sample of 50 items sel. The mean value of the sampled items was found to = $542.50 with a sample stand. dev. of $120.56.(a) what is point estimator...

a) The point estimator of the poulation mean M is the
sample mean m itself.Therefore, $542.50 is the estimated value of the population mean by
point estimation.

b)

If the
sample of size n has a standrd devition of s, then the sample mean ha the standard
deviation of s/sqrt(n).

We know that z = |M-m|/(s/sqrtn) is
a standard normal variate.

Therefore probability that the
estimated value of M is within  limits is 95% confidetial limits  is  as good as
P(z <= |(M-m)|/ (s/sqrtn)  <  t ) =  95% =
0.95.

Therefore from tables t =
2.74.

Threfore |(M-542.5)|/(120.56/sqrt50) <
2.74.

 542.5 - 2.74(120.56/sqrt50) < = M  < =
542+ 2.74 (120/sqrt50)

542.50-46.72 <=  M <=
542.5+46.72

496.78 <=  M <=
589.22.

Therefore the population mean leis between 496.78 
and 589.78 with 95%
confidence.

c)

Similar to the
b.

P(z = |M-m|/(s/sqrtn) <
0.90

P(z = {|M-542.5|/120.52/sqrt50) < x } =
0.90

x = 1.645 from normal distribution
tables.

Therefore  542.5- (1.645)(120.52/sqrt50) < M
< 542.5 + (1.645)(120.52/sqrt50)

 542.5 -28.04
<= M < = 542.5 +28.04

514 .46 <=  M
<= 570.54.

Therefore the 90% confidence interval for
the population mean M is 514.46 to 570.54.