Find the confidence interval specified. Assume that the population is normally distributed.
-A savings 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
$664.14 and a standard deviation of $297.29. Find a 90% confidence interval for the true mean Checking account balance for local customers.
A)$523.43 to $804.85
B)$455.65 to $872.63
C)$493.71 to $834.57
D)$492.52 to $835.76

Question

Find the confidence interval specified. Assume that the population is normally distributed.
-A savings 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
$664.14 and a standard deviation of $297.29. Find a 90% confidence interval for the true mean Checking account balance for local customers. 
A)$523.43 to $804.85
B)$455.65 to $872.63
C)$493.71 to $834.57
D)$492.52 to $835.76

Quizplus · Accepted Answer

To find a 90% confidence interval for the true mean checking account balance, we use the formula for the confidence interval for the mean when the population standard deviation is unknown and the sample size is small (n < 30). This involves using the t-distribution.The formula for the confidence interval is: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$Given:- $\bar{x} = 664.14$ (sample mean)- $s = 297.29$ (sample standard deviation)- $n = 14$ (sample size)- Confidence level = 90%Degrees of freedom (df) = $n - 1 = 14 - 1 = 13$.For a 90% confidence level and 13 degrees of freedom, we look up the t-value from the t-distribution table, which is approximately 1.771.Plugging the values into the formula:Margin of error $= t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 1.771 \cdot \frac{297.29}{\sqrt{14}} \approx 140.71$Thus, the confidence interval is:$\bar{x} \pm \text{Margin of error} = 664.14 \pm 140.71 = (523.43, 804.85)$Therefore, the correct choice is A, which corresponds to the calculated confidence interval of $523.43 to $804.85.

Find the Confidence Interval Specified