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Chapter 8: Problem 84

Organic chemists often purify organic compounds by a method known as fractional crystallization. An experimenter wanted to prepare and purify \(4.85 \mathrm{g}\) of aniline. Ten 4.85 -gram specimens of aniline were prepared and purified to produce acetanilide. The following dry yields were obtained: $$\begin{array}{llllllll} 3.85, & 3.88, & 3.90, & 3.62, & 3.72, & 3.80, & 3.85, & 3.36, & 4.01, & 3.82 \end{array}$$ Construct a \(95 \%\) confidence interval for the mean number of grams of acetanilide that can be recovered from 4.85 grams of aniline.

Short Answer

Expert verified
The 95% confidence interval is (3.6463, 3.9157).

Step by step solution

01

Calculate the Sample Mean

First, add up all the yields to calculate the sample mean. The yields are: 3.85, 3.88, 3.90, 3.62, 3.72, 3.80, 3.85, 3.36, 4.01, and 3.82. The sum is 37.81. There are 10 data points, so the sample mean \( \bar{x} \) is: \[ \bar{x} = \frac{37.81}{10} = 3.781 \].
02

Calculate the Sample Standard Deviation

Calculate the standard deviation using the formula \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \), where \( n \) is the number of samples.Calculate each \((x_i - \bar{x})^2\) value and sum them, then divide by \(n-1=9\), and take the square root. After calculation, the sample standard deviation \( s \) is approximately \(0.1882\).
03

Determine the T-Value for 95% Confidence Level

For a 95% confidence interval with \( n-1 = 9 \) degrees of freedom, use a t-table to find the t-value, which is approximately \( t_{0.025,9} = 2.262 \).
04

Calculate the Margin of Error

The margin of error \( E \) can be calculated using the formula \( E = t \cdot \frac{s}{\sqrt{n}} \). Substitute \( t = 2.262 \), \( s = 0.1882 \), and \( n = 10 \). \[ E = 2.262 \times \frac{0.1882}{\sqrt{10}} \approx 0.1347 \].
05

Construct the Confidence Interval

To find the 95% confidence interval, use the formula \( \bar{x} \pm E \). Substitute \( \bar{x} = 3.781 \) and \( E = 0.1347 \).The confidence interval is: \( 3.781 - 0.1347 \) to \( 3.781 + 0.1347 \), which equals \[ (3.6463, 3.9157) \].

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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 is essentially the average of a set of data points in a sample. In statistical terms, it provides a single value that represents the entire sample. To calculate the sample mean, add up all the values in the dataset and then divide by the number of data points.

In our exercise, the yields of acetanilide were added to get a total of 37.81 grams. With 10 data points, we divided the total by 10 to obtain the sample mean, which was 3.781 grams. This value helps us estimate the average yield of acetanilide recovered. Understanding and calculating the sample mean is essential, as it serves as the starting point for constructing confidence intervals. This statistical measure tells us what we can expect from repeated experiments under the same conditions.
Sample Standard Deviation
The sample standard deviation quantifies the amount of variation or dispersion within a set of data points. It measures how spread out the numbers in the sample are from the mean.

In this context, to find the sample standard deviation, each data point is subtracted from the sample mean, and each result is squared. These squared differences are then summed and divided by the number of data points minus one ( -1), which represents degrees of freedom. Finally, the square root of this division provides the sample standard deviation.
  • A small standard deviation indicates that the data points are close to the sample mean.
  • A large standard deviation suggests a wide range of values.
From the exercise, the sample standard deviation was found to be approximately 0.1882 grams, indicating some variability in the acetanilide yields.
T-Value
The t-value is a critical value used in the creation of confidence intervals, especially when dealing with small sample sizes. It is derived from the t-distribution, which resembles a normal distribution but has thicker tails. This is particularly useful when the population standard deviation is unknown, and the sample size is small.

For a 95% confidence interval, with a sample size of 10, resulting in 9 degrees of freedom (n-1), the t-value from a t-table is 2.262. This t-value adjusts the confidence interval to account for the small sample size, making it wider than it would be if calculated with a z-distribution. This added width helps increase the certainty that the population mean falls within the interval.
Margin of Error
The margin of error determines how much the sample mean is expected to vary from the true population mean. It provides the range that the true mean is expected to fall into a specified percentage of the time.

The margin of error (E) is calculated by multiplying the t-value by the sample standard deviation, divided by the square root of the sample size. This formula provides the range above and below the sample mean.
  • It offers a buffer around the sample mean, making it a crucial element in constructing the confidence interval.
  • The wider the margin of error, the more uncertain we are about the population mean.
In the exercise, the margin of error was calculated to be approximately 0.1347 grams. This means the actual mean yield could vary by this amount on either side of the sample mean, providing a buffer for potential variability in future results.

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Most popular questions from this chapter

In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were firstborn or only children; in a sample of 100 non-graduates of comparable age and socioeconomic background, the number of firstborn or only children was \(54 .\) Estimate the difference in the proportions of firstborn or only children for the two populations from which these samples were drawn. Give a bound for the error of estimation.

Sometimes surveys provide interesting information about issues that did not seem to be the focus of survey initially. Results from two CNN/USA Today/Gallup polls, one conducted in March 2003 and one in November \(2003,\) were recently presented online. \(^{\star}\) Both polls involved samples of 1001 adults, aged 18 years and older. In the March sample, \(45 \%\) of those sampled claimed to be fans of professional baseball whereas \(51 \%\) of those polled in November claimed to be fans. a. Give a point estimate for the difference in the proportions of Americans who claim to be baseball fans in March (at the beginning of the season) and November (after the World Series). Provide a bound for the error of estimation. b. Is there sufficient evidence to conclude that fan support is greater at the end of the season? Explain.

Suppose that independent samples of sizes \(n_{1}\) and \(n_{2}\) are taken from two normally distributed populations with variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2},\) respectively. If \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the respective sample variances, Theorem 7.3 implies that \(\left(n_{1}-1\right) S_{1}^{2} / \sigma_{1}^{2}\) and \(\left(n_{2}-1\right) S_{2}^{2} / \sigma_{2}^{2}\) have \(\chi^{2}\) distributions with \(n_{1}-1\) and \(n_{2}-1\) df, respectively. Further, these \(\chi^{2}\) -distributed random variables are independent because the samples were independently taken. a. Use these quantities to construct a random variable that has an \(F\) distribution with \(n_{1}-1\) numerator degrees of freedom and \(n_{2}-1\) denominator degrees of freedom. b. Use the \(F\) -distributed quantity from part (a) as a pivotal quantity, and derive a formula for a \(100(1-\alpha) \%\) confidence interval for \(\sigma_{2}^{2} / \sigma_{1}^{2}\)

An investigator is interested in the possibility of merging the capabilities of television and the Internet. A random sample of \(n=50\) Internet users yielded that the mean amount of time spent watching television per week was 11.5 hours and that the standard deviation was 3.5 hours. Estimate the population mean time that Internet users spend watching television and place a bound on the error of estimation.

The sample mean \(\bar{Y}\) is a good point estimator of the population mean \(\mu\). It can also be used to predict a future value of \(Y\) independently selected from the population. Assume that you have a sample mean \(\bar{Y}\) and variance \(S^{2}\) based on a random sample of \(n\) measurements from a normal population. Use Student's \(t\) to form a pivotal quantity to find a prediction interval for some new value of \(Y-\) say, \(X_{p}-\) to be observed in the future. [Hint: Start with the quantity \(Y_{p}-\bar{Y} .\) ] Notice the terminology: Parameters are estimated; values of random variables are predicted.
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