Telemarketing: Cold calls. Teachers make good sales reps

Sherry Kelley

P.O. Box 2917

Silver Spring, MD 20915

240-277-3698

301-587-6595(Fax-use my name)

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EDUCATION

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Columbia University, School of Engineering & Applied Sciences

MS Operations Research(1982)

University of San Francisco, School of Business

BS Business Economics(1979)

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COLLEGE TEACHING EXPERIENCE-Adjunct(82-87)

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Sherry Kelley

P.O. Box 2917

Silver Spring, MD 20915

4/8/04

Mall Manager

Ms. Marion Julier

Lake Forest Mall

701 Russell Ave

Gaithersburg,MD 20877

Dear Ms. Julier:

You may find it beneficial to have more than one fast food chain provide breakfast in the mornings

(McDonald's has been rude...).

I've found myself recommending Lake Forest

Mall to clients at work(I am in sales - used

to teach MBA courses). Rude employees who

try to bribe me with extra tea-bags when I

speak up) have dampened that enthusiasm somewhat.

I thank you in advance for your kind attention,

Sincerely,

Sherry Kelley

BUSINESS STATISTICS TUTOR

I can tutor you in business statistics

I used to teach MBA courses

(management science)

Send your questions to

ecommerceaid1@lycos.com

The default value goes here.

I tutor in business statistics. I used to teach college level business maths. Send your questions to ecommerceaid1@lycos.com

EXAMPLE1- ESTIMATING THE VARIANCE CONFIDENCE INTERVAL - NORMAL PUPULATION.

Want to estimate the variance of your customers' ages.
Select a random sample of 20 customers.
Compute a sample variance of 30 years.
This translates into a sample standard deviation of 5.5 years.
You can contruct your confidence interval as follows:
1. Choose a confidence level for your test, such as 95%.
2. Look up the appropriate values in the chi- square distribution table. Chi-square of 0.025 and Chi-square of 0.975.
For a chi-square distribution with 19 degrees of freedom(20-1), these values are 32.852 and 8.907.
3. Compute the confidence interval using the following formula:[(n-1)sigma squared]/[table chi-squared].
Your result is: [(20-1)30]/(32.852) to [(20-1)30]/(8.907)= 17 to 64.
Therefore can be 95% confident that the variance of your customers' ages is between 17 and 64 years.
This translates into a confidence interval of 4 to 8 years for the population standard deviation.
EXAMPLE2: TESTING A VARIANCE - NORMAL POPULATION.
Want: To compare the dispersion in ages of your competitor's customers with yours. Historically, the standard deviation in your customer's ages has been 6 years. The corresponding variance that you use in the test is 36 years.
A sample of 15 of your competitor's customers yields a sample standard deviation of 8 years and a variance of 64 years.
Perform the test as follows:
1. Form your null & alternate hypothesis.
For this test, your null hypothesis would be that the variance in ages is equal for your customers and those of your competitor. The alternate hypothesis is that the variance is greater for your competitor's customers.
2. Choose a confidence interval, such as 95% for your test.
3. Since your alternate hypothesis is that your competitor's variance is greater than yours your test is one-tailed. Look up in the chi-square distribution table the value that has 5% of the curve above it. This value for 14 degrees of freedom(15-1) is (chi-square of 0.05)= 23.685.
4. Use Formula 5.5 to compute the test-statistic:
Chi-square ={[(15 - 1)64]/36}= 24.89.
Since your test statistic is greater than the upper limit of a confience interval, you reject the null hypothesis + infer that the ages of your competitor's customers are more dispersed than the ages of your customers.