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High concentration of the toxic element arsenic is all too common in groundwater. The article "Evaluation of Treatment Systems for the Removal of Arsenic from Groundwater" (Practice Periodical of Hazardous, Toxic, and Radioactive Waste Mgmt., 2005: 152-157) reported that for a sample of \(n=5\) water specimens selected for treatment by coagulation, the sample mean arsenic concentration was \(24.3 \mu \mathrm{g} / \mathrm{L}\), and the sample standard deviation was 4.1. The authors of the cited article used \(t\)-based methods to analyze their data, so hopefully had reason to believe that the distribution of arsenic concentration was normal. a. Calculate and interpret a \(95 \%\) CI for true average arsenic concentration in all such water specimens. b. Calculate a \(90 \%\) upper confidence bound for the standard deviation of the arsenic concentration distribution. c. Predict the arsenic concentration for a single water specimen in a way that conveys information about precision and reliability.

Short Answer

Expert verified
The 95% CI for mean is \([19.22, 29.38]\), the 90% upper bound for standard deviation is \(5.67\), and the prediction interval for a single sample is \([15.532, 33.068]\).

Step by step solution

01

Identify the Parameters and Distributions

Given: Sample size \(n=5\), sample mean \(\bar{x} = 24.3 \mu g/L\), and sample standard deviation \(s = 4.1\). Since the sample size is small, we will use the \(t\)-distribution for calculating the confidence intervals.
02

Calculate the 95% Confidence Interval for the Mean

The formula for a \(95\%\) confidence interval using the \(t\)-distribution is \[\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]. The degrees of freedom are \(df = n - 1 = 4\). Use a \(t\)-table to find \(t_{0.025}\) for \(df = 4\), which is approximately \(2.776\). Substitute into the formula:\[24.3 \pm 2.776 \cdot \frac{4.1}{\sqrt{5}}\].This gives the interval: \[24.3 \pm 5.08\], so the \(95\%\) CI is \([19.22, 29.38]\).
03

Interpret the 95% Confidence Interval for the Mean

The \(95\%\) confidence interval from \(19.22 \mu g/L\) to \(29.38 \mu g/L\) means we are \(95\%\) confident that the true average arsenic concentration lies within this interval.
04

Calculate the 90% Upper Confidence Bound for Standard Deviation

Use the chi-square distribution to estimate the upper bound for the standard deviation. The formula is \[\frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha, df}}\] for the upper bound. Here, \(df = 4\), and \(\alpha = 0.1\), thus \(\chi^2_{0.1, 4} \approx 7.779\). Calculate:\[\sqrt{\frac{4 \times 4.1^2}{7.779}} \approx 5.67\].This is the \(90\%\) upper confidence bound for the standard deviation.
05

Predict the Arsenic Concentration for a Single Specimen

The prediction interval for a single observation is given by:\[\bar{x} \pm t_{\alpha/2} \cdot s \sqrt{1 + \frac{1}{n}}\].Using \(t_{0.025} = 2.776\), the calculation is:\[24.3 \pm 2.776 \cdot 4.1 \sqrt{1 + \frac{1}{5}} = 24.3 \pm 8.768\].Hence, the prediction interval is \([15.532, 33.068]\).
06

Interpret the Prediction Interval

The prediction interval \(15.532 \mu g/L\) to \(33.068 \mu g/L\) suggests that there is a high probability that a single new measurement of arsenic concentration will fall within this specific range.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values within which we expect the true value of a parameter to lie. This range is calculated from sample data and gives an estimate for the unknown population parameter. Let's break it down further with an example:
In our case, we calculate a 95% confidence interval for the true average arsenic concentration in water specimens. The formula used involves the sample mean, standard deviation, sample size, and a critical value from the t-distribution. The result is an interval from 19.22 to 29.38 \(\mu g/L\).
What does this mean? In simple words, we are 95% confident that the actual average amount of arsenic in the groundwater samples lies between 19.22 and 29.38 micrograms per liter. This doesn't mean there's a 95% chance the actual mean lies in this range. Instead, it implies that if we took many samples and calculated the confidence interval for each, about 95% of those intervals would contain the true population mean. This method helps with estimation by acknowledging variability across different samples.
t-distribution
The t-distribution is a key concept when dealing with small sample sizes. Unlike the normal distribution, which is used when the sample size is large, the t-distribution is specifically catered to smaller sample sizes and accounts for additional variability. Here's how it works:

- When calculating the confidence interval or prediction intervals with smaller samples, the exact shape of the t-distribution should be considered.
- The t-distribution is similar to the normal distribution but has heavier tails, meaning there's more probability in the tails compared to a normal distribution.
- With fewer data points (small n), the t-distribution is wider, which captures extra uncertainty due to having less data.

