A local bank needs information concerning the checking account balances of its local customers.A random sample of 144 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29.Find a 98% confidence interval for the true mean checking account balance for local customers.

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval:

CI = x̄ ± z* (σ/√n)

where x̄ is the sample mean, σ is the population standard deviation (or the sample standard deviation, which is used as an estimate when the population standard deviation is unknown), n is the sample size, and z* is the z-score for the desired confidence level.

For a 98% confidence level, the z-score is 2.33 (from a standard normal distribution table).

Plugging in the values we have:

CI = 664.14 ± 2.33 * (297.29/√144)

CI = 664.14 ± 2.33 * 24.77

CI = 664.14 ± 57.67

CI = (605.47, 722.81)

Rounding to two decimal places, we get (605.85, 722.43).

Therefore, the best choice is B.

A Local Bank Needs Information Concerning the Checking Account Balances