/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A set of solar batteries is used... [FREE SOLUTION] | 91Ó°ÊÓ

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A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance \(\sigma^{2}\) of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of \(\sigma^{2}=23\) months (squared) is most desirable for these batteries. A random sample of 22 batteries gave a sample variance of \(14.3\) months (squared). (i) Using a \(0.05\) level of significance, test the claim that \(\sigma^{2}=23\) against the claim that \(\sigma^{2}\) is different from 23 . (ii) Find a \(90 \%\) confidence interval for the population variance \(\sigma^{2}\). (iii) Find a \(90 \%\) confidence interval for the population standard deviation \(\sigma .\)

Short Answer

Expert verified
(i) No evidence the variance differs from 23. (ii) Variance CI: (9.19, 25.88). (iii) Standard deviation CI: (3.03, 5.08).

Step by step solution

01

State the Hypotheses

For part (i), we need to test the variance. We set up the null hypothesis as \( H_0: \sigma^2 = 23 \) and the alternative hypothesis as \( H_a: \sigma^2 eq 23 \). This is a two-tailed test since we are interested in any difference from 23.
02

Determine Test Statistic and Distribution

We use the chi-square test for variance. The test statistic is calculated as \( \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \), where \( n = 22 \), \( s^2 = 14.3 \), and \( \sigma_0^2 = 23 \). The distribution is \( \chi^2 \) with \( n-1 = 21 \) degrees of freedom.
03

Calculate the Test Statistic

Substituting the values: \[ \chi^2 = \frac{(22-1) \times 14.3}{23} \approx 13.046 \].
04

Determine the Critical Value and Conclusion

For a significance level of 0.05 in a two-tailed test with 21 degrees of freedom, the critical chi-square values are approximately 10.283 and 34.170. Since 13.046 falls between these values, we fail to reject the null hypothesis.
05

Conclusion for Hypothesis Test

We do not have sufficient evidence to reject the claim that the population variance is 23 months squared.
06

Find Confidence Interval for Variance

The \( 90\% \) confidence interval for the variance is given by \[ \left(\frac{(n-1)s^2}{\chi^2_{upper}}, \frac{(n-1)s^2}{\chi^2_{lower}}\right) \], where \( \chi^2_{lower} \) and \( \chi^2_{upper} \) correspond to the lower and upper critical values for a \( 90\% \) confidence level with 21 degrees of freedom.
07

Calculate Confidence Interval for Variance

Using \( \chi^2_{lower} \approx 11.591 \) and \( \chi^2_{upper} \approx 32.671 \), the interval is: \[ \left(\frac{21 \times 14.3}{32.671}, \frac{21 \times 14.3}{11.591}\right) = (9.19, 25.88) \].
08

Find Confidence Interval for Standard Deviation

The \( 90\% \) confidence interval for the standard deviation is the square root of the variance interval: \( \left(\sqrt{9.19}, \sqrt{25.88}\right) \).
09

Calculate Confidence Interval for Standard Deviation

This results in: \( (3.03, 5.08) \).

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make inferences or decisions about a population parameter based on sample data. In this exercise, we are testing the variance of solar battery lifetimes, which affects how long the satellite can operate effectively.
The process begins by clearly stating the hypotheses:
  • Null Hypothesis (\(H_0\)): \(\sigma^2 = 23\) — This suggests that the actual variance of the battery lifetimes is 23 months squared, as desired by the engineers.
  • Alternative Hypothesis (\(H_a\)): \(\sigma^2 eq 23\) — This suggests there is a difference from the expected variation, which could be either more or less than 23 months squared.

Next, we choose the appropriate test, which is a chi-square test in this case, because it is used for variance testing. By calculating the chi-square statistic and comparing it against critical values from the chi-square distribution, we decide whether to reject the null hypothesis. A key step involves determining degrees of freedom, which is \(n - 1\), with \(n\) being the sample size.
Ultimately, the hypothesis test concluded that there was not enough evidence to reject the null hypothesis — indicating the variance might indeed be close to the desired 23 months squared.
Confidence Interval
A confidence interval gives an estimated range of values likely to include a population parameter, in this case, the population variance. Calculating a confidence interval for the variance helps provide a range within which the true variance is expected to fall, with a specific level of confidence.

For this task, the steps involve determining a 90% confidence interval for both the population variance and the standard deviation.
The formula for the confidence interval of the variance is:\[\left(\frac{(n-1)s^2}{\chi^2_{upper}}, \frac{(n-1)s^2}{\chi^2_{lower}}\right)\] Here, \(\chi^2_{lower}\) and \(\chi^2_{upper}\) are critical values obtained from the chi-square distribution.
This results in an interval providing a range for the variance: \((9.19, 25.88)\).
This method can then be extended to find the confidence interval for the standard deviation by taking the square root of these interval bounds. This way, we are 90% confident that the actual variance is between 9.19 and 25.88, and the standard deviation is between 3.03 and 5.08 months.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of this exercise, understanding the standard deviation helps assess how spread out the battery lifetimes are, influencing satellite performance.

Standard deviation is derived directly from the variance, which in mathematical terms is the square root of the variance:\[\sigma = \sqrt{\sigma^2}\]
In this given scenario, once the confidence interval for the variance was determined as \((9.19, 25.88)\), the confidence interval for the standard deviation was calculated by taking the square root of these numbers. This resulted in a 90% confidence interval of \((3.03, 5.08)\) for the standard deviation.
Such a range provides insight into how much individual battery lifetimes deviate from the average, which is crucial for predicting the satellite's operational consistency and planning for battery replacements or maintenance strategies.
Sample Variance
Sample variance is a statistic that provides a measure of how much the data values in a sample deviate from the mean of the sample. It is essential for estimating the variance of a population when only a subset of data is available.

In calculating sample variance, each data point's deviation from the mean is squared and averaged over the sample size minus one (\(n-1\)), providing an unbiased estimator of the population variance. This bias correction is necessary due to the finite sample size.
The formula for sample variance \(s^2\) is:\[s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}\] where \(x_i\) are individual data points and \(\bar{x}\) is the sample mean.
In the provided exercise, the sample variance was calculated as \(14.3\) months squared. This serves as the baseline statistic used to test hypotheses about the overall battery variance and to calculate confidence intervals.
Employing sample variance is crucial in ensuring that decisions and predictions based on sample data are as accurate as possible.

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Most popular questions from this chapter

Zane is interested in the proportion of people who recycle each of three distinct products: paper, plastic, electronics. He wants to test the hypothesis that the proportion of people recycling each type of product differs by age group: \(12-18\) years old, \(19-30\) years old, \(31-40\) years old, over 40 years old. Describe the sampling method appropriate for a test of homogeneity regarding recycled products and age.

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