I have several stat questions. Step by step complete answers would be helpful.

1. A mailing was sent to 20,000 people in a Donation database with a
target response rate of 4%. What is the actual response rate if 989 of
those people donate and what is the statistical significance?

2. A manaufacturing company is considered "in control" of the
long-term mean weight of components produced is .20 kil, even though
individual components may vary from this mean.Here are the weights of
the of a random sample of the recently produced components:
     .253, .240, .247, .0183, .247, .223, .252, .195, .235, .241, .251,
    .261,.194, .236, .256, and .241.

Does this process seem to be in control? Justify answer. 

Thank you.

Problem 1:

In this problem, 989/20000 people responded, which implies a response
rate of .04945, or 4.945%.  Since it would be foolish to assume that
this is exactly correct, a confidence interval around this value
can/should be obtained.  In general, this interval has the form:

p+- 1.96*sqrt(p*(1-p)/N)  It should be noted here that 1.96 comes from
a significance level of .05.  Since you didn't specifiy what level to
use, I chose this one as it is the most common.  If you desire another
significance level, you can substitute 1.96 with another value from a
z-table.

Plugging in the values in this forumla gives:

.04945 + or - 1.96*sqrt(.04945*.95055)/20000) = 
.04945 + or - .0030.

This means we are 95% confident that the actual response rate is
somewhere between .04915 and .04975.  Since .04 is not within this
interval, this observed difference is statistically significant.  In
other words the hypothesis that the actual rate is 4% would be
rejected, using a significance level of .05.

Problem 2:

Here, the actual average is .2244.  We want to test the hypothesis
that this is not statistically different from 0.

H0: (null hypothesis) mu=.2
H1: mu does not equal .2.

This can be tested using a standard t-test, however, we'll first need
to obtain  an estimate of the standard deviation of the group.  This
estimate can obtained by summing the squared differences of each
measurement from the actual average, and dividing by the degrees of
freedom.  The formula is:

s^2=[sum(x-.2244)^2]/15  There are 16 measurements, and 16-1 degrees of freedom.
s^2 (variance) = .00343

Which means s=..05828.

The formula for the test statistic is:

t= (x-bar - mu)/(s/root(n)) where x-bar is the average, mu is .2, s we
just calculated and n is 16.

Plugging everything in gives:
t=(.2244-.2)/(.05828/root(16)) = 1.67

From a standard statistical table, you can obtain the proper t-value
to compare this result with, depending on the desired significance
level.

For a=.05 (the standard), and 15 degrees fo freedom, t=2.13.

Since 1.67<2.13, the null hypothesis cannot be rejected.  This means,
we have no evidence that the process is out of control.

Hope this helps!