How large a sample should be selected to provide a \(95 \%\) confidence interval with a margin of error of \(10 ?\) Assume that the population standard deviation is 40

Short Answer

Expert verified
To find the required sample size for a 95% confidence interval with a margin of error of 10 and a population standard deviation of 40, use the formula \( E = Z \cdot \frac{\sigma}{\sqrt{n}} \), where \(Z=1.96\). Rearranging the formula and solving for \(n\), we get \(n = (\frac{1.96 \cdot 40}{10})^2 = 61.47\). Since the sample size must be a whole number, we round up to the nearest whole number, resulting in a required sample size of 62.

Step by step solution

01

Find the Z-score for the given confidence level

The Z-score is a measure of how many standard deviations an element is from the mean. For a 95% confidence level, the Z-score is typically 1.96. This value is found using a standard normal distribution table or through calculations involving the cumulative distribution function for a standard normal distribution.
02

Write down the formula for the margin of error

The margin of error is defined by the formula \( E = Z \cdot \frac{\sigma}{\sqrt{n}} \). In this formula, \( E \) is the margin of error, \( Z \) is the Z-score, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
03

Plug in the given values into the margin of error formula

We are given that the margin of error \( E \) is 10, the Z-score \( Z \) is 1.96 (from Step 1), and the population standard deviation \( \sigma \) is 40. Substitute these values into the formula, we get: \( 10 = 1.96 \cdot \frac{40}{\sqrt{n}} \)
04

Solve for the sample size \( n \)

Now, we have to rearrange the formula to solve for \( n \). Square both sides to eliminate the square root in the denominator: \( 10^2 = (1.96 \cdot \frac{40}{\sqrt{n}})^2 \) Solve for \( n \): \( n = (\frac{1.96 \cdot 40}{10})^2 \)
05

Calculate the sample size \( n \)

Calculate \( n \): \( n = (\frac{78.4}{10})^2 = 61.47 \) However, the sample size must be a whole number, so we round up to the nearest whole number. The smallest sample size that would provide a 95% confidence interval with a margin of error of 10 is therefore 62.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. The range is calculated from a given set of sample data, primarily by using the sample mean and standard deviation.

In the context of the exercise, a 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect about 95 of those intervals to contain the population mean. It’s important to note that the confidence level doesn’t reflect the probability that a particular interval captures the mean, but rather the proportion of intervals that will capture the mean over a large number of samples.

The confidence interval is determined by the chosen confidence level (in the exercise, 95%), the sample size, the population standard deviation, and the Z-score corresponding to the confidence level.
Margin of Error
The margin of error represents the extent to which the estimate obtained from the sample can differ from the true population parameter. In simpler terms, it’s an expression of the level of uncertainty or the maximum expected difference in the study.

A smaller margin of error indicates a more precise measurement and is determined mainly by the size of the sample and the standard deviation of the population. The formula for the margin of error is given by the expression \( E = Z \cdot \frac{\sigma}{\sqrt{n}} \), where \( E \) is the margin of error, \( Z \) is the Z-score, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. The exercise provided a target margin of error of 10, which is then used with a Z-score and the population standard deviation to calculate the required sample size.
Population Standard Deviation
The population standard deviation (\( \sigma \) ) is a measure of the dispersion or variability in a set of data. It quantifies how spread out the numbers are in the population.

For instance, if the population standard deviation is low, it indicates that the values are closely clustered around the mean. Conversely, a high standard deviation suggests that the values are spread out over a wide range.

In the context of sample size determination, the population standard deviation plays a crucial role: a larger standard deviation would typically require a larger sample size to achieve a specific margin of error for a confidence interval. In the exercise, a population standard deviation of 40 indicates variability in the data, which directly impacts the margin of error and, subsequently, the required sample size.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. The standard normal distribution, which the Z-score is based on, allows us to determine how far away a particular point is from the population mean, expressed in standard deviations.

A Z-score of 1.96, as used in this exercise for a 95% confidence level, tells us that the value lies nearly two standard deviations from the mean. This Z-score is a crucial component in determining the confidence interval and the margin of error, as it encapsulates the desired level of confidence in the statistical conclusions. The Z-score contributes to calculating how large a sample size needs to be for the estimated mean to fall within a certain range of the true population mean.

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Most popular questions from this chapter

The mean number of hours of flying time for pilots at Continental Airlines is 49 hours per month (The Wall Street Journal, February 25, 2003). Assume that this mean was based on actual flying times for a sample of 100 Continental pilots and that the sample standard deviation was 8.5 hours. a. \(\quad\) At \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval estimate of the population mean flying time for the pilots? c. The mean number of hours of flying time for pilots at United Airlines is 36 hours per month. Use your results from part (b) to discuss differences between the flying times for the pilots at the two airlines. The Wall Street Journal reported United Airlines as having the highest labor cost among all airlines. Does the information in this exercise provide insight as to why United Airlines might expect higher labor costs?

A survey conducted by the American Automobile Association showed that a family of four spends an average of \(\$ 215.60\) per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of \(\$ 252.45\) per day and a sample standard deviation of \(\$ 74.50\) a. Develop a \(95 \%\) confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls. b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? Explain.

The National Center for Education Statistics reported that \(47 \%\) of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a \(95 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. b. Provide a \(99 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. c. What happens to the margin of error as the confidence is increased from \(95 \%\) to \(99 \% ?\)

A perfectly competitive market is not characterized by a. many small firms. b. a great variety of different products. c. free entry into and exit from the market. d. any of the above.

A simple random sample of 60 items resulted in a sample mean of \(80 .\) The population standard deviation is \(\sigma=15\) a. Compute the \(95 \%\) confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a \(95 \%\) confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?

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