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The Chi-Square Test is a statistical method used to determine if there is a significant difference between observed and expected data. It's widely used in hypothesis testing for categorical data. Here, it's utilized to test the variance of a normally distributed population. When working with the Chi-Square Test, it’s important to understand how the chi-square statistic is computed and interpreted.
The formula for the chi-square test statistic is given by: \[ \text{Test Statistic} = \frac{(n-1)s^2}{\theta^2} \] where:

• n = sample size
• s = sample standard deviation
• \( \theta \)= hypothesized population standard deviation

We calculate this to help determine whether our sample provides enough evidence to reject the null hypothesis that the population standard deviation is equal to the hypothesized value.
In our exercise, with a sample size \( n = 15 \),sample standard deviation \( s = 37.4 \),and hypothesized standard deviation \( \theta = 35 \),we plug these values into our formula to get \( \text{Test Statistic} \text{Test Statistic} \approx 15.99 \). This value will later be compared to a critical value to make a decision about the null hypothesis.

The sample standard deviation \( s \)measures the spread of sample data around the sample mean. It's a crucial part of calculating the test statistic in hypothesis testing involving variance. To compute \( s \), follow these steps:

• Calculate the mean of the sample.
• Subtract the mean from each data point and square the result.
• Sum all squared results.
• Divide the sum by \( n-1 \), where \( n \)is the sample size.
• Take the square root of the result.

In our exercise, the sample standard deviation is given as \( s = 37.4 \)which indicates the extent of variation within our sample data set. This value is then used in the chi-square test statistic formula to determine if the sample provides sufficient evidence to conclude that the population standard deviation is greater than the hypothesized value.

Critical value analysis is a method used in hypothesis testing to determine the threshold for rejecting the null hypothesis. This threshold is based on the significance level \( \alpha \), which is the probability of rejecting the null hypothesis when it is actually true.
In this exercise, we are testing at the \( \alpha = 0.01 \)level of significance with \( df = n-1 = 14 \)degrees of freedom. Using a chi-square distribution table (found in most statistical textbooks or online resources), we find the critical value corresponding to our \( \alpha \).At \( \alpha = 0.01 \),and \( df = 14 \),the critical chi-square value is approximately \( 29.141 \).

This critical value defines the cutoff in the chi-square distribution:

• If the test statistic exceeds this critical value, we reject the null hypothesis.
• Otherwise, we fail to reject the null hypothesis.

In this scenario, our calculated test statistic was \( \approx 15.99 \),which is less than the critical value \( \approx 29.141 \).Therefore, we fail to reject the null hypothesis, suggesting that the sample does not provide sufficient evidence to conclude that the population standard deviation is greater than 35.