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The following problem is based on information taken from Accidents in North American Mountaineering (jointly published by The American Alpine Club and The Alpine Club of Canada). Let \(x\) represent the number of mountain climbers killed each year. The long-term variance of \(x\) is approximately \(\sigma^{2}=136.2 .\) Suppose that for the past 8 years, the variance has been \(s^{2}=115.1 .\) Use a \(1 \%\) level of significance to test the claim that the recent variance for number of mountain climber deaths is less than \(136.2 .\) Find a \(90 \%\) confidence interval for the population variance.

Short Answer

Expert verified
We do not reject the null hypothesis at the 1% significance level. The 90% confidence interval for the population variance is approximately (57.21, 370.40).

Step by step solution

01

Identify Hypotheses

To test the variance, we will use a chi-square test. We first set up the null and alternative hypotheses: - Null Hypothesis (H鈧): The variance is \( 8; 136.2 \)- Alternative Hypothesis (H鈧): The variance is less than \( 8; 136.2 \)This is a one-tailed test because we are testing if the variance is less.
02

Determine the Chi-Square Test Statistic

The test statistic for variance is given by the formula:\[\chi^2 = \frac{(n-1)s^2}{\sigma^2}\]Where:- \(n\) is the number of years, which is 8- \(s^2\) is the sample variance, 115.1- \(\sigma^2\) is the population variance under the null hypothesis, 136.2Substituting the values we get:\[\chi^2 = \frac{(8-1) \times 115.1}{136.2} = \frac{7 \times 115.1}{136.2} \]Calculate \(\chi^2\):\[ \chi^2 \approx 5.909 \]
03

Find the Critical Value from Chi-Square Distribution

The critical value for a chi-square distribution is determined by the desired significance level and degrees of freedom.- Significance level: 1% (0.01)- Degrees of freedom (df) = \((n - 1) = 7\)Using a chi-square distribution table or calculator, find the critical value for a one-tailed test with df = 7 at \(\alpha = 0.01\):\[\chi^{2}_{critical} \approx 2.167\]
04

Make a Decision

Compare the computed test statistic \(\chi^2 \approx 5.909\) with the critical value \(\chi^{2}_{critical} \approx 2.167\).Since \(5.909 > 2.167\), we do not reject the null hypothesis. Thus, there is not enough evidence to claim that the recent variance of deaths is less than 136.2.
05

Calculate the Confidence Interval for Variance

To calculate a 90% confidence interval for the variance, we will use the formula:\[\left( \frac{(n-1)s^2}{\chi^{2}_{\alpha/2}}, \frac{(n-1)s^2}{\chi^{2}_{1-\alpha/2}} \right)\]Where \(\chi^{2}_{\alpha/2}\) and \(\chi^{2}_{1-\alpha/2}\) are the chi-square critical values for the lower and upper tails.- Find \(\chi^{2}_{0.05}\) and \(\chi^{2}_{0.95}\) for df = 7:\[ \chi^{2}_{0.05} \approx 14.067 \]\[ \chi^{2}_{0.95} \approx 2.167 \]Thus the confidence interval is:\[\left( \frac{7 \times 115.1}{14.067}, \frac{7 \times 115.1}{2.167} \right) \]Calculating both ends:\[\left( \approx 57.21, \approx 370.40 \right)\]
06

Interpret the Confidence Interval

The 90% confidence interval for the variance is approximately (57.21, 370.40). This means we are 90% confident that the population variance falls within this interval.

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Key Concepts

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

Hypothesis Testing
In statistics, hypothesis testing is a way to determine if there is enough evidence to support a particular belief or hypothesis about a population parameter. It鈥檚 a key technique used to make scientific decisions based on sample data. In the context of this exercise, we're dealing with a chi-square test for variance. This helps us understand if the variability of a sample like the number of mountain climber deaths aligns with what we expect.

The process begins with setting two hypotheses:
  • Null Hypothesis ( H鈧 ): The sample variance is equal to or greater than a specified value, here 136.2 .
  • Alternative Hypothesis ( H鈧 ): The sample variance is less than that specified value. This is a one-tailed test because we are only checking for a decrease in variance.
Next, determine a level of significance (often denoted as 伪 ) that represents the threshold for rejecting the null hypothesis. Here, a 1% significance level implies that we're willing to accept a 1% chance of incorrectly rejecting the null hypothesis.

