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A precision instrument is guaranteed to read accurately to within 2 units. A sample of four instrument readings on the same object yielded the measurements \(353,351,351,\) and \(355 .\) a. Test the null hypothesis that \(\sigma=.7\) against the alternative \(\sigma>.7\). Use \(\alpha=.05 .\) b. Find a \(90 \%\) confidence interval for the population variance.

Short Answer

Expert verified
Answer: Yes, there is enough evidence to suggest that the true standard deviation is larger than 0.7 because we rejected the null hypothesis which stated the standard deviation is equal to 0.7. The test statistic, \(X^2\), is greater than the critical value and falls within the rejection region.

Step by step solution

01

Define the Null and Alternative Hypotheses

Define the null hypothesis H鈧 and the alternative hypothesis H鈧: H鈧: 蟽 = 0.7 H鈧: 蟽 > 0.7
02

Calculate the Sample Mean and Sample Variance

Calculate the sample mean and sample variance using the formulas: Sample Mean: \(\bar{x} = \frac{\sum x_i}{n}\) Sample Variance: \(s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}\) We have four measurements: \(353, 351, 351,\) and \(355\). \(\bar{x} = \frac{353 + 351 + 351 + 355}{4} = \frac{1410}{4} = 352.5\) \(s^2 = \frac{(353 - 352.5)^2 + (351 - 352.5)^2 + (351 - 352.5)^2 + (355 - 352.5)^2}{4-1} = \frac{5}{3} \approx 1.67\)
03

Calculate the Test Statistic for Chi-Square Distribution

Calculate the test statistic, which follows a Chi-Square distribution with n - 1 degrees of freedom: \(X^2 = \frac{(n-1)s^2}{\sigma_0^2}\) Here, n = 4, \(s^2 = 1.67\), and \(\sigma^2_0 = 0.7^2 = 0.49\). \(X^2 = \frac{(4-1)(1.67)}{0.49} \approx 10.24\)
04

Determine the Rejection Region and Make Conclusion

Using the significance level \(\alpha = 0.05\), we need to reject the null hypothesis if we are in the rejection region. Rejection region: \(X^2 > \chi^2_{1-\alpha, n-1}\) Here, \(\chi^2_{0.95, 3} \approx 7.81\). Since \(10.24 > 7.81\), we reject the null hypothesis H鈧 and conclude that there is evidence to support the alternative hypothesis, which suggests that the true standard deviation is greater than 0.7.
05

Calculate the Confidence Interval for the Population Variance

To find the 90% confidence interval for the population variance, use the Chi-Square distribution and the formulas: \(P \left(\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}, n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}, n-1}}\right) = 0.9\) Here, 伪 = 0.1, n = 4, \(s^2 = 1.67\), and we use the Chi-Square critical values for 3 degrees of freedom, \(\chi^2_{0.95} \approx 0.35\) and \(\chi^2_{0.05} \approx 9.35\). Lower limit = \(\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}, n-1}} = \frac{(4-1)(1.67)}{9.35} \approx 0.537\) Upper limit = \(\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}, n-1}} = \frac{(4-1)(1.67)}{0.35} \approx 14.37\) Thus, the 90% confidence interval for the population variance is approximately (0.537, 14.37).

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Key Concepts

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

Understanding Confidence Intervals
The concept of a confidence interval is crucial in statistics. It gives us a range of values in which we estimate a population parameter, like the mean or variance, to lie. Confidence intervals are associated with a confidence level, often expressed as a percentage like 90%, which indicates the probability that the interval contains the true parameter value.

Confidence intervals help to account for the natural variability in data. They allow us to make educated guesses about population parameters based on sample statistics. For example, if we calculate a 90% confidence interval for a population variance, we are 90% confident that the true variance is within this interval.

To find a confidence interval for the variance, we rely on the chi-square distribution. The procedure involves using sample variance and the chi-square statistic, which varies depending on the desired confidence level. The calculated interval usually has an upper and lower limit, providing a range where the true population variance is likely to be found.
Chi-Square Distribution in Hypothesis Testing
The chi-square distribution is a fundamental concept in hypothesis testing, particularly when dealing with variance. It's a distribution that arises from the summation of the squares of independent standard normal variables. This makes it immensely useful for tests involving sample variance or standard deviation.

In our specific exercise, the chi-square test helps us decide whether to accept or reject the hypothesis about the population standard deviation. We compute the chi-square statistic using the formula:\[X^2 = \frac{(n-1)s^2}{\sigma_0^2}\]Here, \(n\) is the sample size, \(s^2\) is the sample variance, and \(\sigma_0^2\) is the hypothesized variance. The result is then compared to a chi-square critical value determined by the significance level \(\alpha\) and degrees of freedom \(n-1\).

This comparison helps determine whether the sample data provides sufficient evidence to reject the null hypothesis. Understanding the chi-square distribution's properties and critical value tables are keys to conducting successful variance hypothesis tests.
What is Sample Variance?
Sample variance is a measure of data dispersion around the sample mean and provides an estimate of the overall population variance. Calculated from a sample, it is used to quantify the degree of variation within a dataset.

The formula for sample variance \(s^2\) is:\[s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}\]Where \(x_i\) are the sample data points, \(\bar{x}\) is the sample mean, and \(n-1\) represents the degrees of freedom. Using \(n-1\) instead of \(n\) corrects the bias in the estimation of the population variance, due to the fact that we estimate the mean and not use the true mean.

Sample variance is foundational in statistics as it reflects how much individual data points in a sample differ from the sample mean. In our exercise, it plays a critical role in hypothesis testing and constructing confidence intervals, providing an indispensable tool for statistical inference.

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Most popular questions from this chapter

An experiment is conducted to compare two new automobile designs. Twenty people are randomly selected, and each person is asked to rate each design on a scale of 1 (poor) to 10 (excellent). The resulting ratings will be used to test the null hypothesis that the mean level of approval is the same for both designs against the alternative hypothesis that one of the automobile designs is preferred. Do these data satisfy the assumptions required for the Student's \(t\) -test of Section 10.4 ? Explain.

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