# stuck need help

#### mcafeep

##### New Member
THIS IS THE QUESTION
PiggyBank wants to ESTIMATE the MEAN DOLLARS that each card holder will spend EACH MONTH. It would like to be within PLUS OR MINUS $10 of the TRUE MEAN with a 98% CONFIDENCE LEVEL. The standard deviation is thought to be$500. How many card holders should be sampled? After you’ve determined how many card holders should be sampled, PiggyBank comes back and says that it will cost $5 per sample and they were only planning on spending$10,000 on the sample. In a MEMO to the PiggyBank product development team indicate how many card holders should be sampled to meet the original requirements of the sample. Then explain the TRADE-OFFS that will occur when you lower the sample to $10,000 to meet their budget. THIS IS WHAT HAVE SO FAR Step 1) Calculate alpha: 98% = (1- α) * 100% (1-α) = 98%/100% = 0.98 -(1 –α) = -0.98 α -1 = -0.98 α = 1-0.98= 0.02 α = 0.02 Step 2) α/2 = 0.02/2 = 0.01 TABLE II THE Z-SCORE AREA CLOSEST to 0.01 =2.33 Z-score= 2.33 Step 3) n = [(Z-score * σ) ÷ E]^2 n = [(2.33*30) ÷10]^2 = 48.8601 n=49 “n” is the sample size we are determining 49 “Z-score” is the score we found from TABLE II or Excel sheet – 2.326 “σ “ is the specified standard deviation of$30(-this is where I get lost)
“E” is the specified error of margin of $10 “^2” means to multiply [(Z-score * σ) ÷ E]* [(Z-score * σ) ÷ E] 3) After you’ve determined how many card holders should be sampled, PiggyBank comes back and says that it will cost$5 per sample and they were only planning on spending $10,000 on the sample. In an OFFICIAL MEMO to the PiggyBank product development team indicate how many card holders should be sampled to meet the original requirements of the sample 49*5=$245

INSTRUCTOR NOTES: In this assignment we are assuming that the population of card holders has a symmetric bell-shaped relative frequency histogram of the random variable “card holder mean dollars spent per month “OR” the survey sample size “n” is ≥ 30. In essence the variable “card holder mean dollars spent per month” is NORMALLY DISTRIBUTED

#### JohnM

##### TS Contributor
The formula for determining sample size required, without taking into account statistical power, is:

n >= Z^2 * (s^2/d^2)

For 98% confidence, Z = 2.326
d = 10
s = 500

You will get n >= 13,526. At $5 per sample, this comes to$67,630.

If they have a budget limit of \$10,000, they will only be able to sample n = 2000.

Plug 2000 into n and 500 into s, and then play around with various values of d and Z to be able to discuss trade-offs:

(1) a lower confience level (i.e., lower than 98%)
(2) a wider margin of error (i.e., larger than 10)