The Independent Department Store wants to determine the proportion of their char

Question

The Independent Department Store wants to determine the proportion of their charge accounts that have an unpaid balance of $1, 500 or more. A sample of 250 accounts revealed that 100 of them had an unpaid balance of $1, 500 or more. a. What is the 99% confidence interval for the population proportion? Would it be reasonable to conclude that more than half of the account balances are more than $1, 500? b. The store have a total of 900 charge customers. Develop a 99% confidence interval for the proportion of charge customers with an unpaid account balance over $1, 500.

Explanation / Answer

a) here estimated proportion p=100/250 =0.4

hence std error =(p(!-p)/n)1/2 =0.031

for 99% CI, z=2.5758

99% confidence interval =estimated proportion -/+ z*std error =0.3202 ; 0.4798

as above interval contain proportion value below 0.5 ; hence it would not be reasonable to conclude that more then half of the account balances are more then 1500.

b) for n/N =250/900>0.1

hence

std error =((p(!-p)/n)*(N-n/N-1))1/2 =0.0263

for 99% CI, z=2.5758

99% confidence interval =estimated proportion -/+ z*std error =0.3321 ; 0.4679

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