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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

An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be \(95 \%\) confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that \(\sigma=40\) seconds.

Regular consumption of presweetened cereals contributes to tooth decay, heart disease, and other degenerative diseases, according to studies conducted by Dr. W. H. Bowen of the National Institute of Health and Dr. J. Yudben, Professor of Nutrition and Dietetics at the University of London. In a random sample of 20 similar single servings of Alpha-Bits, the average sugar content was 11.3 grams with a standard deviation of 2.45 grams. Assuming that the sugar contents are normally distributed, construct a \(95 \%\) confidence interval for the mean sugar content for single servings of Alpha-Bits.

(a) A random sample of 200 voters is selected and 114 are found to support an annexation suit. Find the \(96 \%\) confidence interval for the fraction of the voting population favoring the suit. (b) What can we assert with \(96 \%\) confidence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be \(0.57 ?\)

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a \(96 \%\) confidence interval for the population mean of all bulbs produced by this firm.

A consumer group is interested in comparing operating costs for two different types of automobile engines. The group is able to find 15 owners whose cars have engine type \(A\) and 15 who have engine type B. All 30 owners bought their cars at roughly the same time and all have kept good records for a certain 12 month period. In addition, owners were found that drove roughly the same mileage. The cost statistics are \(y_{A}=\$ 87.00 / 1,000\) miles, \(y_{B}=\$ 75.00 / 1,000\) miles, \(s_{A}=\$ 5.99,\) and \(S B-\$ 4.85 .\) Compute a \(95 \%\) confidence interval to estimate \(\mu_{A}-\mu_{B_{1}}\) the difference in the mean operating costs. Assume normality and equal variance.
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