**Determining a 98% Confidence Interval for Savings Account Balances**A local bank is interested in gathering information about the savings account balances of its customers. To do this, they have selected a random sample of 15 savings accounts.Here's the collected data:- The mean (average) balance of these 15 accounts was $686.75.- The standard deviation of the sample was $256.20.**Objective:**Calculate a 98% confidence interval for the true mean savings account balance, assuming that the account balances are normally distributed. The final answer should be rounded to the nearest cent.**Method: Follow the PANIC acronym**1. **P** (Parameter): Define the parameter of interest. - The parameter of interest is the true mean savings account balance of the bank’s customers.2. **A** (Assumptions): Check the necessary assumptions. - The sample of 15 accounts is randomly selected. - The distribution of account balances is normal, as stated in the problem.3. **N** (Name the Interval): Specify the name of the interval. - We will use a t-interval for the mean because the population standard deviation is unknown and the sample size is small (n < 30).4. **I** (Interval Calculation): Calculate the interval. - The formula for a confidence interval for the mean using the t-distribution is: \[ \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \] Where: - $\bar{x}$ is the sample mean ($686.75) - $s$ is the sample standard deviation ($256.20) - $n$ is the sample size (15) - $t^*$ is the t-value for 98% confidence and 14 degrees of freedom (since degrees of freedom = n - 1 = 14). Using a t-table or calculator to find $t^*$ for 14 degrees of freedom at 98% confidence, we get approximately 2.977. Plugging the values into the formula: \[ 686.75 \pm 2.977 \left( \frac{256.20}{\sqrt{15}} \right) \] \[ 686.75 \pm 2.977 \left( \frac

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Description: **Determining a 98% Confidence Interval for Savings Account Balances**A local bank is interested in gathering information about the savings account balances of its customers. To do this, they have selected a random sample of 15 savings accounts.Here

The random variable account balance follows normal distribution. The sample size is 15. The sample…

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A local bank needs information concerning the savings 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. Round to the nearest cent.

A local bank needs information concerning the savings 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. Round to the nearest cent.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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