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Stat 31
January 30, 2006
Stat 31 – estimating and testing means
Assume a population of numerical values with mean  and standard deviation , and assume a
random sample of size n consisting of values x1, x2, …, xn with statistics
n
x
x1 
 xn
n
and s 
x  x 
i 1
n 1
2
.
In a collection of such samples, the values of the sample mean x are roughly normally
distributed with mean  and standard deviation (called standard error)
SE = 
.
n
Therefore, x is an unbiased estimator of . (For the sampling distribution to be normal
requires that n be large --- conventionally, n ≥ 30 --- or that the population values themselves
be normal.) It follows that
x 
Z
 n
has a standard normal distribution. Alternatively, if  is not known, one may substitute the
estimator s for  to obtain the statistic
x 
T
s n
which (under strong assumptions) has a t distribution with n – 1 degrees of freedom. When n
≥ 30 this is very close to the standard normal distribution; in this case s is a very good
estimator of  and the T statistic is very similar to the Z statistic.
Confidence intervals for  may be constructed as


or
x   t / 2,n1   s
x   z / 2   


n
n




where z / 2 is such that the fraction /2 of the standard normal distribution falls above z / 2 .
(Here, 1 –  is the confidence level; for confidence 95% choose  = 0.05 and z / 2  1.9600 .)
The value t / 2, n 1 plays the same role for the t distribution.
To test H0:  = 0 vs. HA:  ≠ 0, compute
x
x
( or
)
Z
T
/ n
s/ n
and reject H0 if Z > z / 2 or Z < z / 2 (or for T use t / 2, n 1 ).
To test H0:  = 0 vs. HA:  ≠ 0 (for some conjectured value 0), use the test statistics
x  0
x  0
( or T 
).
Z
s/ n
/ n
(end)
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