/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A postmix beverage machine is ad... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(95 \%\) 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

Identify the Given Data

We are given a sample size of 25 beverages \((n = 25)\), a sample mean syrup content \((\bar{x} = 1.10)\) fluid ounces, and a sample standard deviation \((s = 0.015)\) fluid ounces. We need to find a 95% confidence interval for the true mean syrup content.
02

Determine the t-Value

Since the sample size is small \((n < 30)\), we'll use the t-distribution to calculate the confidence interval. For a 95% confidence level with \(n-1 = 24\) degrees of freedom, look up the t-distribution table or use statistical software to find the critical t-value \(t_{0.025}\), which is approximately 2.064.
03

Calculate the Standard Error

Use the formula for the standard error of the mean (SEM): \(SEM = \frac{s}{\sqrt{n}} = \frac{0.015}{\sqrt{25}} = 0.003\).
04

Calculate the Margin of Error

The margin of error is calculated as \(ME = t \times SEM = 2.064 \times 0.003 = 0.006192\).
05

Find the Confidence Interval

The confidence interval is given by \(\bar{x} \pm ME\). Thus, the 95% confidence interval for the mean syrup content is \(1.10 \pm 0.006192\), which results in the interval \((1.093808, 1.106192)\) fluid ounces.

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Key Concepts

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

t-distribution
When conducting statistical analyses, particularly with small sample sizes, the t-distribution becomes a vital tool. Unlike the normal distribution, which assumes known population variance, the t-distribution accounts for the additional uncertainty brought by estimating this variance from a small sample.
The shape of a t-distribution is determined by degrees of freedom, which are typically the sample size minus one ( -1"). The fewer the degrees of freedom, the fatter and more spread out the tails of the distribution become. This implies a greater likelihood of larger values, reflecting more variability in your data estimate compared to a normal distribution.
  • Use the t-distribution when your sample size is below 30 and the population standard deviation is unknown.
  • The heavier tails of the t-distribution increase the margin for error, adding to the confidence of capturing the true mean.
In the exercise, with a sample size of 25, we have 24 degrees of freedom. Thus, using the t-distribution is appropriate for calculating the confidence interval.
standard error
The standard error of the mean (SEM) measures the variability of sample means from the true population mean. It is critical in estimating how much sample means are likely to fluctuate from one sample to another. The formula for SEM is important to understand:\[SEM = \frac{s}{\sqrt{n}}\]where \(s\) is the sample standard deviation, and \(n\) is the sample size. The standard error decreases as the sample size increases, reflecting more precision in estimating the population mean. Understanding SEM helps in judging how representative a sample is of its population.
  • A smaller SEM indicates a more accurate estimate of the population mean.
  • If the SEM is large, it suggests that the sample mean may not be very close to the true population mean.
In the exercise, the SEM for 25 beverages was calculated as 0.003 fluid ounces, reflecting how the sample means might vary from the actual mean of syrup content.
margin of error
Margin of Error (ME) provides a range within which we expect the true population parameter to lie, with a certain level of confidence—in this case, 95%. This takes into account the SEM and adds the variability captured by our chosen confidence level. The formula is:\[ME = t \times SEM\]where \(t\) is the t-value for a 95% confidence interval. The margin of error shows potential for variation in your results, based on your sample's data.
  • If the ME is small, the precision of your confidence interval is higher.
  • The larger the ME, the less precise the confidence interval is, indicating variability in the mean estimate.
In this scenario, the ME was approximately 0.0062 fluid ounces, indicating how much the sample mean might differ from the true mean syrup content.
sample mean
The sample mean, often represented as \(\bar{x}\), is the average of a sample dataset and is used as an estimate of the population mean. It is the sum of all sampled values divided by the number of samples. In many practical cases, the sample mean is used because it's typically impossible to survey entire populations.The formula for the sample mean is:\[\bar{x} = \frac{\sum{x_i}}{n}\]where \(\sum{x_i}\) represents the sum of all observations in the sample, and \(n\) is the number of observations.
  • The sample mean offers a central value around which other measures, such as the standard error and margin of error, are calculated.
  • In hypothesis testing, the sample mean is crucial in making inferences about the population.
In the given exercise, a sample mean of 1.10 fluid ounces was used to estimate the average syrup content dispensed by the machine.

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Most popular questions from this chapter

For a normal population with known variance \(\sigma^{2}\), answer the following questions: (a) What is the confidence level for the interval \(\bar{x}-2.14 \sigma / \sqrt{n}\) \(\leq \mu \leq \bar{x}+2.14 \sigma / \sqrt{n} ?\) (b) What is the confidence level for the interval \(\bar{x}-2.49 \sigma / \sqrt{n}\). \(\leq \mu \leq \bar{x}+2.49 \sigma / \sqrt{n} ?\) (c) What is the confidence level for the interval \(\bar{x}-1.85 \sigma / \sqrt{n}\). \(\leq \mu \leq \bar{x}+1.85 \sigma / \sqrt{n} ?\)

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During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs that were 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 \(V_{O}\) of the ball to \(V_{L}\). The ratio \(R=V_{O} / V_{L}\) 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 \(99 \% \mathrm{CI}\) 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).

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