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In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of \(\$ 5\) a. At \(95 \%\) confidence, what is the margin of error? b. If the sample mean is \(\$ 24.80\), what is the \(95 \%\) confidence interval for the population mean?

Short Answer

Expert verified
a. Margin of error is \(1.40\). b. Confidence interval is \((23.40, 26.20)\).

Step by step solution

01

Identify Given Information

Before solving the problem, identify the essential information given:- Sample size, \( n = 49 \)- Population standard deviation, \( \sigma = 5 \)- Sample mean, \( \bar{x} = 24.80 \)- Confidence level, \( 95\% \) which means \( \alpha = 0.05 \).
02

Calculate Z-Score for Confidence Level

For a \( 95\% \) confidence level, the Z-score corresponding to \( 1 - (\alpha / 2) = 0.975 \) is found in statistical Z-tables. The Z-score is \( Z = 1.96 \).
03

Calculate the Margin of Error

The margin of error (ME) can be calculated using the formula: \[ ME = Z \times \frac{\sigma}{\sqrt{n}}\]Substitute the known values: \[ ME = 1.96 \times \frac{5}{\sqrt{49}} = 1.96 \times \frac{5}{7} = 1.96 \times 0.7143 \]Calculate:\[ ME \approx 1.40 \]
04

Calculate the Confidence Interval

The \( 95\% \) confidence interval for the population mean is given by \[ \left(\bar{x} - ME, \bar{x} + ME\right)\]Using \( \bar{x} = 24.80 \) and \( ME = 1.40 \), compute:\[ \text{Lower Limit} = 24.80 - 1.40 = 23.40\]\[ \text{Upper Limit} = 24.80 + 1.40 = 26.20\]Thus, the confidence interval is \((23.40, 26.20)\).

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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 an important concept when estimating a population parameter, such as the mean amount spent per customer in a restaurant. It provides a range within which we can be fairly confident that the true population mean falls.
To calculate the margin of error, we use the formula:
  • Margin of Error (ME) = \( Z \times \frac{\sigma}{\sqrt{n}} \)
Here, \( Z \) is the Z-score associated with the desired confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
In our scenario, with a 95% confidence level, the Z-score is 1.96. The margin of error tells us how much we expect the sample mean to vary from the true population mean.
As calculated in the step-by-step solution, the margin of error is approximately 1.40. This means we can be 95% confident that the true mean falls within about 1.40 units above or below the sample mean we observed.
Population Mean
Understanding the population mean is crucial in statistics, especially when it's not feasible to measure an entire population. We often want to estimate what the average, or mean, of a certain measurement is across a whole group.
The population mean is denoted by \( \mu \) and is often unknown because it involves measuring the entire population, which isn't usually possible.
Instead, we use a sample mean, \( \bar{x} \), calculated from a subset of the population. In our original exercise, this sample mean was given as \( 24.80 \).
To estimate the range where the actual population mean might lie, we create a confidence interval.In our case, the calculated confidence interval at a 95% confidence level was \((23.40, 26.20)\), as shown in the original solution. This indicates that with 95% confidence, the average amount spent per customer falls between these values.
Sample Size
Sample size (denoted as \( n \)) is the number of observations or data points collected in a study. It plays a crucial role in determining the reliability of our estimates. The larger the sample size, the more accurate our estimation of the population mean is likely to be, since more data gives a better representation of the entire population.
The formula for the margin of error includes \( n \), in the term \( \sqrt{n} \). This indicates that as the sample size increases, the margin of error typically decreases, leading to a more precise confidence interval. In our exercise, the sample size was 49 customers. The relatively large sample size helps reduce the uncertainty in our estimation of the population mean, making the interval more reliable.
When planning a study, selecting an appropriate sample size is crucial to ensure accurate results without unnecessary data collection efforts.

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Most popular questions from this chapter

According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance (CNBC, February 23,2006 ). Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a \(95 \%\) confidence interval for the population proportion.

An online survey by ShareBuilder, a retirement plan provider, and Harris Interactive reported that \(60 \%\) of female business owners are not confident they are saving enough for retirement (SmallBiz, Winter 2006). Suppose we would like to do a follow-up study to determine how much female business owners are saving each year toward retirement and want to use \(\$ 100\) as the desired margin of error for an interval estimate of the population mean. Use \(\$ 1100\) as a planning value for the standard deviation and recommend a sample size for each of the following situations. a. \(\quad\) A \(90 \%\) confidence interval is desired for the mean amount saved. b. \(\quad\) A \(95 \%\) confidence interval is desired for the mean amount saved. c. \(\quad\) A \(99 \%\) confidence interval is desired for the mean amount saved.

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

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