A local bank needs information concerning the checking account balances of its customers. A random sample of 15 accounts was checked. The mean balance was $686.75 with a standard deviation of $256.20. Find a 98%
Confidence interval for the true mean. Assume that the account balances are normally distributed.
A)($513.17, $860.33)
B)($238.23, $326.41)
C)($326.21, $437.90)
D)($487.31, $563.80)

Question

Quizplus · Accepted Answer

Using the formula: 
CI = x̄ ± z*(σ/√n), 
where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, z is the z-score corresponding to the desired confidence level (98% or 2.33 for a two-tailed test), we get:
CI = 686.75 ± 2.33*(256.20/√15) 
CI = ($513.17, $860.33)
Therefore, the best choice is A because it is the only option that includes the calculated confidence interval. Using the formula for a confidence interval with a given level of confidence:
CI = x̄ ± z*(σ/√n)
where x̄ is the sample mean, σ is the population standard deviation (unknown in this case), n is the sample size, and z is the z-score corresponding to the level of confidence.
For a 98% confidence interval, the z-score is 2.33.
Plugging in the values:
CI = 686.75 ± 2.33*(256.20/√15)
CI = (513.17, 860.33)
Therefore, the best choice is A, since it contains the correct confidence interval.

A Local Bank Needs Information Concerning the Checking Account Balances