Question

the measurements of the diameters (in inches) of 12 randomly chosen golf balls are listed. at α=0.05, is there enough evidence to reject the claim that the standard deviation of the measurements of these diameters is 0.005? assume the population is normally distributed. click the icon to view the chi - square distribution table. (b) find the critical value(s). $\chi_{0}^{2}=3.816,21.920$ (round to three decimal places as needed. use a comma to separate answers as needed.) identify the rejection region(s). choose the correct graph below. (c) find the standardized test statistic for the $\chi^{2}$ - test. $\chi^{2}=\square$ (round to three decimal places as needed.)

Explanation:

Step 1: Recall the formula for the chi - square test statistic

The formula for the chi - square test statistic when testing a hypothesis about a population variance \(\sigma^{2}\) is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s^{2}\) is the sample variance, and \(\sigma^{2}\) is the hypothesized population variance.

Step 2: Identify the values of \(n\), \(\sigma\)

We know that \(n = 12\) (the number of golf balls in the sample) and \(\sigma=0.005\) (the claimed population standard deviation). So, \(\sigma^{2}=(0.005)^{2}=0.000025\) and \(n - 1=11\)

Step 3: Calculate the sample variance \(s^{2}\) (requires sample data)

To calculate \(s^{2}\), we first find the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\) from the sample data of the 12 golf ball diameters. Then we calculate the sum of squared deviations \(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\) and divide by \(n - 1\) to get \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)

Step 4: Substitute into the chi - square formula

Once we have \(s^{2}\), we substitute \(n-1 = 11\), \(s^{2}\) and \(\sigma^{2}=0.000025\) into the formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\) to get the value of the chi - square test statistic.

Answer:

To solve for the chi - square test statistic \(\chi^{2}\), we use the formula for a hypothesis test about a population standard deviation (or variance):

The formula for the chi - square test statistic is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\)

However, since the problem statement does not provide the sample standard deviation \(s\) or the sample data, we assume that the sample data of the 12 golf ball diameters is used to calculate the sample variance \(s^{2}\). Let's assume that we have calculated the sample standard deviation \(s\) from the data.

1. First, we know that \(n = 12\) (the sample size), \(\sigma=0.005\) (the claimed population standard deviation), so \(\sigma^{2}=(0.005)^{2}=0.000025\)

1. Let's assume that from the sample data of the 12 golf ball diameters, we calculate the sample variance \(s^{2}\). Suppose the sample data is: Let's assume the data values are (for example, if we had the data, we would calculate the sample mean \(\bar{x}\) first and then the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\))

But since the problem is about a chi - square test for variance, and we need to find \(\chi^{2}\), let's assume that we have the sample standard deviation \(s\). Let's say after calculating from the data (the 12 golf ball diameters), we find that \(s\) has a certain value.

Let's assume that the sample standard deviation \(s\) is calculated as follows (we need the actual data to calculate it precisely, but since the problem is likely expecting the use of the formula, we will proceed with the formula steps)

The formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\)

\(n=12\), so \(n - 1=11\)

If we assume that we have calculated \(s^{2}\) from the data, and then we can plug into the formula.

For example, if the sample standard deviation \(s = 0.006\) (this is just an example, in a real problem we would calculate it from the data), then \(s^{2}=0.000036\)

\(\chi^{2}=\frac{11\times0.000036}{0.000025}=\frac{0.000396}{0.000025} = 15.84\)

But since we don't have the actual data, we can't calculate the exact value. However, if we assume that the data is such that when we calculate \((n - 1)s^{2}/\sigma^{2}\)

Let's assume that the sample data gives us \(s\) such that:

Suppose the sample variance \(s^{2}\) is calculated as follows:

Let the 12 data points be \(x_1,x_2,\cdots,x_{12}\)

\(\bar{x}=\frac{\sum_{i = 1}^{12}x_{i}}{12}\)

\(s^{2}=\frac{\sum_{i=1}^{12}(x_{i}-\bar{x})^{2}}{11}\)

After calculating \(s^{2}\), we use \(\chi^{2}=\frac{11\times s^{2}}{(0.005)^{2}}\)

If we assume that from the data, the value of \((n - 1)s^{2}\) is \(0.000396\) (as in our earlier example), then \(\chi^{2}=\frac{0.000396}{0.000025}=15.840\)

(Note: Since the actual data is not provided, we can't give the exact value, but the formula and the steps to calculate are as above. If we had the actual data, we would:

1. Calculate the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)

1. Calculate the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)

1. Then calculate \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\) with \(n = 12\) and \(\sigma=0.005\))

If we assume that the sample data gives us a sample standard deviation \(s\) such that when we calculate \(\chi^{2}\), the value is (for example, if the data leads to \(s = 0.006\)) \(\chi^{2}=15.840\)