Question

A saving and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $785 and standard deviation of $289. Find a 96% confidence interval for the true mean checking account balance for local customers.

a) Find the point estimate

b) Find the critical value.

c) Find the margin of error

d) Construct the confidence interval

e) Interpret the confidence interval

Answer 1

(a) The point estimate of the mean = = 785

The point estimate of the standard deviation (SE) = 289 / sqert(14) = 77.2727

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(b) Since population standard deviation is unknown, the t critical (2 tail) for = 0.04, for df = n -1 = 13, is 2.282

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(c) The Margin of error = Critical value * SE = 2.282 * 77.2727 = 176.336

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(d) The Confidence Interval is given by , where

The Lower Limit = 785 - 176..336 = 608.664

The Upper Limit = 785 + 176.336 = 961.336

The 96% Confidence Interval is (608.664 , 961.336)

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(e) We are 96% Confident that the true population mean checking account balance of all the local customers of the saving and loan association lies between the confidence limits from $608.664 to $961.336.

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