (100)^{1/2}) > C.I. [0.36 < < 3.64] = 0.90

Just use the sample standard deviation, s, as an estimator for .
so we get
P{{  1.96 s / n^{1/2} < < + 1.96 s / n^{1/2}} = 0.95
or
Example 3
n = 90 A 99% confidence interval would be: Z_{a/2} = Z_{0.005} = 2.576 810  2.579 (85 / (90)^{1/2} < < 810 + 2.579 (85 / (90)^{1/2} 810  2.579 (8.96) < < 810 + 2.579 (8.96) 786.92 < < 833.08

If both samples are large, and if the confidence coefficient is set equal to (1  ), the confidence interval for the difference between the population means (_{1}  _{2}) is:
where
s^{2}_{1} is the variance of the first sample and
s^{2}_{2} is the variance of the second sample.
Example 4
A Soup Company has two plants and suspects that the mean drained weight of the contents of a can is higher at plant 1 than plant 2. To test this they draw a random sample of 100 cans from each plant and construct a 90% confidence interval for difference of means.
(1  ) = 0.90 thus, = 0.10 and /2 = 0.05 Z_{0.05} = 1.64 so
so the 90% confidence interval is 0.026 to 0.354 oz.

has the tdistribution. The tdistribution is symmetrical, bellshaped and has zero as its mean.
The tdistribution is a sampling distribution of the statistic
It is a family of distributions, each of which corresponds to a particular number of degrees of freedom. In this context, the number of degrees of freedom equals (n  1).
where t_{a/2} is the value of a t variable (with n1 degrees of freedom) that is exceeded with a probability of /2.
Example 5
n = 16 = 20 s = 4 a 95% C.I. is given by: (1  ) = 0.95 thus, = 0.05 and /2 = 0.025 from Table A6 with degrees of freedom = n1 = 161 = 15 we get a tvalue of 2.131. so
The 95% confidence interval is 17.869 to 22.131

Example 6
Mean length of the life of a light bulb? Test 9.
Calculate a 90% confidence interval. solution First calculate the sample mean and standard deviation.
because n = 9 use
(1  ) = 0.90 thus, = 0.10 and /2 = 0.05 Look up t_{0.05} with 8 degrees of freedom from Table A6. t_{0.05} = 1.86 so
or 5,107 to 5,293 hours.
