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Chapter 8: Problem 101

In laboratory work, it is desirable to run careful checks on the variability of readings produced on standard samples. In a study of the amount of calcium in drinking water undertaken as part of a water quality assessment, the same standard sample was run through the laboratory six times at random intervals. The six readings, in parts per million, were \(9.54,9.61,9.32,9.48,9.70,\) and 9.26 Estimate the population variance \(\sigma^{2}\) for readings on this standard, using a \(90 \%\) confidence interval.

Short Answer

Expert verified
The 90% CI for the population variance is (0.01289, 0.12466).

Step by step solution

01

Calculate the Sample Mean

First, we need to calculate the sample mean, \( \bar{x} \), of the given readings. The formula for the sample mean is: \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \] Substituting the given readings: \[ \bar{x} = \frac{1}{6} (9.54 + 9.61 + 9.32 + 9.48 + 9.70 + 9.26) = 9.485 \] The sample mean, \( \bar{x} \), is 9.485.
02

Calculate the Sample Variance

Next, calculate the sample variance, \( s^2 \), using the formula: \[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \] Calculating each squared deviation: \[(9.54 - 9.485)^2 = 0.003025\] \[(9.61 - 9.485)^2 = 0.015625\] \[(9.32 - 9.485)^2 = 0.027225\] \[(9.48 - 9.485)^2 = 0.000225\] \[(9.70 - 9.485)^2 = 0.046225\] \[(9.26 - 9.485)^2 = 0.050225\] Adding them: \[\sum (x_i - \bar{x})^2 = 0.14255\] Now, substitute into the variance formula: \[ s^2 = \frac{0.14255}{5} = 0.02851\] Thus, the sample variance \( s^2 \) is 0.02851.
03

Find the Confidence Interval for Population Variance

To find the 90% confidence interval for the population variance, use the chi-square distribution. Calculate the chi-square values for 5 degrees of freedom: - The lower critical value, \( \chi^2_{0.05,5} \) is 1.145 - The upper critical value, \( \chi^2_{0.95,5} \) is 11.070 The confidence interval for \( \sigma^2 \) is given by: \[ \left( \frac{(n-1)s^2}{\chi^2_{upper}}, \frac{(n-1)s^2}{\chi^2_{lower}} \right) \] Substitute the values: \[ \left( \frac{5 \times 0.02851}{11.070}, \frac{5 \times 0.02851}{1.145} \right) \] \[ \left( 0.01289, 0.12466 \right) \] The 90% confidence interval for the population variance \( \sigma^2 \) is \( (0.01289, 0.12466) \).

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Key Concepts

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

Sample Variance
Sample variance helps us determine how data points diverge from the mean in a sample data set. To calculate it, first, find the sample mean, which is the average of your data. Subtract this mean from each data reading to find deviations. Then, square each deviation. Squaring ensures that negative differences do not cancel out positive differences. Finally, sum the squared deviations and divide by the number of data points minus one, which accounts for the degrees of freedom in the sample. This gives you the sample variance, which quantifies the spread of your sample data.
Confidence Interval
A confidence interval provides a range of values within which we are fairly sure our population parameter lies. Consider it a way of expressing uncertainty about a point estimate, like the sample variance, when inferring to a larger population. For a 90% confidence interval, like in our example, it implies that if we repeatedly took samples and calculated the interval from them, 90% of the intervals would contain the true population variance. A wider interval means more uncertainty, while a narrower interval suggests higher confidence in the estimate.
Chi-Square Distribution
The chi-square distribution is key in statistical tests, including confidence intervals for variance. It's especially useful because it models the sum of squared differences of independent standard normal variables. When we calculate intervals for population variance, we utilize the chi-square distribution to determine critical values. These critical values determine the range of plausible variance values. It's determined by the degrees of freedom, which, in our case, is the number of samples minus one. Knowing and using appropriate chi-square values is crucial to accurately estimating the population variance.
Statistical Analysis
Statistical analysis involves applying mathematical formulas and concepts to data in order to extract information. In our exercise, we utilized statistical analysis to estimate population variance from sample data. It includes calculating measures like mean and variance, and applying methods like confidence intervals. Each part plays a role in understanding data characteristics and powering informed decisions. For the population variance estimate, applying statistical analysis means accurately following methodological steps like computing variance and using distributions, which provides rigor and reliability to conclusions we draw from sample observations.

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Most popular questions from this chapter

Suppose that \(\hat{\theta}\) is an estimator for a parameter \(\theta\) and \(E(\hat{\theta})=a \theta+b\) for some nonzero constants \(a\) and \(b\) a. In terms of \(a, b,\) and \(\theta\), what is \(B(\hat{\theta}) ?\) b. Find a function of \(\hat{\theta}-\operatorname{say}, \hat{\theta}^{*}-\) that is an unbiased estimator for \(\theta\)

The number of breakdowns per week for a type of minicomputer is a random variable \(Y\) with a Poisson distribution and mean \(\lambda\). A random sample \(Y_{1}, Y_{2}, \ldots, Y_{n}\) of observations on the weekly number of breakdowns is available. a. Suggest an unbiased estimator for \(\lambda\). b. The weekly cost of repairing these breakdowns is \(C=3 Y+Y^{2} .\) Show that \(E(C)=4 \lambda+\lambda^{2}\) c. Find a function of \(Y_{1}, Y_{2}, \ldots, Y_{n}\) that is an unbiased estimator of \(E(C)\). [Hint: Use what you know about \(\left.\bar{Y} \text { and }(\bar{Y})^{2} .\right]\)

For a comparison of the rates of defectives produced by two assembly lines, independent random samples of 100 items were selected from each line. Line A yielded 18 defectives in the sample, and line B yielded 12 defectives. a. Find a \(98 \%\) confidence interval for the true difference in proportions of defectives for the two lines. b. Is there evidence here to suggest that one line produces a higher proportion of defectives than the other?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample of size \(n\) from a population whose density is given by $$f(y)=\left\\{\begin{array}{ll} 3 \beta^{3} y^{-4}, & \beta \leq y \\ 0, & \text { elsewhere } \end{array}\right.$$ where \(\beta>0\) is unknown. (This is one of the Pareto distributions introduced in Exercise \(6.18 .\) ) Consider the estimator \(\widehat{\beta}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) a. Derive the bias of the estimator \(\widehat{\beta}\). $$\text { b. Derive } \operatorname{MSE}(\widehat{\beta})$$

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.
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