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In 2014 , the variance of the ages of all workers at a large company that has more than 30,000 workers was \(133 .\) A recent random sample of 25 workers selected from this company showed that the variance of their ages is 112 . a. Using a \(2.5 \%\) significance level, can you conclude that the current variance of the ages of workers at this company is lower than 133 ? Assume that the ages of all current workers at this company are (approximately) normally distributed. b. Construct a \(98 \%\) confidence intervals for the variance and the standard deviation of the ages of all current workers at this company.

Short Answer

Expert verified
a. No, one cannot conclude at a 2.5% significance level that the current variance of the ages is lower than 133. \n b. The 98% confidence interval for the variance and the standard deviation of the ages of all current workers at this company are (67.10, 246.61) and (8.19, 15.70) respectively.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0: \sigma^2 = 133\) states that the variance of the age of all workers is 133. The alternative hypothesis \(H_a: \sigma^2 < 133\) states that the variance of the age of all workers is less than 133.
02

Calculate Test Statistic

Compute the test statistic using chi-square distribution formula: \((n-1) * s^2 / \sigma^2\), where \(n = 25\) is the sample size, \(s^2 = 112\) is the sample variance and \(\sigma^2 = 133\) is the population variance. Substituting the appropriate values gives us a test statistic of \( (\(24*112/133) = 20.17 \).
03

Determine Critical Value

From the chi-squared distribution table with degrees of freedom equals \(n-1=24\), the critical value of chi-square for the 2.5% significance level is 13.85.
04

Hypothesis Test Decision

As the test statistic (20.17) is higher than the critical value (13.85), we do not reject the null hypothesis \(\sigma^2=133\). There is not enough evidence at the 2.5% level of significance to say that the variance of the workers' ages is less than 133.
05

Construct 98% Confidence Interval for Variance

The confidence interval for variance is given by \(\((n-1) * s^2 / \chi^2_{1-\alpha/2, n-1} \leq \sigma^2 \leq (n-1) * s^2 / \chi^2_{\alpha/2, n-1}\)\). The \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) values for 2% significance level (98% confidence level) and 24 degrees of freedom are 10.856 and 40.113. Therefore, the 98% confidence interval for variance is \( (24*112/40.113, 24*112/10.856) = (67.10, 246.61)\)
06

Confidence Interval for Standard Deviation

The confidence interval for standard deviation is obtained by simply taking the square root of confidence interval for variance. Therefore, the 98% confidence interval for standard deviation is \(\sqrt{67.10} to \sqrt{246.61}\), which equals (8.19, 15.70).

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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 statistical tool used primarily for hypothesis testing and estimating population variance. It is particularly useful when analyzing categorical data. In hypothesis testing, the chi-square distribution helps determine if there's a significant difference between observed and expected frequencies in a dataset.
This distribution is positively skewed and its shape depends on the degrees of freedom, which are determined by the sample size minus one \(df = n-1\).
  • For large degrees of freedom, the distribution resembles a normal distribution.
  • For smaller degrees of freedom, it skews to the right.
In this exercise, the chi-square distribution was used to calculate the test statistic in determining the variance of workers' ages. Because the sample size was 25, the degrees of freedom used were 24.
Null Hypothesis
The null hypothesis is a statement used in statistics that proposes no statistical significance exists in a set of given observations. Essentially, it suggests that any kind of difference or significance observed in a dataset is due to chance.
The primary purpose of the null hypothesis is to provide something testable. In hypothesis testing, we either reject or fail to reject the null hypothesis based on the data presented.
  • In this problem, the null hypothesis \(H_0: \sigma^2 = 133\) posits that the variance of the workers' ages is 133.
  • The alternative hypothesis \(H_a: \sigma^2 < 133\) suggests otherwise, that the variance is less than 133.
After conducting the chi-square test, the decision was to not reject the null hypothesis, which means that the sample did not provide enough evidence to state that the variance is actually lower than 133.
Confidence Interval
A confidence interval gives a range of values within which we can be "confident" a population parameter lies. It's a fundamental concept in inferential statistics and provides more information than a simple point estimate.
Confidence intervals quantify uncertainty and allow researchers to make more informed conclusions.
  • In this exercise, a 98% confidence interval was constructed for both the variance and standard deviation of worker ages.
  • This interval means that we are 98% confident that the true population variance of worker ages falls between 67.10 and 246.61.
  • The corresponding interval for standard deviation (the square root of variance) is between 8.19 and 15.70.
The wider the interval, the more uncertainty there is about the estimate. Selecting a 98% confidence level indicates a higher degree of certainty compared to lower confidence levels.
Variance Estimation
In statistics, variance is a measure of how much values in a dataset differ from the mean. Estimating variance is crucial for understanding variability within a dataset.
Variance estimation allows statisticians to understand the spread of data and predict trends.
  • The formula for computing the variance in a sample is \(s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}\), where \(x_i\) represents each data point, \(\bar{x}\) is the sample mean, and \(n\) is the number of observations.
  • The variance in the exercise was calculated to be 112, based on a sample size of 25 workers.
Variance estimation forms the basis for further statistical analysis, including hypothesis testing and constructing confidence intervals. A precise estimate of variance is critical because it influences the results and interpretations in statistical research.

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Most popular questions from this chapter

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