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Chapter 10: Problem 50

A random sample of \(n=15\) observations was selected from a normal population. The sample mean and variance were \(\bar{x}=3.91\) and \(s^{2}=.3214 .\) Find a \(90 \%\) confidence interval for the population variance \(\sigma^{2}\).

Short Answer

Expert verified
The 90% confidence interval for the population variance is approximately 0.1901 to 0.6846.

Step by step solution

01

Determine the degrees of freedom

In this case, we have \(n = 15\) observations. The degrees of freedom for a chi-square distribution concerning population variance is equal to \(n-1\). Therefore, in our case: Degrees of freedom (df) = \(n-1 = 15-1 = 14\)
02

Calculate the chi-square values

In order to find the chi-square values for the \(90\%\) confidence interval, we can use a chi-square table or calculator. We need to calculate the values for \(\chi^2_{0.05,14}\) and \(\chi^2_{0.95,14}\) (since \(\alpha=0.1\), so \(\frac{\alpha}{2}=0.05\) for both tails). From the chi-square table or calculator, we get the following values: \(\chi^2_{0.05,14} = 23.685\) and \(\chi^2_{0.95,14} = 6.571\)
03

Calculate the interval for the population variance

Now that we have the degrees of freedom and chi-square values, we can plug these into the formula for the \(\sigma^2\) confidence interval: Lower bound: \(\frac{(n-1) \cdot s^2}{\chi^2_{0.05,14}}=\frac{ (15-1)\cdot0.3214}{23.685} = \frac{4.4996}{23.685}=0.1901\) Upper bound: \(\frac{(n-1) \cdot s^2}{\chi^2_{0.95,14}}=\frac{ (15-1)\cdot0.3214}{6.571} = \frac{4.4996}{6.571}=0.6846\) Together, this gives us the confidence interval for the population variance: \(0.1901 \leq \sigma^2 \leq 0.6846\) Hence, with a \(90\%\) confidence level, we can say that the population variance (\(\sigma^2\)) lies between \(0.1901\) and \(0.6846\).

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Key Concepts

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

population variance
The population variance \(\sigma^2\) is a measure of how much the individual observations of a population differ from the population mean. It tells us how spread out the data is in the entire population.
  • It is not the same as the sample variance, which is calculated from a subset of the population.
  • In statistics, understanding the population variance helps in making predictive models and gauging the reliability of these predictions.
To find the population variance, you'll usually use the chi-square distribution, especially when you lack the entire population data and are working with a sample. Knowing the population variance can help assess the variability within your data.
If your data points are close to each other, the variance will be low. Conversely, if they vary widely, the variance will be high.
chi-square distribution
A chi-square distribution is commonly used in statistics to assess the variance within a dataset, thus making it crucial for calculating confidence intervals for population variance.
  • It arises when you square standard normal random variables, meaning it's related to normal distributions.
  • The chi-square distribution is always positive, thereby suitable for calculating variances which cannot be negative.
  • It varies depending on the degrees of freedom, which affect the shape of the distribution.
To use a chi-square distribution for confidence intervals, specific points in the distribution are used. These are often found using chi-square tables or statistical software.
In our exercise, the chi-square values \(\chi^2_{0.05, 14}\) and \(\chi^2_{0.95, 14}\) were selected from a chi-square table.
degrees of freedom
Degrees of freedom refer to the number of independent values that can freely vary in a given statistical calculation. In our scenario:
  • When estimating a population parameter, such as variance, the degrees of freedom lead to more precise estimates.
  • For a sample of \(n\) observations, the degrees of freedom for calculating variance are \(n-1\), since one parameter is used for the sample mean.
Degrees of freedom are crucial because they adjust the parameters of various distributions, including the chi-square distribution you use in variance calculations.
Recognizing the correct degrees of freedom ensures you're using the right part of the chi-square distribution. This precision is necessary to form accurate confidence intervals.

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Most popular questions from this chapter

Here are the prices per ounce of \(n=13\) different brands of individually wrapped cheese slices: $$ \begin{array}{lllll} 29.0 & 24.1 & 23.7 & 19.6 & 27.5 \\ 28.7 & 28.0 & 23.8 & 18.9 & 23.9 \\ 21.6 & 25.9 & 27.4 & & \end{array} $$ Construct a \(95 \%\) confidence interval estimate of the underlying average price per ounce of individually wrapped cheese slices.

An experiment published in The American Biology Teacher studied the efficacy of using \(95 \%\) ethanol or \(20 \%\) bleach as a disinfectant in removing bacterial and fungal contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using eggplant as the plant tissue being cultured. \({ }^{8}\) Five cuttings per plant were placed on a petri dish for each disinfectant and stored at \(25^{\circ} \mathrm{C}\) for 4 weeks. The observation reported was the number of uncontaminated eggplant cuttings after the 4 -week storage. $$ \begin{array}{lcl} \text { Disinfectant } & 95 \% \text { Ethanol } & 20 \% \text { Bleach } \\ \hline \text { Mean } & 3.73 & 4.80 \\ \text { Variance } & 2.78095 & .17143 \\ n & 15 & 15 \end{array} $$ a. Are you willing to assume that the underlying variances are equal? b. Using the information from part a, are you willing to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested?

Use Table 4 in Appendix I to find the following critical values: a. An upper one-tailed rejection region with \(\alpha=.05\) and \(11 d f\). b. A two-tailed rejection region with \(\alpha=.05\) and \(7 d f\). c. A lower one-tailed rejection region with \(\alpha=.01\) and \(15 d f\).

In a study to determine which factors predict who will benefit from treatment for bulimia nervosa, an article in the British Journal of Clinical Psychology indicates that self-esteem was one of these important predictors. \({ }^{4}\) The table gives the mean and standard deviation of self-esteem scores prior to treatment, at posttreatment, and during a follow-up: $$ \begin{array}{lccc} & \text { Pretreatment } & \text { Posttreatment } & \text { Follow-up } \\ \hline \text { Sample Mean } \bar{x} & 20.3 & 26.6 & 27.7 \\ \text { Standard Deviation } s & 5.0 & 7.4 & 8.2 \\ \text { Sample Size } n & 21 & 21 & 20 \end{array} $$ a. Use a test of hypothesis to determine whether there is sufficient evidence to conclude that the true pretreatment mean is less than \(25 .\) b. Construct a \(95 \%\) confidence interval for the true posttreatment mean. c. In Section \(10.4,\) we will introduce small-sample techniques for making inferences about the difference between two population means. Without the formality of a statistical test, what are you willing to conclude about the differences among the three sampled population means represented by the results in the table?

To properly treat patients, drugs prescribed by physicians must not only have a mean potency value as specified on the drug's container, but also the variation in potency values must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that his drug has a potency of \(5 \pm .1\) milligram per cubic centimeter \((\mathrm{mg} / \mathrm{cc})\). A random sample of four containers gave potency readings equal to 4.94,5.09 , 5.03, and \(4.90 \mathrm{mg} / \mathrm{cc} .\) a. Do the data present sufficient evidence to indicate that the mean potency differs from \(5 \mathrm{mg} / \mathrm{cc} ?\) b. Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? (HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval \(5 \pm .1 \mathrm{mg} / \mathrm{cc}\) with very high probability - the implication is almost always-let us assume that the range \(.2 ;\) or 4.9 to \(5.1,\) represents \(6 \sigma\) as suggested by the Empirical Rule).
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