Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution.
-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 98% confidence interval for the true mean checking account balance for local Customers.
A) $\$ 453.59

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution.
-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 98% confidence interval for the true mean checking account balance for local Customers. 
A) $\$ 453.59 < \mu < \$ 874.69$
B) $\$ 455.65 < \mu < \$ 872.63$
C) $\$ 493.71 < \mu < \$ 834.57$
D) $\$ 492.52 < \mu < \$ 835.76$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:$$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation, n is the sample size, and $z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the desired confidence level.In this case, we have $\bar{x} = 664.14$, $\sigma = 297.29$, n = 14, and $\alpha = 0.02$ (since we want a 98% confidence level). Using a standard normal distribution table, we find that $z_{\alpha/2} = 2.326$.Plugging these values into the formula, we get:$$664.14 \pm 2.326 \frac{297.29}{\sqrt{14}} = 664.14 \pm 210.55$$Therefore, the 98% confidence interval is $453.59 < \mu < 874.69$.

Use the Given Degree of Confidence and Sample Data to Construct