In general, a confidence interval is given by:
 [sample statistic  Table value like Z/2 * SE],
where SE is the standard deviation of the
sampling distribution of the statistic.
If n>30 or if  is known and the Population being
sampled is normal, a (l) Confidence interval for
the population mean is given by
   x    z /2 ( /  n)
If  is unknown and n>30, sample standard de. s
can be used as an approximation for 
n=100 and x =425,  =900,        99% Confidence
Interval is 425 ± 2.58 * (900/10)
or 192.8 to 657.2 or P(192.8  657.2)=0.99
Interval Estimation of the Population Mean for a
Normal Population with  Unknown
If n 30 ,,  is unknown and the sample is
drawn from a normal population, a (1)
confidence interval for the population mean is
given by
   x    t /2, n1 [s /  n]
where t /2, n1 is the critical value for a t-
distribution with n-1 degrees of freedom which
captures an area of ( / 2) in the right tail of the
t distribution. (t has FATTER tails than
normal). The sample mean and standard
deviation are 29.86 and 7.08, degrees of
freedom associated with the problem is
d.f.= n1 = 71 = 6.
The t-value corresponding to 6 degrees of
freedom and 95% confidence is given in the
table as (t 0.025, 6 )=2.447. The corresponding
confidence interval is
29.86 ± 2.447 * [7.08/  7 ]
29.86 ± 6.55.

             Precision and Sample Size
The sample size should be selected in relation to
the size of the maximum Positive or negative
error the decision maker is willing to accept.
This can be achieved by setting the error equal
to W/2 or one-half the confidence interval
width, Error = (z /2 ) /  n
This equation can be solved for the sample size,
 n = [ (z /2 ) / Error ] 2
(Remember, always round up when finding n).

Estimating Population Attributes (e.g.
Estimate the fraction of defectives from a
sample of size n=800. If X=5 defectives are
observed, then p = (5/800)= 0.00625

Estimation of an Interval for Population Attribute
In order to develop the confidence interval for a
population proportion the sampling distribution
of the point estimate must be developed. The
random variable p has a binomial distribution,
which is approximated with a normal random
The sample proportion, p is distributed
normally with mean=p, and variance=[p(1p)/n]
The SE or the standard deviation of the sample
proportion p is denoted symbolically as  p
             ˆ                               ˆ

 p = [ p(1p) / n ] or approx = [ p (1 p ) / n ]
  ˆ                                  ˆ     ˆ

 If the sample size n is sufficiently large, both
 np 5 and n(1 p) 5, then 1- confidence
 interval for the population proportion is
 given by the expression p  z/2  p
                             ˆ         ˆ

where z/2 is the distance from the point
estimate to the end of the interval in standard
deviation units, and  p is the standard deviation

of p or standard error (SE).
This is a special case of statistic  Z/2 * SE]

Precision and Sample Size for Population
If W is the width of the interval, W/2 should
obviously equal the maximum allowable error.
Hence Error = z/2*  p = z/2* [ p(1p) /n]

this is solved for n in the formula below.
How large a sample would be required to
estimate the proportion of buyers with an
accuracy of 0.002 and a 95% degree of
confidence if the true proportion is
approximately 0.008? Use the formula:
n = (z/2 )2 p ( 1- p) / Error 2

p =0.008, z/2 = 1.96 for 95% confidence, and
error = 0.002. The above formula yields:
n=(1.96)2 (0.008)(10.008)/ (0.002)2 =7662
(Remember, always round up when finding n).
Thus, to be 95% confident that the proportion is
estimated with an error of at most 0.002 requires
a sample size of 7,622.

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