The measurements of the diameters​ (in inches) of 12 randomly chosen golf balls are listed. At alpha equals 0.05α=0.05​, is there enough evidence to reject the claim that the standard deviation of the measurements of these diameters is 0.0050.005​? Assume the population is normally distributed.
LOADING...  Click the icon to view the​ Chi-Square Distribution Table.
1.6771.677
1.6821.682
1.6791.679

1.6791.679
1.6791.679
1.6781.678
1.6821.682
1.6791.679
1.6831.683
1.6811.681
1.6771.677
1.6791.679
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Part 1
​(a) Write the claim mathematically and identify Upper H 0H0 and Upper H Subscript aHa. Choose the correct answer below.
A.
Upper H 0 : sigma greater than 0.005H0: σ>0.005​; Upper H Subscript a Baseline : sigma less than or equals 0.005Ha: σ≤0.005 ​(Claim)
B.
Upper H 0 : sigma greater than or equals 0.005H0: σ≥0.005​; Upper H Subscript a Baseline : sigma less than 0.005Ha: σ<0.005 ​(Claim)
C.
Upper H 0 : sigma equals 0.005H0: σ=0.005 ​(Claim); Upper H Subscript a Baseline : sigma not equals 0.005Ha: σ≠0.005
Your answer is correct.D.
Upper H 0 : sigma less than or equals 0.005H0: σ≤0.005 ​(Claim); Upper H Subscript a Baseline : sigma greater than 0.005Ha: σ>0.005
Part 2
​(b) Find the critical​ value(s).
chi Subscript 0 Superscript 2χ20equals=enter your response here
​(Round to three decimal places as needed. Use a comma to separate answers as​ needed.)

### Part 1: Hypothesis Testing Setup

The problem asks to test if there is enough evidence to reject the claim that the standard deviation of the measurements of the golf ball diameters is 0.005 at the \(\alpha = 0.05\) significance level. Let's identify the null and alternative hypotheses:

- **Claim:** The standard deviation \(\sigma = 0.005\).
- **Null Hypothesis (\(H_0\)):** \(\sigma = 0.005\)
- **Alternative Hypothesis (\(H_a\)):** \(\sigma \neq 0.005\)

This corresponds to a two-tailed test, so the correct answer for Part 1 is:
- **Option C:** \(H_0: \sigma = 0.005\) (Claim); \(H_a: \sigma \neq 0.005\)

### Part 2: Finding the Critical Value(s)

To find the critical value(s) for this test, we will use the chi-square distribution. The formula for the chi-square statistic in a test for the variance is:

\[
\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2}
\]

Where:
- \(n\) is the sample size
- \(s\) is the sample standard deviation
- \(\sigma_0\) is the population standard deviation under the null hypothesis

**Calculations:**

1. **Sample Size (\(n\))**: 12
2. **Sample Standard Deviation (\(s\))**: We need to calculate this from the provided data.

#### Calculation of Sample Standard Deviation \(s\):

The data provided is:

1.677, 1.682, 1.679, 1.679, 1.679, 1.678, 1.682, 1.679, 1.683, 1.681, 1.677, 1.679

To calculate the sample standard deviation, \(s\), we first find the sample mean, \(\bar{x}\), and then use the formula:

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]

Let's compute this step-by-step.

1. **Calculate the sample mean (\(\bar{x}\)):**
\[
\bar{x} = \frac{\sum x_i}{n} = \frac{1.677 + 1.682 + 1.679 + 1.679 + 1.679 + 1.678 + 1.682 + 1.679 + 1.683 + 1.681 + 1.677 + 1.679}{12}
\]

2. **Calculate the sample variance and then \(s\)**:

\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\]
\[
s = \sqrt{s^2}
\]

We will calculate these values now.The sample mean \(\bar{x}\) is approximately \(1.6796\) and the sample standard deviation \(s\) is approximately \(0.001975\).

### Critical Value(s) Calculation:

Given:
- \(\sigma_0 = 0.005\)
- \(n = 12\)
- \(\alpha = 0.05\) (two-tailed test)

The degrees of freedom for the chi-square distribution are \(df = n - 1 = 11\). We need to find the critical values for \(\chi^2\) using the chi-square distribution table for a two-tailed test with \(\alpha/2 = 0.025\) for each tail.

#### Calculating Critical Values:

The critical values will be:
- \(\chi^2_{0.025,11}\)
- \(\chi^2_{0.975,11}\)

Let’s find the critical values from the chi-square distribution table.The critical values for the chi-square distribution with \(df = 11\) and \(\alpha = 0.05\) (two-tailed) are:

- Lower critical value: \(\chi^2_{0.025, 11} = 3.816\)
- Upper critical value: \(\chi^2_{0.975, 11} = 21.920\)

### Conclusion

To test the hypothesis, we will calculate the chi-square statistic using the formula:

\[
\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2}
\]

Substitute the values to find the test statistic and compare it with the critical values to determine whether to reject the null hypothesis.

Would you like to proceed with the calculation of the test statistic and make a decision regarding the hypothesis test?

Explore hypothesis testing for the standard deviation of golf ball diameters with a detailed solution. Test if there is enough evidence to reject the claim that the standard deviation is 0.005 at alpha = 0.05 level. Includes calculations and critical values.

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