In our exercise, since there were only 5 samples (n=5), we used the t-distribution to find the appropriate critical value needed for our confidence interval calculation. This critical value is 2.776 for 4 degrees of freedom (n-1). Using this value, we adjust our confidence interval so that it appropriately reflects the uncertainty in our sample analysis.
chi-square distribution
The chi-square distribution helps in assessing the variability or spread of a dataset, particularly when estimating variances or standard deviations. It's widely used in statistics, especially when dealing with confidence bounds.
To establish a 90% upper confidence bound for the standard deviation of arsenic concentrations, we use the chi-square distribution because the variability in the population needs careful understanding. Here's the process:

- First, recognize that chi-square distribution is used to understand variances, which naturally involves squared data values.
- We consider the degrees of freedom, which is calculated as the sample size minus one (here, it's 4).
- Using the chi-square table, we find the critical value associated with a specific confidence level, which is 7.779 for our calculation.

By substituting these values into our formula, we obtain an upper confidence bound of approximately 5.67 \(\mu g/L\) that tells us there's a 90% confidence that the true population standard deviation isn't exceeding this value, hence understanding the extent of variability in groundwater arsenic concentrations.
Prediction Interval
Prediction intervals are similar to confidence intervals but serve a different purpose: they predict the range in which future individual observations will fall.
Unlike a confidence interval, which estimates a population parameter (like a mean), a prediction interval looks at the expected variation in individual data points. Let’s illustrate this concept:

- Prediction intervals rely on the same kind of distribution (in our case, the t-distribution) used in creating confidence intervals.
- The formula incorporates both the sample standard deviation and accounts for estimating a single observation by adding a bit more variability to the interval.

In our exercise, a prediction interval was calculated for an individual water specimen's arsenic concentration, which was found to be between 15.532 and 33.068 \(\mu g/L\). This broad range reflects the higher uncertainty compared to a parameter estimate and indicates that there's a high probability that any single new water specimen tested will have an arsenic concentration within this range.
This result emphasizes the broader range of possible individual outcomes, making it a strong tool in forecasting and planning for individual data rather than inferential statistics about means.

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Most popular questions from this chapter

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

A study of the ability of individuals to walk in a straight line ("Can We Really Walk Straight?" Amer. J. of Physical Anthro., 1992: 19-27) reported the accompanying data on cadence (strides per second) for a sample of \(n=20\) randomly selected healthy men. \(\begin{array}{rrrrrrrrrr}.95 & .85 & .92 & .95 & .93 & .86 & 1.00 & .92 & .85 & .81 \\ .78 & .93 & .93 & 1.05 & .93 & 1.06 & 1.06 & .96 & .81 & .96\end{array}\) A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from MINITAB follows: \(\begin{array}{lccccc}\text { Variable } \mathbb{N} & \text { Mean } & \text { Median } & \text { TrMean } & \text { StDev } & \text { SEMean } \\ \text { cadence } 20 & 0.9255 & 0.9300 & 0.9261 & 0.0809 & 0.0181 \\ \text { Variable } & \text { Min } & \text { Max } & \text { Q1 } & \text { Q3 } & \\ \text { cadence } & 0.7800 & 1.0600 & 0.8525 & 0.9600 & \end{array}\) a. Calculate and interpret a \(95 \%\) confidence interval for population mean cadence. b. Calculate and interpret a \(95 \%\) prediction interval for the cadence of a single individual randomly selected from this population. c. Calculate an interval that includes at least \(99 \%\) of the cadences in the population distribution using a confidence level of \(95 \%\).

The article "Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers" (IEEE Electrical Insul. Mag., 1996: 28-34) contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: \(\begin{array}{lllllll}418 & 421 & 421 & 422 & 425 & 427 & 431 \\ 434 & 437 & 439 & 446 & 447 & 448 & 453 \\ 454 & 463 & 465 & & & & \end{array}\) a. Construct a boxplot of the data and comment on any interesting features. b. Is it plausible that the given sample observations were selected from a normal distribution? c. Calculate a two-sided \(95 \%\) confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that 440 is a plausible value for true average degree of polymerization? What about 450 ?

By how much must the sample size \(n\) be increased if the width of the CI \((7.5)\) is to be halved? If the sample size is increased by a factor of 25 , what effect will this have on the width of the interval? Justify your assertions.

A random sample of \(n=8\) E-glass fiber test specimens of a certain type yielded a sample mean interfacial shear yield stress of \(30.2\) and a sample standard deviation of \(3.1\) ("On Interfacial Failure in Notched Unidirectional Glass/Epoxy Composites," J. of Composite Materials, 1985: 276-286). Assuming that interfacial shear yield stress is normally distributed, compute a \(95 \%\) CI for true average stress (as did the authors of the cited article).

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