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Chapter 10: Problem 11

Organic chemists often purify organic compounds by a method known as fractional crystallization. One chemist prepared ten \(4.85-\mathrm{g}\) quantities of aniline and purified it to acetanilide. The following dry yields were recorded: $$ \begin{array}{lllllll} 3.85 & 3.80 & 3.88 & 3.85 & 3.90 \\ 3.36 & 3.62 & 4.01 & 3.72 & 3.83 \end{array} $$ Estimate the mean grams of acetanilide that can be recovered from an initial amount of \(4.85 \mathrm{~g}\) of aniline. Use a \(95 \%\) confidence interval.

Short Answer

Expert verified
Based on our analysis, we estimate that the mean grams of acetanilide that can be recovered from an initial amount of 4.85g of aniline falls between 3.6593 and 3.9247 grams, with a 95% confidence level.

Step by step solution

01

Calculate Sample Mean

To calculate the sample mean, we sum all the dry yields and divide by the number of samples (10). $$ \bar{x} = \frac{3.85 + 3.80 + 3.88 + 3.85 + 3.90 + 3.36 + 3.62 + 4.01 + 3.72 + 3.83}{10} $$ Compute \(\bar{x}\): $$ \bar{x} = 3.792 $$
02

Find Sample Standard Deviation

To find the sample standard deviation, compute the average squared difference between each yield and the sample mean, and take the square root of that value. $$ s = \sqrt{\frac{\Sigma{(x_{i} - \bar{x})^2}}{n - 1}} $$ Compute s: $$ s \approx 0.1857 $$
03

Calculate Standard Error

The standard error is the sample standard deviation divided by the square root of the number of samples. $$ SE = \frac{s}{\sqrt{n}} $$ Compute SE: $$ SE \approx 0.0587 $$
04

Determine Critical Value

The critical value is the t-score corresponding to the \(95\%\) confidence level and \(9\) degrees of freedom. Using the t-table, or a calculator, we find that the critical value is approximately \(2.262\).
05

Calculate Margin of Error and Confidence Interval

The margin of error (ME) is the product of the critical value and the standard error. $$ ME = t \cdot SE $$ Compute ME: $$ ME \approx 0.1327 $$ The \(95\%\) confidence interval is given by the sample mean plus/minus the margin of error. $$ CI = (\bar{x} - ME, \bar{x} + ME) $$ Compute the CI: $$ CI = (3.792 - 0.1327, 3.792 + 0.1327) \approx (3.6593, 3.9247) $$ So, we estimate that the mean grams of acetanilide that can be recovered from an initial amount of \(4.85 \mathrm{~g}\) of aniline is between \(3.6593\) and \(3.9247\) grams, with a \(95\%\) confidence level.

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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 a crucial part of statistical analysis, especially when estimating the characteristics of a larger population. Here, we are using the sample mean to estimate how much acetanilide can be recovered from an initial amount of aniline.
To calculate the sample mean, simply add up all the yields from the samples and divide by the number of samples to get an average.
This calculation helps us represent the central point of the data we collected, and it is often denoted by \(\bar{x}\).
Understanding the sample mean:
  • It acts as a simple estimate of the true population mean.
  • Provides a measure of center or central tendency for the data.
  • In our exercise, the sample mean, \(\bar{x} = 3.792\), suggests the typical yield from aniline to acetanilide.
It's essential to always ensure the data collected is a random representative sample from the population to accurately use the sample mean for estimation.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values.
It tells us how much the yields varied around the sample mean in the chemist's experiment.
The more spread out the yields are, the higher the standard deviation will be, indicating less consistency in the purification process.
Key points about standard deviation:
  • Standard deviation is represented as \(s\) in sample data.
  • It is calculated as the square root of the variance.
  • In our exercise, \(s \approx 0.1857\) shows a moderate amount of variability in the yields.
Standard deviation is indispensable for understanding data reliability and guiding enhancements in processes like purification.
Standard Error
The standard error provides insight into the accuracy of the sample mean as an estimation of the true population mean.
It is derived from the standard deviation divided by the square root of the sample size \(n\).
A smaller standard error suggests that the sample mean is more precise as a representation of the population mean.
Understanding standard error:
  • It indicates how far the sample mean is expected to stray from the true population mean due to random sample variability.
  • It鈥檚 represented as \(SE\) and in our context, \(SE \approx 0.0587\).
  • The smaller the \(SE\), the more reliable the sample mean calculation becomes.
This metric aids in forming the confidence interval, which estimates the range in which the actual population mean is likely to fall.
Critical Value
The critical value is a factor used to calculate the margin of error in confidence intervals.
It depends on the confidence level desired and the degrees of freedom in the data set.
For a common confidence level of 95%, we use this value to stretch or shrink the confidence interval accordingly.
Key aspects of the critical value:
  • This value is typically derived from t-distribution when standard deviation is unknown and sample size is small.
  • It represents the t-score that corresponds to the desired confidence level, ensuring the interval captures the true mean a certain percentage of the time.
  • In our example, the critical value was approximately 2.262 based on the t-table for 9 degrees of freedom (as we are using 10 samples).
Understanding the critical value is essential for constructing confidence intervals, which provide insights into potential variability or the precision of our estimations.

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Most popular questions from this chapter

Refer to Exercise 10.94. Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: $$ \begin{array}{ccc} \text { Person } & \text { Stimulus } 1 & \text { Stimulus 2 } \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{array} $$ Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

Want to attend a pro-basketball finals game? The average prices for the NBA rematch of the Boston Celtics and the LA Lakers in 2010 compared to the average ticket prices in 2008 are given in the table that follows. \({ }^{18}\) $$ \begin{array}{lcc} \text { Game } & \text { 2008 (\$) } & \text { 2010 (\$) } \\ \hline 1 & 593 & 532 \\ 2 & 684 & 855 \\ 3 & 727 & 541 \\ 4 & 907 & 458 \\ 5 & 769 & 621 \\ 6 & 753 & 681 \\ 7 & 533 & 890 \end{array} $$ a. If we were to assume that the prices given in the table have been randomly selected, test for a significant difference between the 2008 and 2010 prices. Use \(\alpha=.01\) b. Find a \(98 \%\) confidence interval for the mean difference, \(\mu_{d}=\mu_{08}-\mu_{10} .\) Does this estimate confirm the results of part a?

Independent random samples from two normal populations produced the variances listed here: $$ \begin{array}{cc} \text { Sample Size } & \text { Sample Variance } \\ \hline 16 & 55.7 \\ 20 & 31.4 \end{array} $$ a. Do the data provide sufficient evidence to indicate that \(\sigma_{1}^{2}\) differs from \(\sigma_{2}^{2}\) ? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value.

Under what assumptions can the \(F\) distribution be used in making inferences about the ratio of population variances?

A production plant has two complex fabricating systems, both of which are maintained at 2 -week intervals. However, one system is twice as old as the other. The number of finished products fabricated daily by each of the systems is recorded for 30 working days, with the results given in the table. Do these data present sufficient evidence to conclude that the variability in daily production warrants increased maintenance of the older fabricating system? Use the \(p\) -value approach. $$ \begin{array}{ll} \text { New System } & \text { Old System } \\ \hline \bar{x}_{1}=246 & \bar{x}_{2}=240 \\ s_{1}=15.6 & s_{2}=28.2 \end{array} $$
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