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Chapter 8: Problem 31

A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of \(\bar{x}=1.10\) fluid ounce and a standard deviation of \(s=0.015\) fluid ounce. Find a \(95 \% \mathrm{CI}\) on the mean volume of syrup dispensed.

Short Answer

Expert verified
The 95% confidence interval is \([1.0938, 1.1062]\) fluid ounces.

Step by step solution

01

Understanding Confidence Interval

A confidence interval (CI) gives an estimated range believed to contain the population mean. Here, the machine releases syrup, and we know each sample's syrup content. We will use the sample data to estimate the average syrup content in all beverages made by this machine.
02

Determine the Variables

Given data: \(\bar{x} = 1.10\) fluid ounces is the sample mean, \(s = 0.015\) fluid ounces is the standard deviation, and \(n = 25\) is the sample size. The confidence interval we need to calculate is for the mean syrup content with a 95% confidence level.
03

Find the T-Value

Since the sample size is small (less than 30), we use the t-distribution. For 24 degrees of freedom (since \(n - 1 = 24\)) and a 95% confidence level, we find the t-value (often denoted as \(t_{\alpha/2}\)). From the t-distribution table, \(t_{0.025, 24} = 2.064\).
04

Calculate the Standard Error

The standard error (SE) of the mean is calculated using the formula \( SE = \frac{s}{\sqrt{n}} \). Substituting the given values: \[ SE = \frac{0.015}{\sqrt{25}} = 0.003 \] fluid ounces.
05

Compute the Margin of Error

The margin of error (E) can be computed using the formula \( E = t_{\alpha/2} \times SE \). Thus, \[ E = 2.064 \times 0.003 = 0.006192 \] fluid ounces.
06

Estimate the Confidence Interval

The 95% confidence interval for the mean is \( \bar{x} \pm E \). Therefore, the interval is \[ 1.10 \pm 0.006192, \] resulting in the range \( [1.0938, 1.1062] \) fluid ounces.

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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 (often represented as \( \bar{x} \)) is the average value obtained from a sample set. In the context of our beverage machine, the sample mean is calculated from the syrup content across 25 randomly selected drinks. This gives us a snapshot of how much syrup is on average being dispensed per drink, in our sample. For this exercise, the sample mean is \(1.10\) fluid ounces.

To calculate the sample mean, you sum all the observed individual values (syrup contents in our case) in the sample and divide by the number of observations. This measure is significant because it serves as an estimate for the population mean that we're trying to understand. However, note that while the sample mean provides an estimate, it may not exactly match the population mean due to random sampling variability.
T-Distribution
In statistics, the t-distribution is often used when dealing with small sample sizes, typically less than 30. It is similar to the normal distribution but has heavier tails, meaning there is a greater chance of extreme values. This accounts for the increased variability expected with smaller samples.

In our problem, the sample size is 25, which is considered small, prompting the use of the t-distribution. With 24 degrees of freedom (calculated as the sample size minus one), the t-value for a 95% confidence interval is found from statistical tables. Here, we use the value \(t_{0.025, 24} = 2.064\). The use of t-distribution ensures that the resulting confidence interval accurately reflects the uncertainty associated with the small sample size.
Standard Error
The standard error (SE) measures the spread of the sample mean from the population mean. It provides an estimation of how much variability one might expect among different sample means if we were to repeatedly sample from the population. This concept is crucial for constructing a confidence interval.

The standard error is calculated using the formula:
  • \( SE = \frac{s}{\sqrt{n}} \)
where \(s\) is the standard deviation of the sample and \(n\) is the sample size. In this exercise, substituting \(s = 0.015\) and \(n = 25\) gives us \(SE = 0.003\) fluid ounces. A smaller standard error indicates that the sample mean is a more accurate reflection of the population mean, which helps in creating a precise confidence interval.
Margin of Error
The margin of error (E) quantifies the uncertainty around the sample mean and is a key component in determining the confidence interval. It tells us the range within which we expect the population mean to lie, with a certain level of confidence (in this case, 95%).

The margin of error is calculated using the formula:
  • \( E = t_{\alpha/2} \times SE \)
For our problem, using the t-value of \(2.064\) and the standard error of \(0.003\), the margin of error is \(0.006192\) fluid ounces. This margin is then used to add and subtract from the sample mean (\(1.10\)), providing the confidence interval \([1.0938, 1.1062]\).

