/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A sample survey of 54 discount b... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at \(\$ 50\) per share was \(\$ 33.77\) (AAII Journal, February 2006 ). The survey is conducted annually. With the historical data available, assume a known population standard deviation of \(\$ 15\) a. Using the sample data, what is the margin of error associated with a \(95 \%\) confidence interval? b. Develop a \(95 \%\) confidence interval for the mean price charged by discount brokers for a trade of 100 shares at \(\$ 50\) per share.

Short Answer

Expert verified
The margin of error associated with a \(95\%\) confidence interval is approximately \(\$4.00\). The \(95\%\) confidence interval for the mean price charged by discount brokers for a trade of 100 shares at \(\$50\) per share is \(\$29.77\) to \(\$37.77\).

Step by step solution

01

Identify the given values

The given values in this problem are as follows: Sample mean: \(\bar{x} = \$33.77\) Sample size: \(n = 54\) Population standard deviation: \(\sigma = \$15\)
02

Calculate the standard error

The standard error of the sample is calculated using the population standard deviation and sample size. Standard Error (SE) = \(\frac{\sigma}{\sqrt{n}}\) Plug in the given values and calculate the standard error: SE = \(\frac{\$15}{\sqrt{54}}\approx \$ 2.04\)
03

Find the Z-score for a \(95\%\) confidence interval

For a \(95\%\) confidence interval, the Z-score (critical value) is found by identifying the value which attracts \(2.5\%\) in each tail of the standard normal distribution (as the total \(5\%\) is divided equally between the two tails). Therefore, we want the Z-score such that: \(\mathbf{P}(-Z_{\alpha/2} < Z < Z_{\alpha/2}) = 0.95\) Using a Z-table or calculator, we find the value of \(Z_{\alpha/2}\) to be: \(Z_{\alpha/2} = 1.96\)
04

Calculate the margin of error

Now, we can calculate the margin of error associated with a 95% confidence interval using the formula: Margin of Error (ME) = \(Z_{\alpha/2} * SE\) Plug in the values from Steps 2 and 3: ME = \(1.96 * \$ 2.04\approx \$ 4.00\)
05

Develop the 95% confidence interval for the mean price

With the margin of error calculated, we can now develop the 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share. Lower Limit = \(\bar{x} - ME\) Upper Limit = \(\bar{x} + ME\) Plug in the values from Step 1 and Step 4: Lower Limit = \(\$33.77 - \$4.00\approx \$29.77\) Upper Limit = \(\$33.77 + \$4.00\approx \$37.77\)
06

Interpret the results

The 95% confidence interval for the mean price charged for a trade of 100 shares at $50 per share is \(\$29.77\) to \(\$37.77\). This means we are confident that, if the survey were conducted repeatedly, the true population mean price would fall within this range 95% of the time.

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Key Concepts

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

Margin of Error
The margin of error is a vital component in statistics when we deal with confidence intervals. It represents the amount by which the sample estimate might differ from the true population parameter. Think of it as the wiggle room around our estimate.
When we say we have a margin of error, we imply that our confidence interval is plus or minus a certain range around the sample mean.
To calculate it, we multiply the Z-score by the standard error:
  • Formula: Margin of Error (ME) = \( Z_{\alpha/2} \times \text{Standard Error} \)
  • This range helps us capture the true population mean with a certain level of confidence.
Hence, knowing the margin of error is crucial as it determines the range of our confidence interval.
Standard Error
Standard Error (SE) indicates how much the sample mean is expected to vary from the true population mean. It's a way of measuring the accuracy of our sample estimate.
This is particularly important for a confidence interval, which is computed using the standard error.
To compute it, you use the following formula:
  • Standard Error (SE) = \( \frac{\sigma}{\sqrt{n}} \)
  • Where \( \sigma \) represents the population standard deviation and \( n \) is the sample size.
Using this formula, a larger sample size results in a smaller standard error, meaning more precision in our estimates.
This concept is essential as it measures the likely deviation of the sample mean from the population mean.
Z-score
The Z-score is crucial when calculating confidence intervals, as it helps us determine how many standard deviations a data point is from the mean.
For a 95% confidence interval, the critical Z-score is 1.96, representing the cut-off for the middle 95% of the data in a standard normal distribution.

Why is the Z-score important?

  • It helps define the width of our confidence interval.
  • It's a fixed value drawn from a standard normal distribution curve.
  • Different confidence levels will have different Z-scores.
Understanding the Z-score helps us determine how confident we are that the population parameter lies within a particular interval.
Population Standard Deviation
Population Standard Deviation (\(\sigma\)) quantifies the amount of variation or dispersion in a dataset. It measures the average distance of each data point from the mean of the population.
In your problem, it's provided as a known value: \(\$15\).

Why is it important?

  • Having a known population standard deviation enables us to use the standard normal distribution (Z-distribution) for calculating the standard error and confidence intervals, making our results more accurate.
  • It helps in understanding the spread of the entire population's data.
The population standard deviation is integral in forming the standard error formula, further affecting the margin of error and the construction of the confidence interval.

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Most popular questions from this chapter

The 92 million Americans of age 50 and over control \(50 \%\) of all discretionary income \((A A R P \text { Bulletin, March } 2008) .\) AARP estimated that the average annual expenditure on restaurants and carryout food was \(\$ 1873\) for individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is \(\$ 550\) a. At \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval for the population mean amount spent on restaurants and carryout food? c. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food? d. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater or less than \(\$ 1873 ?\)

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For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that \(52 \%\) of U.S. employers were likely to require higher employee contributions for health care coverage in 2009 (Business Week, February 16,2009 ). Suppose the survey was based on a sample of \(800 \mathrm{com}\) panies. Compute the margin of error and a \(95 \%\) confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009

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