Application of sampling distribution
Joe Greene, a new manager at Pilgrim Bank wants to better understand profitability data for bank’s customers. Joe is able to obtain a random sample of 31,634 customers on the following variables – Profitability (in $, for the most recent completed year, i.e. 2006), whether or not the customer uses the online banking channel, customer tenure, age and income where available, as well as the customer’s residential area. Descriptive statistics for Profits indicates that the average profit per customer is $111.50 with a standard deviation of $272.84.
a. Is Joe justified in assuming that this is a “large” sample? (see slide 7-14) YES, BECAUSE 31,634 IS A LOT MORE THAN 30 OBVERSATIONS. GENERALLY, THE LARGER THE SAMPLE, THE MORE RELIABLE ARE THE ESTIMATES. THE KEY IS TO HAVE A RANDOMLY SELECTED SAMPLE TO REDUCE THE RISK OF BIASED ESTIMATES
b. Joe has been informed that the bank serves approximately 5 million customers nationwide – should he worry about using finite population correction factor (slide 7-9). FINITE POPULATION CORRECTION (FPC) IS NOT NECESSARY, SINCE THE POPULATION OF APPROXIMATELY 5 MILLION IS QUITE LARGE. FPC IS ONLY RECOMMENDED FOR SMALL POPULATIONS.
c. Joe wants to estimate average profit for the entire population based on his sample. He knows that the “point estimate” for average profit would be $111.50, but, he will need to calculate the margin of error. The first step for this is to calculate the standard error (see slide 7-6); provide the value below. 272.84 / SQRT(31,634) = $1.53
d. The second step in calculating the margin or error (often simply called error) is to multiply the standard error by 1.96; provide the value below. 1.96*$1.53=$3. THUS THE MARGIN OF ERROR FOR OUR ESTIMATE OF AVERAGE PROFIT PER CUSTOMER, FOR THE ENTIRE CUSTOMER BASE WILL BE $3
e. Joe can combine responses from c and d to report the estimated average profit customer for all the bank’s customers as...
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