Finally, the test statistic is compared to a critical value from the chi-square distribution table. If our test statistic is smaller than the critical value, we reject the null hypothesis. Otherwise, as in this problem, we do not reject it, indicating insufficient evidence to prove that the variance is less than 136.2.
Variance
Variance is a statistical measure that represents the spread or dispersion of a set of data points in a dataset. It's calculated by averaging the squared differences from the mean. Variance provides insights into how much the data points differ from the average value.
In the context of mountain climber deaths, variance helps us understand the variability in the number of deaths each year. A high variance signifies that the numbers fluctuate widely from year to year, while a low variance indicates more consistency.

In this exercise, we compare a long-term population variance ( 蟽虏 = 136.2 ) with a sample variance ( s虏 = 115.1 ) over 8 years. By analyzing the sample variance, we can infer if the recent years have shown lesser variability than what we've seen over the long term. If a significant decrease in variance is observed, it could signify more predictable patterns, possibly due to improved safety measures.

Variance is not only useful for understanding dispersion but is also foundational for other statistical analyses, like hypothesis testing and building confidence intervals.
Confidence Interval
A confidence interval is a range of values, derived from sample data, that is likely to include the value of an unknown population parameter. It provides a way to estimate the population parameter with a certain level of confidence. Here, we are constructing a 90% confidence interval for the population variance.

The interval is calculated using the chi-square distribution. The chi-square values are selected based on degrees of freedom, which is derived from the sample size minus one (n - 1). In our example with 8 years of data, there are 7 degrees of freedom.
To construct the interval:
  • Find the chi-square critical values for the lower and upper tails of the distribution (蠂虏0.05 and 蠂虏0.95, respectively).
  • Use the formula to calculate the confidence limits: \[(\frac{(n-1)s虏}{蠂虏_{\alpha/2}}, \frac{(n-1)s虏}{蠂虏_{1-\alpha/2}})\]
In this exercise, the 90% confidence interval is approximately (57.21, 370.40). This means we're 90% confident that the true variance falls within this range. However, a wide interval might imply more sample variability, meaning estimates might not be very precise. Keeping a smaller interval can lead to more accurate estimates.

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Most popular questions from this chapter

In general, how do the hypotheses for chi-square tests of independence differ from those for chi-square tests of homogeneity? Explain.

Vegetation Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by R. A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179-188). Measurements of sepal length in centimeters from andom samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows:$$\begin{array}{ccc}\mathrm{I} & \mathrm{II} & \mathrm{III} \\\5.4 & 5.5 & 6.3 \\\4.9 & 6.5 & 5.8 \\\5.0 & 6.3 & 4.9 \\\5.4 & 4.9 & 7.2 \\\4.4 & 5.2 & 5.7 \\\5.8 & 6.7 & 6.4 \\\5.7 & 5.5 & \\\& 6.1 &\end{array}$$,Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a \(5 \%\) level of significance.

An economist wonders if corporate productivity in some countries is more volatile than that in other countries. One measure of a company's productivity is annual percentage yield based on total company assets. Data for this problem are based on information taken from Forbes Top Companies, edited by J. T. Davis. A random sample of leading companies in France gave the following percentage yields based on assets: $$\begin{aligned} &\begin{array}{cccccccccc} 4.4 & 5.2 & 3.7 & 3.1 & 2.5 & 3.5 & 2.8 & 4.4 & 5.7 & 3.4 & 4.1 \\ 6.8 & 2.9 & 3.2 & 7.2 & 6.5 & 5.0 & 3.3 & 2.8 & 2.5 & 4.5 \end{array}\\\ \end{aligned}$$ Use a calculator to verify that \(s^{2} \approx 2.044\) for this sample of French companies. Another random sample of leading companies in Germany gave the following percentage yields based on assets: $$\begin{array}{lllllllll} 3.0 & 3.6 & 3.7 & 4.5 & 5.1 & 5.5 & 5.0 & 5.4 & 3.2 \\ 3.5 & 3.7 & 2.6 & 2.8 & 3.0 & 3.0 & 2.2 & 4.7 & 3.2 \end{array}$$ Use a calculator to verify that \(s^{2} \approx 1.038\) for this sample of German companies. Test the claim that there is a difference (either way) in the population variance of percentage yields for leading companies in France and Germany. Use a \(5 \%\) level of significance. How could your test conclusion relate to the economist's question regarding volatility (data spread) of corporate productivity of large companies in France compared with large companies in Germany?

A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 31 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was \(58.4 .\) Another random sample of 25 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was \(31.6 .\) Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?

For chi-square tests of independence and of homogeneity, do we use a right- tailed, left-tailed, or two-tailed test?

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