Question

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 15) 15) 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) ($238.23, $326.41) B) ($487 31, $563.80) © ($513.17, $860.33) D) ($326.21, $437.90)

Answer 1

15 Given data; Sample size (n) = 15 gample mean (7) 686.75 Sample Standard deviation (S) - 256-20 The confidence interval for the population mean de is, et I t x 5 5 where t, the t distrobution with C15-1=14) degrees of freedom & 98% contidence level. Ci.e, 0.98) here, tvalue at df-14 98% Confidence level = 2.624 ( from t-distubution table Required contidence interval is, - z t x 5 256.20 > 686.75 t 2624 x 256.20 686.75 † 2.624 X 3.87298 686.75 +9.624 X 66.150 6 686.75 + 173.5792 (5 13.1708 860.3292) ~ (513.17., 860.33 )

. Lower limit :(513.17) Upper limit : (860.33) for the true mean i Therefore, The 98% confidence interval is 8513.17, $860.33) Answeg : option (c) 1 513.17, $860.33