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In hypothesis testing, the chi-square distribution is critical for various types of tests, especially tests involving the population variance. It is essential when you want to understand the variability within your data based on a sample. The chi-square distribution is asymmetric and its shape depends on the degrees of freedom (df). The more degrees of freedom, the more the distribution looks like a normal distribution. For our problem, we use the chi-square distribution because we are comparing our sample standard deviation to a known population standard deviation. The formula to compute the test statistic is: $$\chi^2 = \frac{(n-1)s^2}{\theta^2}$$ where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\theta\) is the population standard deviation. The test statistic follows a chi-square distribution with \(n-1\) degrees of freedom.

The sample standard deviation measures the dispersion of sample data points around the sample mean. It is crucial when estimating and making conclusions about the population. The formula for sample standard deviation (s) is: $$s = \sqrt{\frac{\sum (X_i - \overline{X})^2}{n-1}}$$ where \(X_i\) are the sample data points, \(\overline{X}\) is the sample mean, and \(n\) is the sample size. In our exercise, the sample standard deviation is 7500 psi. We use this value in our test statistic formula to compare against the population standard deviation of 7000 psi.

The level of significance (\(\alpha\)) is the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by the researcher before conducting the hypothesis test. Common levels of significance are \(0.05\), \(0.01\), and \(0.10\). For our exercise, the level of significance is \(0.01\). This means there is a 1% risk of concluding that the population standard deviation is greater than 7000 psi when it actually is not. We use this \(\alpha\) value to determine the critical value from the chi-square distribution table.

The alternative hypothesis (\(H_a\)) represents what we are trying to prove. It contradicts the null hypothesis (\(H_0\)). In our scenario, the null hypothesis is that the population standard deviation is 7000 psi (\(\sigma = 7000\)). The alternative hypothesis states that the population standard deviation exceeds 7000 psi (\(\sigma > 7000\)). We seek to gather evidence to support this claim using our sample data. If our test statistic exceeds the critical value from the chi-square distribution, we reject the null hypothesis in favor of the alternative. However, in this problem, the computed chi-square is less than the critical chi-square value, leading us to not reject the null hypothesis.