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Chapter 10: Problem 14

Pollution Control The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter \((\mathrm{mg} / \mathrm{L})\) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is \(48 \mathrm{mg} / \mathrm{L},\) but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for \(n=20\) days and found \(s^{2}=39(\mathrm{mg} / \mathrm{L})^{2}\). Find a \(90 \%\) confidence interval for \(\sigma^{2}\). Interpret your results.

Short Answer

Expert verified
Answer: (24.67, 77.65) (mg/L)^2

Step by step solution

01

Identify Relevant Formulas

To find the interval estimate for the population variance, we can use the chi-square distribution formula. Recall that the chi-square statistic is given by: \( \chi^{2} = \frac{(n-1)s^2}{\sigma^2} \) We will also use the formula for the \(100(1 - \alpha)\%\) confidence interval, which is given by: \( \frac{(n-1)s^2}{\chi^{2}_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^{2}_{1 - \frac{\alpha}{2},n-1}} \) Our goal is to find the interval estimate for the population variance \(\sigma^2\) using the given sample variance and a \(90\%\) confidence level.
02

Calculate Degrees of Freedom and Critical Values

Calculate the degrees of freedom, which is \(n-1\) where \(n\) is the sample size: \( df = n - 1 = 20 - 1 = 19\) Next, find the critical values for the chi-square distribution for a \(90\%\) confidence level (\(1 - \alpha = 0.9\), so \(\alpha = 0.1\)). Look up the values in the chi-square table, or use a chi-square calculator: \( \chi^{2}_{0.05,19} = 30.1435\) \( \chi^{2}_{0.95,19} = 9.5908\)
03

Compute the Confidence Interval

Now use the chi-square critical values and the formula for the confidence interval to find the range for \(\sigma^2\): \( \frac{(19)(39)}{30.1435} < \sigma^2 < \frac{(19)(39)}{9.5908} \) \( 24.67 < \sigma^2 < 77.65 \)
04

Interpret the Results

The \(90\%\) confidence interval for the population variance of the suspended solid discharge rates is \((24.67, 77.65)\ (\text{mg}/\text{L})^2\). This means that we can be \(90\%\) confident that the true variance of the suspended solid discharge rates lies within this interval.

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Key Concepts

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

Population Variance
Population variance, denoted as \( \sigma^2 \), is a measure of how individual values of a population differ from the population mean. It provides insight into the spread or variability of a data set. In many cases, like environmental measures from the given exercise, determining the exact variance of an entire population is not feasible due to the sheer size or constraints in data collection.
Degrees of Freedom
Degrees of freedom, often represented as \( df \), refer to the number of independent values or scores that can vary in an analysis without breaking any constraints. It is crucial for statistical analyses like calculating sample variance or establishing confidence intervals. In the exercise, the degrees of freedom are calculated as \( df = n - 1 \), where \( n \) is the number of observations in the sample. For instance, with 20 days of discharge rates measured, the degrees of freedom would be 19. This value impacts the shape of the sampling distribution and the critical values obtained from the chi-square distribution needed to estimate confidence intervals.
Sampling Distribution
A sampling distribution is a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. In our solved problem, the chi-square distribution is used as the sampling distribution for the variance. It is skewed to the right, meaning most of the values are to the left with a long tail extending to the right. The shape of the chi-square distribution depends on the degrees of freedom. We utilize this distribution to create confidence intervals for parameters like the population variance by capturing the range where the true parameter is believed to fall with a certain level of confidence, in this case, 90%.

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Most popular questions from this chapter

Use the information in Exercises \(7-8\) to test \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) versus the given alternate hypothesis. Construct a \((1-\alpha)\) 100\% confidence interval for \(\sigma^{2}\) using the \(\chi^{2}\) statistic. $$ n=15, \bar{x}=3.91, s^{2}=.3214, H_{\mathrm{a}}: \sigma^{2} \neq .5, \alpha=.10 $$

The stability of measurements on a manufactured product is important in maintaining product quality. A manufacturer of lithium batteries suspected that one of the production lines was producing batteries with a wide variation in length of life. To test this theory, he randomly selected \(n=50\) batteries from the suspect line and \(n=50\) from a line that was judged to be "in control." He then measured the length of time (in hours) until depletion for both samples. The sample means and variances for the two samples were as follows: $$\begin{array}{ll}\hline \text { Suspect Line } & \text { Line }^{\prime \prime} \text { in Control }^{\prime \prime} \\\\\hline \bar{x}_{1}=9.40 & \bar{x}_{2}=9.25 \\\s_{1}=0.25 & s_{2}=0.12\end{array}$$ a. Do the data provide sufficient evidence to indicate that batteries produced by the "suspect line" have a larger variance in length of life than those produced by the line that is assumed to be in control? Test using \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test and interpret its value. c. Construct a \(90 \%\) confidence interval for the variance ratio.

An experiment was conducted to compare the use of iPads versus regular textbooks in teaching algebra to two classes of middle school students. \({ }^{6}\) To remove teacher-to-teacher variation, the same teacher taught both classes, and all teaching materials were provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow. $$ \begin{array}{lcc} \hline & \text { iPad } & \text { Textbook } \\ \hline \text { Mean } & 86.4 & 79.7 \\ \text { Standard Deviation } & 8.95 & 10.7 \\ \text { Sample Size } & 10 & 10 \\ \hline \end{array} $$ a. Use the summary data to test for a significant difference in advancement scores for the two groups using $$ \alpha=.05 . $$ b. Find a \(95 \%\) confidence interval for the difference in mean scores for the two groups. c. In light of parts a and b, what can we say about using iPads versus traditional textbooks in teaching algebra at the middle school level?

Calculate the number of degrees of freedom for a paired-difference test in Exercises \(2-4,\)with \(n_{1}=n_{2}=\) number of observations in each sample and \(n=\) number of pairs. $$n=12$$

Use the information given in Exercises \(2-7\) to find the tabled value for an \(F\) variable based on \(n_{1}-\) I numerator degrees of freedom, \(n_{2}-1\) denominator degrees of freedom with an area of a to its right. \(n_{1}=16, n_{2}=19, a=.005\)
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