91Ó°ÊÓ

Thirty fast-food restaurants including Wendy's, McDonald's, and Burger King were visited during the summer of 2000 (The Cincinnati Enquirer, July 9,2000 ). During each visit, the customer went to the drive-through and ordered a basic meal such as a "combo" meal or a sandwich, fries, and shake. The time between pulling up to the menu board and receiving the filled order was recorded. The times in minutes for the 30 visits are as follows: $$\begin{array}{cccccccc} 0.9 & 1.0 & 1.2 & 2.2 & 1.9 & 3.6 & 2.8 & 5.2 & 1.8 & 2.1 \\\6.8 & 1.3 & 3.0 & 4.5 & 2.8 & 2.3 & 2.7 & 5.7 & 4.8 & 3.5 \\ 2.6 & 3.3 & 5.0 & 4.0 & 7.2 & 9.1 & 2.8 & 3.6 & 7.3 & 9.0\end{array}$$ a. Provide a point estimate of the population mean drive-through time at fast- food restaurants. b. At \(95 \%\) confidence, what is the margin of error? c. What is the \(95 \%\) confidence interval estimate of the population mean? d. Discuss skewness that may be present in this population. What suggestion would you make for a repeat of this study?

Step by step solution

01

Calculate the Sample Mean

To estimate the population mean, we need to calculate the sample mean from the provided data. The sample mean is given by:\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]where \(n\) is the number of observations (30 in this case) and \(x_i\) are the given times. Adding up all times:\[0.9 + 1.0 + 1.2 + 2.2 + 1.9 + 3.6 + 2.8 + 5.2 + 1.8 + 2.1 + 6.8 + 1.3 + 3.0 + 4.5 + 2.8 + 2.3 + 2.7 + 5.7 + 4.8 + 3.5 + 2.6 + 3.3 + 5.0 + 4.0 + 7.2 + 9.1 + 2.8 + 3.6 + 7.3 + 9.0 = 101.2\]\[\bar{x} = \frac{101.2}{30} = 3.3733\]The point estimate of the population mean is approximately 3.37 minutes.

02

Calculate the Standard Deviation

The sample standard deviation helps us compute the margin of error. It is calculated using:\[s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}\]Calculating each \((x_i - \bar{x})^2\), summing them up, and taking their average gives the variance. Square root gives us \(s\). For this dataset, \\[s \approx 2.15\]

03

Compute the Margin of Error at 95% Confidence Level

The margin of error (ME) is calculated as:\[ME = z\frac{s}{\sqrt{n}}\]where \(z\) is the z-value for a 95% confidence level (1.96), \(s\) is the standard deviation, and \(n\) is the sample size. \Substituting the values:\[ME = 1.96 \cdot \frac{2.15}{\sqrt{30}} \approx 0.7693\]Thus, the margin of error is approximately 0.77 minutes.

04

Determine the Confidence Interval

The 95% confidence interval for the population mean is given by \((\bar{x} - ME, \bar{x} + ME)\):\[(3.3733 - 0.7693, 3.3733 + 0.7693) = (2.6040, 4.1426)\]Therefore, the 95% confidence interval is approximately (2.60, 4.14) minutes.

05

Discuss Skewness and Suggestions for Future Studies

By inspecting the data, we see higher values skewing the times (e.g., 9.1, 9.0), indicating positive skewness. This may imply that the average waiting time could be influenced by outliers. In a future study, considering the median or transforming the data may provide a more accurate representation.

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Key Concepts

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

Point Estimate

A point estimate provides a single, straightforward value that serves as an approximation of an unknown population parameter. In the context of our exercise, the point estimate of the population mean drive-through time is the sample mean calculated from the collected data. The formula for the point estimate is: \[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]where \( n \) is the number of observations, and \( x_i \) are the observed times.

• In our exercise, the sum of all drive-through times is 101.2 minutes.
• The number of observations (or visits) is 30.

Calculating the point estimate, we have:\[ \bar{x} = \frac{101.2}{30} = 3.3733\]Therefore, the point estimate of the population mean is approximately 3.37 minutes.

Sample Mean

The sample mean is key in statistics as it gives us a straightforward summary of data. It is used to estimate the central tendency of the population from which our sample is drawn. In our example, the sample mean was calculated by dividing the sum of all drive-through times by the number of observations. It represents the "average" wait time based on our sample data. Keep in mind that it is just one way of measuring central tendency, and it assumes that the data is symmetrically distributed. The sample mean is also used as a foundation for calculating other statistical metrics, such as variance and standard deviation, which help to quantify the variability in the data. Calculating the sample mean involves adding all data points and dividing by the sample size, as we saw in our exercise, resulting in a mean of approximately 3.37 minutes. This "average" helps summarize our dataset in a single value that reflects the overall trend of the observed data.

Standard Deviation

Standard deviation measures the extent of variation or dispersion in a set of values. A smaller standard deviation indicates that the values tend to be closer to the mean, while a larger standard deviation suggests greater variability. It is calculated by taking the square root of the variance.Here's the formula for standard deviation:\[s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}\]

• \( x_i \) refers to each individual data point.
• \( \bar{x} \) is the sample mean.
• \( n \) is the number of observations.

In the exercise, the standard deviation is approximately \(2.15\) minutes. This value shows how much individual drive-through times deviate, on average, from the mean drive-through time.

Margin of Error

The margin of error provides a range of values within which we can confidently say the true population parameter will fall. It accounts for random sampling variability and is associated with confidence intervals. For a 95% confidence level, the margin of error formula is:\[ME = z \frac{s}{\sqrt{n}}\]

• \( z \) is the z-value corresponding to the desired level of confidence (1.96 for 95% confidence).
• \( s \) is the standard deviation.
• \( n \) is the sample size.

In the exercise, the margin of error was calculated as approximately \(0.77\) minutes. It implies that if our sample accurately represents the population, the true mean drive-through time is expected to fall within \(0.77\) minutes of the sample mean, 95% of the time.

Skewness

Skewness reveals the asymmetry or bias in a dataset. Positive skewness indicates that the tail on the right side (higher values) is longer or fatter than the left side. Conversely, negative skewness shows a longer or fatter tail on the left side.In our drive-through time data, there are some notably high values, such as \(9.1\) and \(7.3\), suggesting a positive skew. This means most values are lower, but a few high numbers pull the mean upwards which results in a "tail" extending to the right. Skewness can impact the accuracy of statistical measures that assume normal distribution, like the mean.To address skewness in future studies, one could:

• Consider using the median, which is less sensitive to extreme values, as a measure of central tendency.
• Use data transformations to reduce skewness, such as logarithmic, square root, or other functions.

Address these options for a clearer picture of the dataset and minimize misleading influences of skewness.