Hi elshana!
These are the answers to your questions:
- What will be the population mean with 99% confidence with normal
population of 410 observations randomly picked, SD = 7.8 and Xbar =
23.7?
I will assume here that you want to define a 99% confidence interval
around the sample mean, which is 23.7. I will also assume that the SD
you provide is the population SD. If any of these assumptions is
wrong, please let me know through a clarification request.
Given that the population distribution is normal, this will make
things a bit easier, because we can use the result that if the
population is normal, then the sample mean will also follow a normal
distribution. In particular, we know that the sample mean (Xbar)
follows a normal distribution with mean equal to the population mean
('u') and SD equal to the population SD divided by the square root of
the sample size. That is, the SD of Xbar is 7.8/sqrt(410)=0.38. From
now on, let's call the sample mean X instead of Xbar.
Sampling Distribution of the Sample Mean
http://www.ms.uky.edu/~viele/sta417s00/sampdist/sampdist.html
In order to find this confidence interval, we must find a number 'a'
such that:
Prob(u < X-a) = 0.005
and
Prob(u > X+a) = 0.005
The intuition behind this is that we want to find 'a' such that the
probability that u is outside the interval [X-a,X+a] is 0.01. Now,
since X follows a normal distribution (approximately), which is
symmetric around its mean, it turns out that you can use either
equation to calculate 'a' and you will get the same result. Let's use
the first one and rearrange it:
Prob(u < X-a)
=Prob(X-u > a)
=Prob( (X-u)/0.38 > a/0.38 )
Now, the left hand side of the inequality is basically X minus its
mean divided its SD. Since we know that X is normally distributed,
this new variable ((X-u)/0.38) is also normally distributed; in
particular, it's standard normally distributed. This is so because if
you take a normally distributed random variable, substract its mean
and divide it by its SD, you get a standard normally distributed
random variable. That is, a normal distribution with mean 0 and
variance (and SD) 1. Let's call this new variable Z. We have then
that:
Prob( Z > a/0.38) = 0.005
In order to find 'a' from this equation we have to use a normal
distribution table. You can find one at:
z-Distribution Table
http://www.math2.org/math/stat/distributions/z-dist.htm
Looking up the table, we find that
Prob( Z > 2.57 ) = 0.005
Therefore,
a/0.38 = 2.57
a=2.57*0.38
a=0.97
Therefore, a 99% confidence interval for the population mean is
[23.7-0.97,23.7+0.97], or [22.73,24.67].
- In a normally distributed situation 20 cars were selected to find
the
number of km they have done so far. Sample mean = 236,500 and sample
SD = 23,000. What would be the mean value of all cars with 90%
confidence level in a normally distributed situation?
This question is very similar to the last one, but with a small twist.
We now have the *sample* SD instead of the population SD. The setup of
the problem is quite similar. We have to find 'a' such that:
Prob(u < X-a) = 0.05
and
Prob(u > X+a) = 0.05
(now it's 0.05 instead of 0.005 because here you ask for a 90%
confidence interval)
Rewriting as before, and using the fact that 23000/sqrt(20)=5142
Prob(u < X-a) = 0.05
=Prob(X-u > a) = 0.05
=Prob( (X-u)/5142 > a/5142 ) = 0.05
While it's true that u is the mean of X, it's not necessarily true
that 23,000 is the SD of X. It's the *sample* variance, not the
population variance. Fortunately, this equation can still be solved.
We use the fact that if we take a normally distributed random variable
(X), substract its mean (u) and divide it by the sample variance
divided sqrt(n); we get a random variable that follows the t
distribution with n-1 degrees of freedom (df). In this case, it's 19
df. Therefore, we have
Prob( T > a/5142) = 0.05
Now we have to look up a t-distribution table.
t Distribution Table
http://www.stat.ucla.edu/~dinov/courses_students.dir/Applets.dir/T-table.html
Looking up 0.05 at the 19 df row, we get the number 1.729. This means
that:
Prob( T > 1.729) = 0.05
Therefore,
a/5142=1.729
a=1.729*5142
a=8890.51
Thus a 90% confidence interval for the population mean is
[236500-8890.51,236500+8890.51] or [227609.49,245390.51]
Google search strategy
hypothesis testing
://www.google.com.ar/search?q=hypothesis+testing&ie=UTF-8&oe=UTF-8&hl=es&meta=
normal distribution table
://www.google.com.ar/search?hl=es&ie=UTF-8&oe=UTF-8&q=normal+distribution+table&meta=
t distribution table
://www.google.com.ar/search?hl=es&lr=&ie=UTF-8&oe=UTF-8&q=t+distribution+table&spell=1
I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.
Best wishes!
elmarto |
Clarification of Answer by
elmarto-ga
on
21 Oct 2003 06:17 PDT
Hi elshana!
I got the value from a normal distribution table in the following way.
Let's use the one I included the link to. This table gives the
probability of Z being less than a number 'a'. For example, if you
want to know the probability of Z being less than 3.82, you'd look up
the row '-3.8', column '0.02', and get the number from there. As you
can see, it's 0.00007. The question here is different. What's the
number 'a' such that
Prob(Z > a) = 0.005?
Since the table gives us the probability of Z being *less* than
something, we use the symmetry property of the standard normal
distribution to rewrite this as
Prob(Z > a) = Prob(Z < -a) = 0.005
And now, in order to find -a, we find the number 0.005. As you can
see, it's not in the table, but it's very close to 0.00508, which is
in the table. Now, 0.00508 is at the intersection of row '-2.5' and
'.07', indicating that
Prob(Z < -2.57) = 0.00508 ~=0.005
Therefore, -a=-2.57, so a=2.57.
Best wishes!
elmarto
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