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Chapter 9: Problem 72

A random sample of 20 students obtained a mean of \(\bar{x}=72\) and a variance of \(s \sim^{2}=16\) on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a \(98 \%\) confidence interval for \(\sigma^{2}\).

Short Answer

Expert verified
The 98% confidence interval for the population variance \(\sigma^{2}\) is approximately (8.89, 36.94).

Step by step solution

01

Identify the Confidence Level and Degrees of Freedom

First, recognize that the confidence level is \(98\%\) which gives an alpha level of \(\alpha=1-0.98=0.02\). A Chi-square distribution will be used in this context. Therefore, also identify degrees of freedom for this distribution which is \(df=n-1=20-1=19\).
02

Obtain Chi-square Critical Values

The critical values for chi-square distribution can be determined using statistical tables or a calculator, for a two-tailed test divide \(\alpha\) by 2. Look up the critical Chi-square value for \(\alpha=0.01\) in the lower tail (denoted as \(\chi^2_{0.01}\)) and \(\alpha=0.01\) in the upper tail (denoted as \(\chi^2_{0.99}\)). For \(19\) degrees of freedom, you'd find that \(\chi^2_{0.01} = 8.231\) and \(\chi^2_{0.99} = 34.170\).
03

Compute the Confidence Interval

Denote \(s^2\) as the sample variance. A \(98\%\) confidence interval for \(\sigma^2\) can be found using the following formulas: \(\frac{(n - 1)*s^2}{\chi^2_{0.99}}\) and \(\frac{(n - 1)*s^2}{\chi^2_{0.01}}\). Substituting given values gives: \(\frac{19*16}{34.170}\) and \(\frac{19*16}{8.231}\). Calculating these gives an interval of approx. \(8.89\) to \(36.94\).

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Key Concepts

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

Chi-Square Distribution
The Chi-Square distribution is a crucial concept in statistics, particularly when dealing with sample variance. It is a continuous probability distribution that is used to test hypotheses about variances and to construct confidence intervals for variance, as is needed in problems involving normal distributions.
  • Key Properties: The Chi-Square distribution is defined by its degrees of freedom, which influences its shape. More degrees of freedom result in a distribution that becomes more symmetric.
  • Applications: It is often used in hypothesis testing, especially in tests of independence and goodness-of-fit tests.
  • Shape: The distribution's shape is skewed to the right, but as the degrees of freedom increase, it becomes more symmetrical.
In the given exercise, because we're looking at variances, the Chi-Square distribution helps in finding the bounds of the confidence interval.
Sample Variance
Sample variance, denoted as \(s^2\), measures the spread of sample data points around their mean. It is an estimate of the population variance, which allows us to understand variability within a data set.
  • Calculation: To calculate sample variance, subtract each data point from the sample mean, square these differences, sum them, and divide by the number of data points minus one.
  • Role in Confidence Intervals: Sample variance is used in the formula for constructing confidence intervals for the population variance, as in the original problem.
  • Importance: Understanding variance provides insights into data reliability and volatility, and is pivotal in quality control and risk management.
In our scenario, the sample variance is given as 16, which feeds directly into the Chi-Square formula for finding the confidence interval.
Degrees of Freedom
Degrees of freedom (df) is a concept that describes the number of values in a calculation that are free to vary. It is essential for identifying the correct distribution, especially in the context of the Chi-Square distribution.
  • Formula: In the context of sample variance, the degrees of freedom are calculated as \(df = n - 1\), where \(n\) is the sample size.
  • Application: Degrees of freedom affect the critical values in statistical tables, which are necessary to compute confidence intervals.
In the exercise, with a sample size of 20, the degrees of freedom is 19, which guides us in selecting the correct Chi-Square critical values.
Statistical Tables
Statistical tables are essential tools in statistics, providing critical values needed for hypothesis testing and constructing confidence intervals. They help identify the right values for specific distributions like the Chi-Square.
  • Purpose: These tables allow statisticians to find the critical values of statistical tests based on the chosen confidence level and degrees of freedom.
  • Usage: For the Chi-Square distribution, statistical tables list the critical values needed to find the upper and lower limits of confidence intervals.
In the given problem, such tables help locate the critical values \(\chi^2_{0.01}\) and \(\chi^2_{0.99}\), which are 8.231 and 34.170 respectively, for 19 degrees of freedom. This information is pivotal for determining the confidence interval for the population variance.

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Most popular questions from this chapter

A study was conducted to determine if a certain metal treatment has any effect on the amount of metal removed in a pickling operation. A random sample of 100 pieces was immersed in a bath for 24 hours without the treatment, yielding an average of 12.2 millimeters of metal removed and a sample standard deviation of 1.1 millimeters. A second sample of 200 pieces was exposed to the treatment, followed by the 24 -hour immersion in the bath, resulting in an average: removal of 9.1 millimeters of metal with a sample standard deviation of 0.9 millimeter. Compute a \(98 \%\) confidence interval estimate for the difference between the population means. Docs the treatment appear to reduce the mean amount of metal removed?

In a batch chemical process, two catalysts arc being compared for their effect on the output of the process reaction. A sample of \(\lfloor 2\) batches was prepared using catalyst 1 and a sample of 10 batches was obtained using catalyst \(2 .\) The 12 batches for which catalyst 1 was used gave an average yield of 85 with a sample standard deviation of \(4,\) and the second sample gave an average of 81 and a sample standard deviation of \(5 .\) Find a \(90 \%\) confidence interval for the difference between the population means, assuming that the: populations art: approximately normally distributed with equal variances.

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a) Construct a \(98 \%\) confidence interval for the mean height of all college students. (b) What can we assert with \(98 \%\) confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters?

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming a normal distribution for the number of words typed per minute, find a \(95 \%\) confidence interval for the average number of words typed by all graduates of this school.

A machine is used to fill boxes of product in an assembly line operation. Much concern centers around the variability in the number of ounces of product in the box. The standard deviation in weight of product is known to be 0.3 ounces. An improvement is implemented after which a random sample of 20 boxes are selected and the sample variance is found to be 0.045 ounces. Find a \(95 \%\) confidence interval on the variance in the weight of the product. Does it appear from the range of the confidence interval that the improvement of the process enhanced quality as far as variability is concerned? Assume normality on the distribution of weight of product.
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