Mean Value of Credit Card Expenditure - Essay Example

It will have to have an error of (+ or -) $10 with a reduced confidence level of only 67%. This shows that the estimate that could be obtained within the existing budget is not sufficient to prepare accurate revenue forecasts. For fairly accurate revenue forecasts, accuracy in the survey of $10 error and a confidence level of 95% can be considered adequate. For this, a minimum of 9,600 cardholders would have to be sampled. Therefore it is recommended to increase the budget to $48,000 which would give the above confidence level and error. If this is not feasible, refer to the below table which lists all other possible ways forward. The next best alternative would be to upgrade the budget by $5,000 and gain a survey result with an error of (+ or -) $15 and a confidence level of 90%.
Calculations
Assuming that the true mean value of the total number of cardholders of Piggy bank is  and the standard deviation is given $ 500.
The sample should contain n cardholders to give an estimated mean value that would lie between ( – 10) and (+10 ) with a confidence level of 98%.
The amount each cardholder spends in a month can be considered a normal distribution.
X ~ N( , σ2)
The Mean of the sample also is a normal distribution
X ~ N( , σ2 /n )
It is given that the standard deviation is $ 500. Therefore σ =500
By transforming this to standard normal distribution
X ~ N( , 5002 /n )
Z = (X-) / ( σ/√n) ~ N(0, 12)
If the confidence level expected is 98%, 98% of the distribution must lie between Z=–2.3263 and Z=+ 2.3263 of the standard normal curve. (From the table Percentage points of the normal distribution. for p = 0.01, z=2.3263) p= (100-98/100)/2
In other words for a confidence level of 98% (for the sample mean lying within a limit of (+ or-) $10 with a probability of 0.98)
(–2.3263) < Z<(+ 2.3263)
i.e (–2.3263) < (Ẍ-) / ( σ/√n) <(+ 2.3263)
as Z = (Ẍ-) / ( σ/√n)
because of the symmetry of the distribution let us consider the positive sides only
(Ẍ-) / ( σ/√n) <(+ 2.3263)
(Ẍ-) < (+ 2.3263) ( σ/√n)
(It is given that the standard deviation is $ 500. Therefore σ =500)
If the estimated value is to lie between (+ or -) $10, (-10) < (Ẍ- ) <(+10)
(Ẍ- ) <(+10) This is the tolerable error and let us call it ε
(Ẍ-) < (+ 2.3263) ( σ/√n)………………(i)
(Ẍ- ) <(+10) ……………………………….(ii)
2.3263 ( 500/√n) =10
n = 13529 say (13500)
z( 500/√n) = ε
√n = (500 x z)/ ε
By finding the corresponding z values for different p (confidence levels) the tolerable error values and the number of cardholders to be sampled are calculated and tabulated. Read More