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Chapter 14: Problem 489

The lengths of a random sample of 10 staples have a sample variance of 32 centimeters squared. Find a \(.95\) confidence interval estimate for the variance of all staple lengths.

Short Answer

Expert verified
The 95% confidence interval estimate for the variance of all staple lengths is approximately [15.05, 106.45] cm虏.

Step by step solution

01

Identify the given information

Here, we have the following information: - Sample size (n) = 10 - Sample variance (s^2) = 32 cm^2 - Confidence level = 95% (or 0.95)
02

Find the degrees of freedom (df)

Degrees of freedom (df) is calculated using this formula: df = n - 1 So in this case, we have df = 10 - 1 = 9
03

Find the chi-square values (蠂虏)

Using the chi-square distribution table, we will find the upper and lower critical values corresponding to the 95% level of confidence and the 9 degrees of freedom: - 蠂虏(伪/2, df) = 蠂虏(0.025, 9) = 19.023 - 蠂虏(1-伪/2, df) = 蠂虏(0.975, 9) = 2.700
04

Calculate the confidence interval for the population variance

Now, we will use the following formulas to find the confidence interval estimate for the population variance: Lower limit (L): \(L = \frac{(n-1) \times s^2}{蠂虏(1-伪/2, df)}\) Upper limit (U): \(U = \frac{(n-1) \times s^2}{蠂虏(伪/2, df)}\) Plugging in the values: - Lower limit (L): \(L = \frac{(10-1) \times 32}{2.700} \approx 106.45\) - Upper limit (U): \(U = \frac{(10-1) \times 32}{19.023} \approx 15.05\) Therefore, the confidence interval estimate for the population variance is [15.05, 106.45] cm虏.

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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 key tool in statistics used to make inferences about the variability of a population based on a sample. It is often used when estimating a population variance.
Unlike the normal distribution, the chi-square distribution is not symmetrical, having more pronounced skewness especially with lower degrees of freedom.
The shape of the distribution depends on the degrees of freedom, meaning the parameters of the chi-square distribution change with the size of the sample used.
  • The chi-square distribution is helpful in forming confidence intervals for variances as shown in the exercise.
  • It allows us to determine critical values that are needed for finding confidence bounds.
  • Understanding the chi-square distribution helps in assessing how spread out our data is in a statistical context.
Degrees of Freedom
Degrees of freedom often referred to as 'df', is a concept that describes the number of independent values or quantities which can be assigned to a statistical distribution.
In the context of confidence intervals for variances, the degrees of freedom (df) is generally calculated as the sample size minus one (df = n - 1).
  • In the exercise, with a sample size of 10, the degrees of freedom were 9.
  • This number dictates the exact shape of the chi-square distribution being used to evaluate the sample variance.
  • Higher degrees of freedom result in a chi-square distribution that looks more like the regular normal distribution.
This concept is crucial as it impacts statistical calculations that rely on knowing the variability or spread of data, like calculating a confidence interval for variance.
Sample Variance
Sample variance is an estimate of the variance of a population, calculated from a sample. This is a measure of how much the data points in a sample differ from the mean of the sample.
In statistics, variance is essential to quantify the extent of variation or how much the data points differ among themselves.
  • Sample variance is symbolized by \(s^2\) and forms the basis for calculating other measures such as the confidence interval for variance.
  • For the exercise, the sample variance \(s^2\) was calculated to be 32 centimeters squared.
  • Using the sample variance, we can make educated guesses about the spread and variation present in the overall population from which the sample is drawn.
Understanding sample variance is vital for data analysis and helps convey the diversity within the data set.

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Most popular questions from this chapter

During the \(1976-77\) season Coach Jerry Tarkanian outfitted his University of Nevada at Las Vegas basketball team with new sneakers. The 16 member team had an average size of \(14.5\) and a standard deviation of \(5 .\) Find a 90 percent confidence interval for the mean sneaker size of all collegiate basketball players. Assume the population is normal and the variance is not known.

In the previous problem, assume \(\mathrm{n}=10, \mu=0\), \({ }^{10} \sum_{1} \mathrm{X}_{\mathrm{i}}^{2}=106.6\) and \(\alpha=.05\). Find a \(95 \%\) confidence interval for \(\sigma^{2}\)

The weights of 15 New York models had a sample mean of 107 lbs. and a sample standard deviation of 10 lbs. Twelve Philadelphia models had a mean weight of 112 and a standard deviation of 8 . Make a \(.90\) confidence interval estimate of the difference of the mean weights between the two model populations.

Given \(\mathrm{X}=60\), construct a central \(95 \%\) confidence interval for the parameter \(\mathrm{p}\) of a binomial distribution for which \(\mathrm{n}=100\)

We are looking for a comparison between the financial states of people leaving Las Vegas and Reno after their vacations. 23 tourists leaving Vegas lost an average \(\underline{\mathrm{X}}\) of \$551.49. Similarly 23 Reno vacationers came up short an average, \(Y\), of \(\$ 549.93\). It is assumed that the losses in each city are independent, normally distributed random variables with common unknown standard deviation. The quantity \(\left[\left\\{\sum\left(\mathrm{X}_{\mathrm{i}}-\underline{\mathrm{X}}\right)^{2}+\sum\left(\mathrm{Y}_{\mathrm{i}}-\underline{\mathrm{Y}}\right)^{2}\right\\} /(\mathrm{n}+\mathrm{m}-2)\right]\) was found to be \(87.683 .\) Find a \(90 \%\) confidence interval for the difference in means.
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