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An economist wishes to estimate, to within 2 minutes, the mean time that employed persons spend commuting each day, with \(95 \%\) confidence. On the assumption that the standard deviation of commuting times is 8 minutes, estimate the minimum size sample required.

Short Answer

Expert verified
The minimum sample size required is 62.

Step by step solution

01

Identify the Required Information

To solve this problem, we need the confidence level, the margin of error (E), and the population standard deviation (\( \sigma \)). Here, the confidence level is 95%, the margin of error is 2 minutes, and the standard deviation is 8 minutes.
02

Locate the Z-Score for the Confidence Level

For a 95% confidence level, we need to find the Z-score, which corresponds to the critical value of the standard normal distribution. The Z-score for a 95% confidence level is 1.96.
03

Apply the Sample Size Formula

Use the formula for the minimum sample size for estimating a population mean: \[ n = \left(\frac{Z \cdot \sigma}{E}\right)^2 \]Where:- \( Z = 1.96 \) (the Z-score for 95% confidence)- \( \sigma = 8 \) minutes (standard deviation)- \( E = 2 \) minutes (margin of error).Substitute these values into the formula.
04

Calculate the Sample Size

Substituting the values into the formula from Step 3:\[ n = \left(\frac{1.96 \cdot 8}{2}\right)^2 \]\[ n = \left(\frac{15.68}{2}\right)^2 \]\[ n = 7.84^2 \]\[ n = 61.4656 \]Since the sample size must be a whole number, round up to the nearest whole number.
05

Interpret the Result

Rounding 61.4656 to the next whole number gives 62. Therefore, a sample size of at least 62 is needed to estimate the mean commuting time with a margin of error of 2 minutes and 95% confidence.

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Key Concepts

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

Sample Size Calculation
Understanding how to calculate sample size is crucial when you want to make accurate estimates about a population based on sample data. In our exercise, the goal was to determine how many employed persons to include in the study to accurately estimate their commute times with a high level of confidence and within a specific margin of error.
To get started with sample size calculation, there are several key elements you need to have:
  • Confidence Level: The probability that the sample accurately reflects the population.
  • Margin of Error (E): The maximum acceptable difference between the sample mean and the population mean.
  • Population Standard Deviation (σ): A measure of variability within the population.
By combining these components, you can calculate the necessary sample size using the formula:\[ n = \left(\frac{Z \cdot \sigma}{E}\right)^2 \]Where \( n \) is the sample size, \( Z \) is the Z-score corresponding to your confidence level, \( \sigma \) is the standard deviation, and \( E \) is the margin of error. Plugging in the values from the exercise—\( Z = 1.96 \), \( \sigma = 8 \), and \( E = 2 \)—calculates the minimum sample size you need for reliable results.
Margin of Error
The margin of error reflects the range within which the true population parameter is expected to fall. It is crucial when interpreting the reliability of estimates from sample data. In simple terms, it's how much you can expect your sample estimate to differ from the actual population value.
The margin of error gives you an indication of the precision of your estimate. A smaller margin of error means more confidence in your estimate, while a larger margin might suggest less precision. However, reducing the margin of error requires you to increase the sample size, or sometimes to accept a lower confidence level.
To calculate the margin of error, you can use this formula in the context of population mean estimates:\[ E = \frac{Z \cdot \sigma}{\sqrt{n}} \]Where \( E \) is the margin of error, \( Z \) is the critical value, \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size. In our exercise, having a set margin of error helped us backtrack and calculate the sample size necessary for maintaining that precision level with a 95% confidence interval.
Standard Deviation
Standard deviation is a crucial statistical measure that quantifies the amount of variation or dispersion in a set of data values. In the context of our exercise, it tells us about the variability in commuting times among the employed people.
Knowing the standard deviation is essential because it directly impacts both your sample size and your margin of error. A larger standard deviation suggests more diversity in the data, which may require a larger sample size to achieve the same level of confidence and margin of error.
Mathematically, the standard deviation (\( \sigma \)) is calculated through the following steps:
  • Compute the mean of the data set.
  • Subtract the mean from each data point and square the result.
  • Find the average of those squared differences.
  • Take the square root of that average to get the standard deviation.
A standard deviation should be high enough to reveal variability, but manageable enough not to demand extremely large samples for a feasible level of accuracy. In our case, we used a standard deviation of 8 minutes to determine the necessary sample size for estimating mean commuting times.

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Most popular questions from this chapter

The variation in time for a baked good to go through a conveyor oven at a large scale bakery has standard deviation 0.017 minute at every time setting. To check the bake time of the oven periodically four batches of goods are carefully timed. The recent check gave a mean of 27.2 minutes with sample standard deviation 0.012 minute. Construct a \(99.8 \%\) confidence interval for the mean bake time of all batches baked in this oven. Assume bake times are normally distributed. Hint: Not all the numbers given in the problem are used.

In a random sample of 900 adults, 42 defined themselves as vegetarians. a. Give a point estimate of the proportion of all adults who would define themselves as vegetarians. b. Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion. c. Construct an \(80 \%\) confidence interval for the proportion of all adults who would define themselves as vegetarians.

Large Data Sets 4 and 4 Alist the results of 500 tosses of a die. Let \(p\) denote the proportion of all tosses of this die that would result in a four. Use the sample data to construct a \(90 \%\) confidence interval for \(p\).

Wild life researchers trapped and measured six adult male collared lemmings. The data (in millimeters) are: \(104,99,112,115,96,109 .\) Assume the lengths of all lemmings are normally distributed. a. Construct a \(90 \%\) confidence interval for the mean length of all adult male collared lemmings using these data. b. Convert the six lengths in millimeters to lengths in inches using the conversion \(1 \mathrm{~mm}=0.039\) in (so the first datum changes to \((104)(0.039)=4.06\) ) . Use the converted data to construct a \(90 \%\) confidence interval for the mean length of all adult male collared lemmings expressed in inches. c. Convert your answer in part (a) into inches directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.

You wish to estimate the current mean birth weight of all newborns in a certain region, to within 1 ounce (1/16 pound) and with \(95 \%\) confidence. A sample will cost \(\$ 400\) plus \(\$ 1.50\) for every newborn weighed. You believe the standard deviations of weight to be no more than 1.25 pounds. You have \(\$ 2,500\) to spend on the study. a. Can you afford the sample required? b. If not, what are your options?

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