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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% CI is [234.20, 270.70]. The Niagara Falls average spending differs from the average reported by AAA.

Step by step solution

01

Identify the Problem

We need to develop a 95% confidence interval for the mean amount spent per day by a family of four visiting Niagara Falls based on the sample provided.
02

Determine the Formula for Confidence Interval

The formula for a confidence interval for the mean when the sample size is large (n > 30) is given by: \( \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \). Here, \( \bar{x} \) is the sample mean, \( z_{\alpha/2} \) is the z-score corresponding to the desired confidence level, \( \sigma \) is the sample standard deviation, and \( n \) is the sample size.
03

Identify Values from the Problem

From the problem, we know: \( \bar{x} = 252.45 \), \( \sigma = 74.50 \), \( n = 64 \), and the confidence level is 95%. Therefore, \( z_{\alpha/2} \) for a 95% confidence level is 1.96.
04

Calculate the Standard Error

The standard error is calculated as \( \frac{\sigma}{\sqrt{n}} = \frac{74.50}{\sqrt{64}} = 9.3125 \).
05

Calculate the Margin of Error

The margin of error is \( 1.96 \times 9.3125 = 18.25 \).
06

Establish the Confidence Interval

The 95% confidence interval is \( 252.45 \pm 18.25 \), resulting in \( [234.20, 270.70] \).
07

Analyze the Confidence Interval Result

The confidence interval \( [234.20, 270.70] \) does not contain the mean reported by the American Automobile Association (\$215.60), suggesting that the mean amount spent by families visiting Niagara Falls differs from the reported average.

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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 is a central concept in statistics. It is simply the average of a set of values or observations drawn from a subset, known as a sample, of a larger population. When we talk about the sample mean, we denote it as \( \bar{x} \). In our context here, the sample mean \( \bar{x} \) is calculated as \( \$252.45 \) per day, obtained from the group of families visiting Niagara Falls.
This number represents an estimate of the average daily spending for a family of four based on the observations from these 64 families.
  • The sample mean is essential because it gives us a point estimate of the population mean, which is what we aim to make inferences about.
  • The sample mean provides a snapshot of what the actual mean could be for the whole population, though it is subject to variability.

Using a sample mean allows researchers to infer characteristics of the larger population efficiently without the need to collect data from every family visiting the destination. It's a powerful tool in statistics because even a relatively small, well-chosen sample can provide informative insights into the broader trends and averages of an entire population.
Standard Deviation
Standard deviation is another crucial element in statistics. It measures the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the values are around the mean of the dataset. For our study on families visiting Niagara Falls, the standard deviation of the sample is \( \\(74.50 \).
This value indicates the degree to which each family's spending per day varies from the sample mean of \( \\)252.45 \).
  • A smaller standard deviation would suggest that the amounts spent per day by different families are close to the mean, indicating consistency among the data points.
  • A larger standard deviation, as seen here, means more variability in spending habits, suggesting less consistency and greater differences among the families.

Standard deviation is used in calculating the standard error, which is critical for constructing confidence intervals.
  • The standard error itself is the standard deviation of the sampling distribution of a statistic, commonly the mean.
  • It helps quantify the uncertainty associated with our sample mean estimate.
Z-Score
A Z-score, also known as a standard score, measures how many standard deviations an element is from the mean of the population. In the case of confidence intervals, the Z-score is particularly important because it helps determine the critical value that defines the interval's boundaries.
For a 95% confidence level, the Z-score is commonly set at 1.96.
  • The Z-score is derived from the standard normal distribution, specifically targeting the tails of the normal curve.
  • When we talk about a 95% confidence interval, it means there is a 95% probability that the interval contains the true population mean we are estimating.

This is vital because the Z-score scales our standard deviation to account for variations across the entire data set from our survey.
  • It helps us create margins around the sample mean, which effectively forms our confidence interval.
  • The wider the interval, the more confident we are that our sample mean approximates the population mean.
To summarize, the Z-score translates into a margin of error when multiplied by the standard error, determining how precise our confidence estimate is.

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Most popular questions from this chapter

Although airline schedules and cost are important factors for business travelers when choosing an airline carrier, a \(U S A\) Today survey found that business travelers list an airline's frequent flyer program as the most important factor. From a sample of \(n=1993\) business travelers who responded to the survey, 618 listed a frequent flyer program as the most important factor. a. What is the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier? b. Develop a \(95 \%\) confidence interval estimate of the population proportion. c. How large a sample would be required to report the margin of error of .01 at \(95 \%\) confidence? Would you recommend that \(U S A\) Today attempt to provide this degree of precision? Why or why not?

For a \(t\) distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746

An Employee Benefits Research Institute survey of 1250 workers over the age of 25 collected opinions on the health care system in America and on retirement planning \((A A R P\) Bulletin, January 2007 ). a. The American health care system was rated as poor by 388 of the respondents. Construct a \(95 \%\) confidence interval for the proportion of workers over 25 who rate the American health care system as poor. b. Eighty-two percent of the respondents reported being confident of having enough money to meet basic retirement expenses. Construct a \(95 \%\) confidence interval for the proportion of workers who are confident of having enough money to meet basic retirement expenses. c. Compare the margin of error in part (a) to the margin of error in part (b). The sample size is 1250 in both cases, but the margin of error is different. Explain why.

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. 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?

The National Quality Research Center at the University of Michigan provides a quarterly measure of consumer opinions about products and services (The Wall Street Journal, February 18,2003 ). A survey of 10 restaurants in the Fast Food/Pizza group showed a sample mean customer satisfaction index of \(71 .\) Past data indicate that the population standard deviation of the index has been relatively stable with \(\sigma=5\) a. What assumption should the researcher be willing to make if a margin of error is desired? b. Using \(95 \%\) confidence, what is the margin of error? c. What is the margin of error if \(99 \%\) confidence is desired?

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