91Ó°ÊÓ

The quality-control manager of a Long John Silver's restaurant wants to analyze the length of time that a car spends at the drive-through window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right (data based on information provided by Danica Williams, student at Joliet Junior College). (a) To obtain probabilities regarding a sample mean using the normal model, what size sample is required? (b) The quality-control manager wishes to use a new delivery system designed to get cars through the drive-through system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less, assuming that the population mean is 59.3 seconds? Do you think that the new system is effective? (c) Treat the next 50 cars that arrive as a simple random sample. There is a \(15 \%\) chance the mean time will be at or below ____ seconds

Step by step solution

01

Title - Define sample size requirement

To answer part (a), use the Central Limit Theorem which states that regardless of the population distribution, the sampling distribution of the sample mean will be approximately normal for a sufficiently large sample size. Generally, a sample size of 30 or more is considered large enough.

02

Title - Computing the sampling distribution

Calculate the sampling distribution's mean and standard deviation (\(\frac{\text{population standard deviation}}{\text{square root of sample size}}\)).

03

Title - Calculating the probability

To answer part (b), the formula for the standard error is: \[SE = \frac{13.1}{\sqrt{40}}\.\] Calculate using: \[SE = \frac{13.1}{6.32} \approx 2.073\.\]

04

Title - Convert to Z-score

To find the probability, convert the sample mean to a standard normal Z-score: \[Z = \frac{\text{Sample Mean} - \text{Population Mean}}{SE} \] Calculate using: \[Z = \frac{56.8 - 59.3}{2.073} \approx -1.208\.\]

05

Title - Find the probability from Z-score

Check the Z-table for the corresponding probability of Z = -1.208, results in a probability of approximately 0.113.

06

Title - Determine significance

Since the probability of obtaining a sample mean of 56.8 seconds or less is not very low (approximately 11.3%), it suggests that the new system may not be significantly faster. Hence, it may not be effective.

07

Title - Calculating probability for next 50 cars

For part (c), we are given a 15% chance of the mean time being below a certain value. Using the Z-table, find that a Z-score of -1.036 corresponds to the 15th percentile.

08

Title - Solving for the 15th percentile

Using the Z-score formula:\[Z = \frac{X - \mu}{SE}\] solve for X:\[ -1.036 = \frac{X - 59.3}{\frac{13.1}{\sqrt{50}}}\]Solve: \[ -1.036 * \frac{13.1}{7.07}\ + 59.3 \approx 57\.\] seconds

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Key Concepts

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

Normal Distribution

The concept of a normal distribution is essential in statistics. It describes a symmetrical, bell-shaped curve where most observations cluster around the central peak. In real-world terms, if you were to measure something like the time a car spends at a drive-through, the times would generally hover around the average.
Even when the actual data distribution isn't perfect, the Central Limit Theorem tells us that with a large enough sample size, the distribution of sample means will follow a normal distribution.
If you're trying to figure out probabilities or other statistics from real data, assuming a normal distribution can simplify your calculations and help draw meaningful conclusions.

Sample Size Determination

Sample size is crucial for accurate statistical analysis. The Central Limit Theorem helps us here. It shows that the mean of a sufficiently large sample size from any population will approximate a normal distribution, even if the population itself is not normally distributed.
For our calculations, a sample size of 30 or more is generally considered large enough. This is why you often see this number used as a benchmark in statistics. Whether you're analyzing drive-through times or any other data, starting with at least 30 data points can offer more reliable insights.

Standard Error

Standard error measures the accuracy of a sample mean by indicating how much the sample mean differs from the true population mean. It's calculated as:
\[ SE = \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \]
For example, in our case, the population standard deviation is 13.1 seconds and the sample size is 40:
\[ SE = \frac{13.1}{\sqrt{40}} \approx 2.073 \]
Smaller standard errors suggest that the sample mean is closer to the true population mean. This is key for making accurate inferences from sample data.

Z-score Calculation

Z-scores are a way to standardize individual data points relative to the mean and standard deviation. The formula to convert a sample mean to a Z-score is:
\[ Z = \frac{\text{Sample Mean} - \text{Population Mean}}{SE} \]
Let's say the sample mean time at the window is 56.8 seconds, with a population mean of 59.3 seconds and an SE of 2.073:
\[ Z = \frac{56.8 - 59.3}{2.073} \approx -1.208 \]
The Z-score helps you find how that sample mean relates to the overall population, giving you an idea of how likely it is to obtain such a mean by chance.

Probability Calculation

Probability calculations using the Z-score help assess how unusual or usual a sample mean is. For example, a Z-score of -1.208 corresponds to a probability of approximately 0.113.
This means there's about an 11.3% chance of obtaining a sample mean of 56.8 seconds or less, assuming the population mean is 59.3 seconds. If we wanted to determine the cutoff time where 15% of the times fall below it for the next 50 cars, we would use a Z-score of -1.036:
\[ -1.036 = \frac{X - 59.3}{\frac{13.1}{\sqrt{50}}} \approx 57 \text{ seconds} \]
Such calculations help decision-makers determine if changes like a new delivery system are materially significant.