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A consumer agency that proposes that lawyers' rates are too high wanted to estimate the mean hourly rate for all lawyers in New York City. A sample of 70 lawyers taken from New York City showed that the mean hourly rate charged by them is \(\$ 420\). The population standard deviation of hourly charges for all lawyers in New York City is \(\$ 110\). a. Construct a \(99 \%\) confidence interval for the mean hourly charges for all lawyers in New York City. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

Short Answer

Expert verified
A. The 99% confidence interval for the mean hourly rate charged by lawyers in New York City is approximately $(386.69,453.31). B. The width of the interval can be reduced by either increasing the sample size, decreasing the confidence level, or decreasing the variability within the data. Increasing the sample size is often the best method, assuming additional resources are available.

Step by step solution

01

Calculating the Confidence Interval

First, find the corresponding Z-value for a 99% confidence interval in a standard Z-table or online, which is 2.58. Then substitute all the values into the formula to get the confidence interval: \(420 \pm 2.58 \times \frac{110}{\sqrt{70}}\)
02

Calculating the Range

Now, calculate the margin of error which is \(2.58 \frac{110}{\sqrt{70}}\). Use this to find the low and high ends of the confidence interval.
03

Discussing Alternatives

Several ways can be taken into consideration to reduce the width of the confidence interval: increase the sample size, decrease the confidence level, or decrease the variability within the data. While the best alternative depends on the situation, generally, increasing the sample size is the most practical and commonly used method, provided the increased cost and time are manageable.

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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 crucial concept when working with confidence intervals. It represents the extent to which the sample estimate might differ from the actual population parameter.
The formula used to calculate the margin of error is: \[ ME = Z \times \frac{\sigma}{\sqrt{n}} \] where:
  • \( Z \) is the Z-score corresponding to the desired confidence level
  • \( \sigma \) is the population standard deviation
  • \( n \) is the sample size.
In the example of New York City lawyers, the margin of error helps to determine how precisely the sample mean \( \$420 \) approximates the actual average lawyer rate. By understanding and calculating the margin of error, researchers can communicate the potential variance from the true population mean, ensuring that conclusions drawn from the data are adequately grounded. If the margin of error seems large, it indicates a broader range in which the true mean might fall, whereas a smaller margin suggests a more tightened estimate.
Sample Size
Sample size refers to the number of individual data points or observations collected in a study. It plays a critical role in determining the reliability and accuracy of statistical results.
Larger sample sizes tend to yield more precise estimates of the population parameter because they provide more data, reducing the impact of outliers. In our lawyer rate example, a sample size of 70 was chosen. The formula for adjusting the width of a confidence interval with sample size is: \[ CI_{width} = 2 \times Z_{critical} \times \frac{\sigma}{\sqrt{n}} \] This implies:
  • If you increase the sample size, the denominator \( \sqrt{n} \) becomes larger, thus reducing the width of the confidence interval.
  • This narrower interval leads to more precise estimates, but only if the increase in sample size is feasible in terms of cost and time.
Thus, if the existing confidence interval seems too wide, like in the lawyer rate study, increasing the sample size is often the best approach to obtaining a more accurate estimate.
Confidence Level
The confidence level indicates how certain we are that the true population parameter falls within the calculated confidence interval. A common confidence level is 99%, which suggests that if the same population were sampled 100 times, roughly 99 of those intervals would contain the population mean.
Choosing a higher confidence level means more certainty about including the true mean, but the interval becomes wider. This is because a larger portion of possible values is covered to ensure the population mean is captured. In the scenario of New York City lawyers, they used a 99% confidence level, resulting in a relatively wide interval. If they aimed to narrow it, reducing the confidence level to 95% could be a viable option. However, this trade-off results in less certainty, as covering a smaller range reduces the likelihood that the interval contains the true mean.
Therefore, often the confidence level is selected based on the balance between the level of certainty desired and the practical width of the confidence interval.
Population Standard Deviation
The population standard deviation (\( \sigma \)) measures the average deviation of each data point from the population mean. It provides insight into the overall variability within a set of data.
In the lawyer rate example, the standard deviation was \( \$110 \), reflecting the spread in hourly charges among lawyers in New York City.High variability means the data points are spread out over a larger range, potentially widening the confidence interval, as seen in the formula:\[ CI_{width} \propto \frac{\sigma}{\sqrt{n}} \] Here are some details:
  • Larger \( \sigma \) increases the margin of error, leading to a wider confidence interval.
  • Lower \( \sigma \) results in a narrower interval, providing a more precise estimate.
While researchers can't reduce \( \sigma \) themselves, understanding its effect helps them interpret the width of the confidence interval and manage expectations when reporting findings.

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Most popular questions from this chapter

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the amounts of milk in all such cartons is always equal to \(.15\) ounce. The quality control department takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(32.15\) ounces or the lower limit of this confidence interval is less than \(31.85\) ounces, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.94\) ounces. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the amounts of milk put in all such cartons have a normal distribution.

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