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.

Short Answer

Expert verified
The 95% confidence interval for the mean amount spent per day by families of four visiting Niagara Falls is \(\$ 234.14\) to \(\$ 270.76\). Since the American Automobile Association's reported average of \(\$ 215.60\) is not within this interval, it appears that the population mean amount spent per day by families visiting Niagara Falls does differ from the mean reported by the American Automobile Association.

Step by step solution

01

Identify Given Data

We are given the following data: - Sample size: \(n = 64\) - Sample mean: \(\bar{x} = \$ 252.45\) - Sample standard deviation: \(s = \$ 74.50\) - Confidence level: \(95\%\) Next, we will find the critical value using a standard normal distribution table since the sample size is large enough.
02

Find the Critical Value

To find the critical value for a \(95\%\) confidence level, we will need to find the \(Z\) value that corresponds to the \(97.5\%\) percentile because \(95\%\) of the data is within \(\pm 1.96\) standard deviations of the mean. So, we have: \(Z_{\alpha/2} = 1.96\)
03

Calculate Margin of Error

The margin of error formula is: \(E = Z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\) Using the given variables and the critical value, \(E = 1.96 \cdot \frac{74.50}{\sqrt{64}}\) Calculating the margin of error, we get: \(E \approx 18.31\)
04

Calculate Confidence Interval

Now, we have everything we need to calculate the \(95\%\) confidence interval for the mean amount spent per day by families of four visiting Niagara Falls: \((\bar{x} - E, \bar{x} + E) = (252.45 - 18.31, 252.45 + 18.31) = (234.14, 270.76)\) The confidence interval is \(\$ 234.14\) to \(\$ 270.76\).
05

Compare Confidence Interval to American Automobile Association's Value

The American Automobile Association's reported average amount spent per day is \(\$ 215.60\). Since this value is not included within the confidence interval we just calculated, it appears that the population mean amount spent per day by families visiting Niagara Falls does differ from the mean reported by the American Automobile Association.

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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, often represented by the symbol \(\bar{x}\), is a critical statistic used to estimate the average of a population. When collecting data from a population, it's usually impractical or impossible to measure every individual member. Instead, researchers take a subset, or sample, and use the average of this group to estimate the population mean. For instance, in a survey where the focus is on the amount families spend on vacation per day, the sample mean of \$252.45 was calculated from the daily expenses reported by 64 families of four at Niagara Falls. This single number provides a central value that is expected to be a close approximation to the true population mean. It's essential to accurately compute and understand the sample mean, as it forms the basis for many inferential statistics, like confidence intervals.
Sample Standard Deviation
In statistics, the sample standard deviation \(s\) is a measure that tells us how much the values in our sample vary, on average, from the sample mean. It's an indicator of the spread or dispersion within the dataset. If the sample standard deviation is small, it means the data points are clustered closely around the mean, implying less variability. A larger standard deviation suggests that the data are more spread out. For example, \$74.50 as the sample standard deviation means that the daily vacation expenses of different families can vary widely from the average. When calculating confidence intervals, understanding the sample standard deviation is crucial because it influences the margin of error and, consequently, the width of the confidence interval.
Confidence Level
The confidence level is a percentage that reflects how certain we can be about the range within which the true population parameter lies. It's an expression of probability that the interval we're computing includes the population mean. A standard confidence level used is 95%. In our example, when we say we are constructing a 95% confidence interval, it implies that if we were to take many samples and construct intervals in the same way, we would expect about 95% of these intervals to contain the true population mean. It's essentially telling us how confident we can be in our estimate and offers a buffer for the uncertainty inherent in using samples to make inferences about the population.
Margin of Error
The margin of error is a statistic that tells us how much we can expect the estimate from our sample to differ from the true population parameter. It is influenced by both the variability of the data, expressed by the standard deviation, and the size of the sample. As a rule of thumb, a larger sample will reduce the margin of error, giving us a more precise estimate of the population mean. In the calculation, \(E = 1.96 \cdot \frac{74.50}{\sqrt{64}}\), the result approximately \$18.31 is the margin of error which builds on our confidence interval. It shows that the true population mean is likely to fall within \$18.31 above or below the sample mean. Understanding margin of error is key because it directly informs us how wide the confidence interval is, affecting the interpretation of data and subsequent decisions based on the analysis.
Normal Distribution
The normal distribution, also known as the Gaussian distribution or 'bell curve,' is a symmetric distribution where most of the observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions. Many natural phenomena and human-made processes conform to a normal distribution, so it's a foundational concept in standard statistical analyses. In the context of confidence intervals, the normal distribution allows us to determine the critical value (in this case \(Z_{\alpha/2} = 1.96\)) used to calculate the margin of error. The properties of the normal distribution, namely the 68-95-99.7 (empirical) rule, inform us that for a 95% confidence level, we include the mean plus or minus approximately 1.96 standard deviations to encompass 95% of the data. Recognizing the role of normal distribution in statistics is valuable because it underpins many inferential procedures and assumes that sample means will be normally distributed if the sample size is large enough (Central Limit Theorem).

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Most popular questions from this chapter

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.

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Which of the following are the same at all levels of output under perfect competition? a. Marginal cost and marginal revenue b. Price and marginal revenue c. Price and marginal cost d. All of the above

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