# Business mats

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

grading.You

must use comments and text withinthe

spreadsheet to explain your responses. Please print out your assignment to

ensure that it meets the formatting requirements and is

readable.

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?

b. What is your answer in (a) if X

(bar) = 37 minutes and s = 27 minutes?

c. Look at your answers for a and b

above and discuss what you can learn from the results about the effect of a

large standard deviation.

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4.Peter’s NEW IT

Help company is concerned that the mean wait time of their phone customers for

a customer service agent is not greater than 15 minutes. It can be assumed that

the population variance is 9 minutes 6 seconds based on past experience. A

sample of 563 customers is selected and the sample mean is 16 minutes 30

seconds. Using a level of significance of .10, is there evidence that the

population mean wait time is greater than 15 minutes? Fully explain your

answer.

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