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Chapter 14: Problem 449

In Grand Central Station, there is a coffee machine which is regulated so that the amount of coffee dispensed is normally distributed with a standard deviation of \(.5\) ounces per cup. A random sample of 50 cups had an average of 5 ounces per cup. (a) Find the maximum likelihood estimate for the average amount of coffee in each cup dispensed by the machine. (b) Find a \(95 \%\) confidence interval for the mean of all cups dispensed.

Short Answer

Expert verified
(a) The maximum likelihood estimate for the average amount of coffee in each cup dispensed by the machine is 5 ounces per cup. (b) The 95% confidence interval for the mean of all cups dispensed is approximately (4.8609, 5.1391) ounces.

Step by step solution

01

Part (a): Maximum Likelihood Estimate

The maximum likelihood estimate for the average amount of coffee in each cup dispensed by the machine is the sample mean, which we are given as 5 ounces per cup. This means that our best estimate for the average amount of coffee in each cup dispensed is 5 ounces.
02

Part (b): Computing Z Score for 95% Confidence Interval

To obtain the \(Z_{\frac{\alpha}{2}}\) value for a 95% confidence interval, we need to look at the standard normal table (Z-table). For a 95% confidence level, \(\alpha = 0.05\), so we want to look for the Z-score that corresponds to an area of \(1 - \frac{\alpha}{2} = 0.975\). Looking at the Z-table, we find that \(Z_{\frac{\alpha}{2}} \approx 1.96\).
03

Part (b): Calculating the 95% Confidence Interval

Now, we can use the formula for the confidence interval of the mean: \( CI = (\bar{x} - Z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}, \bar{x} + Z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}) \) Plugging in the given information and the calculated \(Z_{\frac{\alpha}{2}}\) value: \( CI = (5 - 1.96 \cdot \frac{0.5}{\sqrt{50}}, 5 + 1.96 \cdot \frac{0.5}{\sqrt{50}}) \) Calculating the values inside the parentheses: \( CI = (5 - 0.1391, 5 + 0.1391) \) \( CI = (4.8609, 5.1391) \) The 95% confidence interval for the mean of all cups dispensed is approximately (4.8609, 5.1391) ounces.

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Key Concepts

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

Maximum Likelihood Estimate
When we talk about the maximum likelihood estimate (MLE), we're referring to a statistical method used to find the most probable value of a parameter that would give rise to the observed data. In simpler terms, it's our best guess at an unknown quantity based on the sample data.

In the context of the Grand Central Station coffee machine, the MLE for the average amount of coffee was determined to be 5 ounces per cup. This was calculated from a random sample of 50 cups, giving us a clue about the general performance of the coffee machine. It's worth noting that the MLE can be particularly effective when we have a larger sample size, which tends to give a more accurate estimate. Furthermore, the MLE is a point estimate, meaning it gives us a single value rather than a range.
Standard Deviation
Moving on to the concept of standard deviation, this is a measure that tells us how spread out the values in a data set are. You can think of it like the average 'distance' each data point is from the mean. A low standard deviation indicates that the data points tend to be very close to the mean, while a high standard deviation shows that the data points are spread out over a wider range of values.

For the coffee machine, a standard deviation of 0.5 ounces indicates that, on average, the amount of coffee dispensed doesn't vary too much cup to cup. This reliability is crucial for quality control and ensures customer satisfaction with consistent product delivery. The standard deviation helps us quantify this consistency鈥攐r lack thereof.
Normal Distribution
Lastly, the normal distribution is a fundamental concept in statistics, often symbolized by a bell curve when graphed. Many natural phenomena and measurement observations, including errors, human characteristics, and in our case, the dispensed coffee, follow this type of distribution, assuming that a random variable is equally likely to lie above or below the mean.

When data is normally distributed, we can make powerful inferences about a population using the sample data. In the coffee machine scenario, the assumption of normal distribution allows us to apply the Z-score and calculate a confidence interval for the average amount dispensed by the machine. A key characteristic of the normal distribution is that about 68% of values are within one standard deviation of the mean, and 95% are within two standard deviations, which relates to the confidence interval we compute for making probabilistic statements about the population mean.

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Most popular questions from this chapter

A group of experts feel that polishing would have a positive effect on the average endurance limit of steel. Eight specimens of polished steel were measured and their average, \(\underline{X}\), was computed to be 86,375 psi. Similarly, 8 specimens of unpolished steel were measured and their average, \(\underline{Y}\), was computed to be \(79,838 \mathrm{psi}\). Assume all measurements to be normally distributed with standard deviation 4,000 psi (both samples). Find a \(90 \%\) lower one-sided confidence limit for \(\mu_{\mathrm{X}}-\mu_{\mathrm{Y}}\), the difference in means.

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In the Idaho State Home for Runaway Girls, 25 residents were polled as to what age they ran away from home. The sample mean was 16 years old with a standard deviation of \(1.8\) years. Establish a \(95 \%\) confidence interval for \(\mu\), the mean age at which runaway girls leave home in Idaho.

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