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Chapter 9: Problem 58

As part of an annual review of its accounts, a discount brokerage selected a random sample of 36 customers and reviewed the value of their accounts. The mean was 32,000 with a sample standard deviation of 8,200 . What is a \(90 \%\) confidence interval for the mean account value of the population of customers?

Short Answer

Expert verified
The 90% confidence interval for the mean account value is (29,753.16, 34,246.84).

Step by step solution

01

Understand the problem

We want to find the 90% confidence interval for the mean account value of the population of customers. The sample mean is 32,000, the sample standard deviation is 8,200, and the sample size is 36.
02

Identify the appropriate formula

Since the sample size is large (n = 36), we can use the formula for the confidence interval of the mean using the z-distribution: \[CI = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(\sigma\) is the sample standard deviation, and \(n\) is the sample size. The z-score for a 90% confidence interval is approximately 1.645.
03

Calculate the standard error

The standard error of the mean (SEM) is given by \(\frac{\sigma}{\sqrt{n}}\). Substitute into the formula:\[SEM = \frac{8,200}{\sqrt{36}} = \frac{8,200}{6} = 1,366.67\]
04

Find the margin of error

The margin of error (ME) is calculated using the z-score and the standard error:\[ ME = z \times SEM = 1.645 \times 1,366.67 = 2,246.84 \]
05

Calculate the confidence interval

The confidence interval is found by subtracting and adding the margin of error from the sample mean:\[CI = 32,000 \pm 2,246.84\]So, the confidence interval will be: \[(32,000 - 2,246.84, 32,000 + 2,246.84) = (29,753.16, 34,246.84)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean
The sample mean is a basic concept in statistics, representing the average value of a sample set. In the context of our exercise, the sample mean is the average account value found in a sample of 36 customers. Here, the sample mean is recorded as 32,000.
  • The sample mean is calculated by summing up all the individual data points (account values in this case) and then dividing the total by the number of data points (customers) in the sample.
  • The formula for the sample mean is given by: \( \bar{x} = \frac{\sum x_i}{n} \), where \( x_i \) is each individual data point, and \( n \) is the total number of data points.
The sample mean provides an estimate of the population mean, making it a vital part of constructing confidence intervals. It serves as the center point in our confidence interval calculation.
Standard Error of the Mean
The Standard Error of the Mean (SEM) indicates how much the sample mean is expected to vary if we took multiple samples from the population. It provides insight into the precision of the sample mean as an estimate of the population mean. In our example, the SEM is calculated as 1,366.67.
  • SEM is found by dividing the sample standard deviation by the square root of the sample size.
  • The formula is: \( SEM = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the sample standard deviation, and \( n \) is the sample size.
A smaller SEM suggests that the sample mean is a more accurate reflection of the population mean, which is crucial for confidence interval accuracy. In our case, it helps determine the range of possible account values within the population.
Margin of Error
The margin of error quantifies the uncertainty or potential error in the estimate of the population mean. It's the amount added and subtracted from the sample mean to create the confidence interval. For our exercise, the margin of error is 2,246.84.
  • The margin of error depends on the desired confidence level (here, 90%) and the standard error.
  • It is calculated by multiplying the z-score corresponding to the desired confidence level by the SEM: \( ME = z \times SEM \).
The margin of error helps us understand the degree of confidence we have in our estimate. In this exercise, it allows us to assert, with a 90% confidence level, that the true mean account value lies within our calculated interval from 29,753.16 to 34,246.84.

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Most popular questions from this chapter

A recent study by the American Automobile Dealers Association surveyed a random sample of 20 dealers. The data revealed a mean amount of profit per car sold was 290, with a standard deviation of 125 . Develop a \(95 \%\) confidence interval for the population mean of profit per car.

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Passenger comfort is influenced by the amount of pressurization in an airline cabin. Higher pressurization permits a closer-to-normal environment and a more relaxed flight. A study by an airline user group recorded the equivalent air pressure on 30 randomly chosen flights. The study revealed a mean equivalent air pressure of 8,000 feet with a standard deviation of 300 feet. a. Develop a \(99 \%\) confidence interval for the population mean equivalent air pressure. b. How large a sample is needed to find the population mean within 25 feet at \(95 \%\) confidence?

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged 10,979 . The standard deviation of the sample was 1,000 . a. Based on this sample information, develop a \(90 \%\) confidence interval for the population mean yearly premium. b. How large a sample is needed to find the population mean within 250$ at 99% confidence?

Dr. Patton is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 a.m. section of 40 students, the mean number of misspelled words was \(6.05 .\) Construct a \(95 \%\) confidence interval for the mean number of misspelled words in the population of student essays.
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