/*! 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 27 This wine stinks, Sulfur compoun... [FREE SOLUTION] | 91Ó°ÊÓ

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This wine stinks, Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine \((\mu \mathrm{g} / \mathrm{L})\). The untrained noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10 untrained students: II WnNer \(\begin{array}{llllllllll}30 & 30 & 42 & 35 & 22 & 33 & 31 & 29 & 19 & 23\end{array}\) (a) Assume that the standard deviation of the odor threshold for untrained noses is known to be \(\sigma=7 \mu \mathrm{g} / \mathrm{L}\). Briefly discuss the other two "simple conditions," using a stemplot to verify that the distribution is roughly symmetric with no outliers. (b) Give a \(95 \%\) confidence interval for the mean DMS odor threshold among all students.

Short Answer

Expert verified
The 95% confidence interval for the mean DMS odor threshold is (25.07, 33.73) µg/L.

Step by step solution

01

Verify Conditions

To use the normal distribution to find a confidence interval for the mean, we must check three conditions: (1) Random Sampling, (2) Independence, and (3) Normality. Assuming the sample is random and each student's response is independent, the third condition requires visual confirmation of normality or approximate symmetry without outliers using a stem plot.
02

Create a Stemplot

Make a stem-and-leaf plot using the given data to check the distribution for symmetry and outliers. Here is the data as a stemplot: ``` 1 | 9 2 | 2 3 9 3 | 0 0 1 3 5 4 | 2 ``` The plot shows a roughly symmetric distribution centered around the 30s, with no obvious outliers.
03

Calculate Standard Error

The standard error (SE) is calculated using the formula \[ SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma = 7 \mu \text{g}/L \) and \( n = 10 \). Thus, \[ SE = \frac{7}{\sqrt{10}} \approx 2.21 \].
04

Find the Mean

Calculate the mean by summing all values and dividing by the number of observations: \[ \bar{x} = \frac{30 + 30 + 42 + 35 + 22 + 33 + 31 + 29 + 19 + 23}{10} = 29.4 \mu \text{g}/L \].
05

Construct Confidence Interval

With \( \bar{x} = 29.4 \), and using the standard normal value for a 95% confidence level (\( z \approx 1.96\)), compute the confidence interval:\[ \bar{x} \pm z \times SE = 29.4 \pm 1.96 \times 2.21 \].\[ \text{Confidence interval: } (29.4 - 4.33, 29.4 + 4.33) = (25.07, 33.73) \].

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Key Concepts

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

Standard Error
The standard error (SE) plays a crucial role when estimating the accuracy of a sample mean as a representation of the entire population mean. It provides a measure of how much variability exists between the sample mean and the actual population mean. This is especially useful in statistics, as it helps determine the reliability of a sample-based estimate.

To calculate the standard error, we use the formula:
  • \( SE = \frac{\sigma}{\sqrt{n}} \)
  • Where \(\sigma\) is the population standard deviation.
  • And \(n\) is the number of observations in the sample.
In our specific example regarding wine tasters, the population standard deviation \(\sigma\) is given as 7 µg/L, and the sample size \(n\) is 10. Thus, the standard error is calculated as \(SE = \frac{7}{\sqrt{10}} \approx 2.21\).

Understanding the standard error allows us to construct confidence intervals, providing a range where we can confidently say the population mean likely lies. As the standard error decreases, which typically happens when sample sizes increase, our confidence in the sample mean being close to the actual population mean increases.
Normal Distribution
The concept of normal distribution is fundamental when it comes to understanding data setting and hypothesis testing. When a dataset follows a normal distribution, it is symmetric around the mean - meaning that most observations cluster around the mean and the probabilities for values to be extreme (very high or very low) decrease symmetrically in both directions.

For confidence interval estimations and many other statistical procedures, assuming that data are approximately normally distributed lets us model and simplify data behavior efficiently. With normal distribution, the mean, median, and mode align, giving rise to the classic "bell curve" shape. This characteristic makes it easier to predict the probability of occurrences within a dataset.
  • A normal distribution is shaped by the parameters \(\mu\) (mean) and \(\sigma\) (standard deviation).
In exercises like the one we have, confirming that data adhere to a normal distribution means checking these criteria and ensuring no skewness or outliers can pose issues.

