October 22, 2018

This week
you answer a second series of problems on statistical inference. You will use
Excel to solve the statistics problems. It is EXTREMELY important that you
prepare a written discussion of each problem to accompany the numerical
calculations. You should attach the Excel file in your Assignment folder.

Please make sure that you use only one
Workbook for your response. You can create several spreadsheets or tabs under
the one Workbook. This will allow me to download one file with all your
responses. It would also help if you could make sure your response is contained
within the 8.5 by 11 layout. This means that none of your response will cross
over several pages. You can also highlight your final answer to assist with the
must use comments and text withinthe
ensure that it meets the formatting requirements and is

Note forproblem 1:

In working with a standard normal distribution there are several variables and
statistics that you will use. You always start with an underlying distribution
of a random variable that you assume is a standard normal distribution. In
problem 1 of the assignment this distribution is the distribution of the annual
return of common stocks from 1936 to 2010. A standard normal distribution is
summarized by its mean and standard deviation.
To show you how to tackle this problem, I
will use slightly different numbers: Mean and SD of 12.4% and 20.6%,
respectively and probability of more than 10%. Assuming a standard normal
distribution for these annual returns for common stocks, you would like to know
what the probability would be if an actual return for common stocks between
1936 and 2010 was greater than or less than a particular return. You must
convert the 10% into a “z statistic” using the mean and standard
deviation of the actual distribution (12.4% and 20.6%). After you calculate the
“z” value equivalent to the 10% return, you go to a table for a
cumulative standard normal distribution. The table provides values that
represent the area under the standard normal distribution curve for specific
values of “z.” Half of the area of a standard normal distribution is
above “z” = 0 and half is below “z” =0. Therefore, the
probability of an actual return being above the mean is 50% and the probability
of an actual return being below the mean is 50%. The value in the standard
normal distribution table at the web site below gives you the probability
between 0 and your z-value. Remember, the mean of the standard normal
distribution is 0. This is equivalent to the mean for the distribution of
common stock returns, which is 12.4%.

Here’s an example. Assume that you want to know the probability that the actual
return could be less than 33%. The z-value would be 1 [(33%-12.4%)/20.6%]. You
look for the value in the table where z = 1.00, which is .3413. This is the
probability that the actual return will be between 12.4% (z=0) and 33% (z=1),
which is the shaded area on the picture at the top of the table. You are
looking for the area under the curve to the left of z=1 (33%). To calculate
this, you can take the value in the table, which is .3413 and add it to .5000
(this is area under the curve to the left of z =0. The result is .8413 or
84.13%. This is the probability that the actual return will be less than 33%.

Here’s another example of using the z-table in connection with problem 1. Let’s
assume that you want to find the probability that the actual return is greater
than 5%, you must first calculate the z value for 5%. This is (5% –
12.4%)/20.6%, which is equal to -.3592 or -.36. The negative number means that
the z value is below the mean (z = 0 is the mean for a standard normal
distribution). This makes sense because the expected return (mean) is 12.4%,
and 5% is less than this. The z-table that we are using shows probability
values between the z value calculated based on the return given in the problem
and z=0. This is the shaded area in the picture. The value in the table for z =
.36 is .1406. This means that the probability that an actual return will be
between 5% (z = -.36) and 12.4% (z = 0) is 14.06% (to change a value from
decimal to percent you multiply by 100; to change a percent to a decimal you
divide by 100). You want to find the probability that the actual return will be
greater than 5%. This the entire area under the standard normal curve to the
right of z = -.36. This is the sum of the areas from z = -.36 to z = 0 and the
area to the right of z = 0. You add the value given in the table, .1406, and
.50, which is the value for the area to the right of z = 0. The answer is .1406
+ .50 = .6406 or 64.06%. The probability that the actual return is greater than
5 % is 64.06%
Probability
Problem Set #2
1. Suppose that
the mean of the annual return for common stocks from 2000 to 2012 was 14.37%,
and the standard deviation of the annual return was 35.14%. Suppose also that
during the same 12-year time span, the mean of the annual return for long-term
government bonds was 0.6%, and the standard deviation was 2.1%. The
distributions of annual returns for both common stocks and long-term government
bonds are bell-shaped and approximately symmetric in thisscenario.
Assume that these distributions are distributed as normal random variables with
the means and standard deviations given previously.
a. Find the probability that the return for
common stocks will be greater than 16.32%.
b.
Find the probability that the return for
common stocks will be greater than 5.89%.
c. Find the probability that the return for
common stocks will be less than 14.37%.
Hint: There are many ways to attack
this problem in the HW. If you would like the normal distribution table so you
can draw the pictures (my preferred way of learning) then I suggest you
bookmark this site:
http://www.statsoft.com/textbook/sttable.html
Confidence Interval Estimation
2.Compute a 95%
confidence interval for the population mean, based on the sample 1.5, 1.54,
1.55, 1.51, 0.09, 0.08, 1.55, 0.07, 0.99, 0.98, 1.12, 1.13, 1.00, 1.56, and
1.53. Change the last number from 1.53 to 50 and recalculate the confidence
interval. Using the results, describe the effect of an outlier or extreme value
on the confidence interval.
Hypothesis Testing
3.The management
of the Ceebler Fairy Corporation is considering relocating the corporate office
to a new location outside HisWood Forest. Management is concerned that the
commute times of the employees to the new office might be too long. The company
decides to survey a sample of employees at other companies in the same office
forest to see how long these employees are commuting to the office. A sample of
23 employees indicated that the employees are commuting X (bar) = 33 minutes
and s = 1 minute, 45 seconds.
a. Using the 0.01 level of
significance, is there evidence that the population mean is above 32 minutes?
(bar) = 37 minutes and s = 27 minutes?