Understanding the margin of error helps you gauge the precision of the sample mean as an estimate of the population mean.

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Most popular questions from this chapter

Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years. (a) Calculate a \(95 \%\) two-sided confidence interval on the death rate from lung cancer. (b) Using the point estimate of \(p\) obtained from the preliminary sample, what sample size is needed to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.03 ?\) (c) How large must the sample be if you wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than \(0.03,\) regardless of the true value of \(p ?\)

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs hit was due at least in part to a livelier ball. One way to test the "liveliness" of a baseball is to launch the ball at a vertical surface with a known velocity \(V_{L}\) and measure the ratio of the outgoing velocity \(V_{0}\) of the ball to \(V_{L} .\) The ratio \(R=V_{0} / V_{L}\) is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. \(\begin{array}{llllll}0.6248 & 0.6237 & 0.6118 & 0.6159 & 0.6298 & 0.6192 \\\ 0.6520 & 0.6368 & 0.6220 & 0.6151 & 0.6121 & 0.6548 \\ 0.6226 & 0.6280 & 0.6096 & 0.6300 & 0.6107 & 0.6392 \\ 0.6230 & 0.6131 & 0.6223 & 0.6297 & 0.6435 & 0.5978 \\ 0.6351 & 0.6275 & 0.6261 & 0.6262 & 0.6262 & 0.6314 \\\ 0.6128 & 0.6403 & 0.6521 & 0.6049 & 0.6170 & \\ 0.6134 & 0.6310 & 0.6065 & 0.6214 & 0.6141 & \end{array}\) (a) Is there evidence to support the assumption that the coefficient of restitution is normally distributed? (b) Find a \(99 \%\) CI on the mean coefficient of restitution. (c) Find a \(99 \%\) prediction interval on the coefficient of restitution for the next baseball that will be tested. (d) Find an interval that will contain \(99 \%\) of the values of the coefficient of restitution with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

An article in the Journal of Agricultural Science ["The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects" (1997, Vol. 128, pp. \(135-142\) ) ] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the United Kingdom. The analysis used a variety of nitrogen fertilizer applications \((\mathrm{kg} \mathrm{N} / \mathrm{ha}),\) temperature \(\left({ }^{\circ} \mathrm{C}\right),\) and total monthly rainfall \((\mathrm{mm})\). The following data below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and \(1993 .\) The temperatures measured in June were obtained as follows: $$ \begin{array}{llllll} 15.2 & 14.2 & 14.0 & 12.2 & 14.4 & 12.5 \\ 14.3 & 14.2 & 13.5 & 11.8 & 15.2 & \end{array} $$ Assume that the standard deviation is known to be \(\sigma=0.5\) (a) Construct a \(99 \%\) two-sided confidence interval on the mean temperature. (b) Construct a \(95 \%\) lower-confidence bound on the mean temperature. (c) Suppose that you wanted to be \(95 \%\) confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used? (d) Suppose that you wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at \(95 \%\) confidence. What sample size should be used?

The confidence interval for a population proportion depends on the central limit theorem. A common rule of thumb is that to use the normal approximation for the sampling distribution for \(\hat{p},\) you should have at least 10 "successes" and 10 "failures." However, Agresti and Coull developed a method that can be used for smaller samples and increases the accuracy of all confidence intervals for proportions. The idea is simply to add 4 "pseudo observations" to the data set- 2 successes and 2 failures. That is, if you have \(X\) successes from of \(n\) trials, use \(\tilde{p}=\frac{X+2}{n+4}\) instead of the usual \(\hat{p}=\frac{X}{n}\) in the formulas for the confidence interval. A quality control engineer is inspecting defects on a newly designed printed circuit board. She inspects 50 boards and finds no defects. The usual estimate would be \(\hat{p}=0,\) but she does not believe that there will ever be a nodefects situation for this product. Use this Agresi-Coull estimate to come up with a \(95 \%\) confidence interval for the true proportion of defects. [Hint: The lower bound may be negative, which is not feasible, so simply use the upper limit and report that she is \(95 \%\) confident that the true defect proportion is no higher than this number.

The wall thickness of 25 glass 2 -liter bottles was measured by a quality- control engineer. The sample mean was \(\bar{x}=4.05\) millimeters, and the sample standard deviation was \(s=0.08\) millimeter. Find a \(95 \%\) lower confidence bound for mean wall thickness. Interpret the interval obtained.
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