The initial assumption of the untrained students' odor threshold following a normal distribution is validated by creating visual aids such as stemplots, which help in quickly checking for symmetry and absence of outliers, ensuring we can confidently use the normal model for further statistical analysis.
Stemplot
Stemplots, also known as stem-and-leaf plots, are a simple, yet powerful, tool for visualizing the distribution of small datasets. They allow us to see each data point individually, while also providing a visual summary of how data are spread and distributed.

To create a stemplot, each number in the dataset is divided into a "stem" (consisting of all but the final digit) and a "leaf" (the final digit). For instance, for the number 42, the stem is 4 and the leaf is 2. This format helps in organizing data where both the spread and individual data points can be easily interpreted.

A stemplot for the given dataset:
  • 1 | 9
  • 2 | 2 3 9
  • 3 | 0 0 1 3 5
  • 4 | 2
This representation shows that the data is roughly symmetric around the 30s, and there are no significant outliers disrupting the pattern. Such details are critical when checking for conditions like normality, which are necessary for applying many statistical tests including confidence intervals.

This visual check confirms that our data set does not have extreme skewness or unusual values, paving the way for accurate statistical analysis and estimation efforts.

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Most popular questions from this chapter

Why are larger samples better? Statisticians prefer large samples. Describe briefly the effect of increasing the size of a sample on the margin of error of a \(95 \%\) confidence interval.

You want a \(99 \%\) confidence interval for the true weight of this specimen. The margin of error for this interval will be Fill weIGHTs (a) smaller than the margin of error for \(95 \%\) confidence. (b) greater than the margin of error for \(95 \%\) confidence. (c) about the same as the margin of error for \(95 \%\) confidence.

To give a \(99.9 \%\) confidence interval for a population mean \(\mu\), you would use the critical value (a) \(z^{*}=1.960\). (b) \(z^{*}=2.576\). (c) \(x^{+}=3.291\). Use the following information for Exercises \(16.12\) through 16.14. A laboratory scale is known to have a standard deviation of \(\sigma=0.001\) gram in repeated weighings. Scale readings in repeated weighings are Normally distributed, with mean equal to the true weight of the specimen. Three weighings of a specimen on this scale give \(3.412,3.416\), and \(3.414\) grams.

Pulling wood apart. How heavy a load (in pounds) is needed to pull apart päeces of Douglas fir 4 inches long and \(1.5\) inches square? Here are data from students doing a laboratory exercise: wood $$ \begin{array}{lllll} 33,190 & 31,860 & 32,590 & 26,520 & 33,280 \\ 32,320 & 33,020 & 32,030 & 30,460 & 32,700 \\ 23,040 & 30,930 & 32,720 & 33,650 & 32,340 \\ 24,050 & 30,170 & 31,300 & 28,730 & 31,920 \end{array} $$ (a) We are willing to regard the wood pleces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that characteristics of materials vary Normally. Make a graph to show the shape of the distribution for these data. Does it appear safe to assume that the Normality condition is satisfied? Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds. (b) Give a \(95 \%\) confidence interval for the mean load required to pull the wood apart.

I want more muscle. Many young men in North America and Europe (but not in Asia) tend to think they need more muscle to be attractive. One study presented 200 young American men with 100 images of men with various levels of muscle. \({ }^{7}\) Researchers measure level of muscle in kilograms per square meter \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\) of fat-free body mass. Typical young men have about \(20 \mathrm{~kg} / \mathrm{m}^{2}\). Each subject chose two images, one that represented his own level of body muscle and one that he thought represented "what women prefer." 'The mean gap between self- image and "what women prefer" was \(2.35 \mathrm{~kg} / \mathrm{m}^{2}\). Suppose that the "muscle gap" in the population of all young men has a Normal distribution with standard deviation \(2.5 \mathrm{~kg} / \mathrm{m}^{2}\). Give a \(90 \%\) confidence interval for the mean amount of muscle young men think they should add to be attractive to women. (They are wrong: women actually prefer a level close to that of typical men.)

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