A recently collected data set on coin weights gives weights of pennies minted after 1983. Here are the statistics for those weights sample size equals 37 simple mean equals 2.49910 g simple standard deviation equals 0.01648 g. Use a 0.05 significance level to test a claim that the sample is from a population of pennies with weights having a standard deviation greater than 0.01000 g. 

To test the claim that the standard deviation of the population of pennies is greater than 0.01000 g, we can use a **chi-square test** for the population variance (which is related to the standard deviation).

### Step-by-step process:

1. **Define the null and alternative hypotheses:**

   - \( H_0 \) (null hypothesis): The population standard deviation is **less than or equal to** 0.01000 g.
     \[
     H_0: \sigma \leq 0.01000
     \]
   - \( H_1 \) (alternative hypothesis): The population standard deviation is **greater than** 0.01000 g.
     \[
     H_1: \sigma > 0.01000
     \]

2. **Test statistic:**
   The chi-square test statistic for variance is given by:
   \[
   \chi^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2}
   \]
   Where:
   - \( n \) is the sample size.
   - \( s \) is the sample standard deviation.
   - \( \sigma_0 \) is the hypothesized population standard deviation (0.01000 g).

   Plugging in the values:
   - \( n = 37 \)
   - \( s = 0.01648 \, \text{g} \)
   - \( \sigma_0 = 0.01000 \, \text{g} \)

   First, calculate \( s^2 \) and \( \sigma_0^2 \):
   \[
   s^2 = (0.01648)^2 = 0.000271635 \quad \text{and} \quad \sigma_0^2 = (0.01000)^2 = 0.00010000
   \]
   
   Now, compute the chi-square statistic:
   \[
   \chi^2 = \frac{(37 - 1) \cdot 0.000271635}{0.00010000} = \frac{36 \cdot 0.000271635}{0.00010000} = 97.7886
   \]

3. **Degrees of freedom (df):**
   The degrees of freedom for the chi-square test are \( n - 1 \). So:
   \[
   df = 37 - 1 = 36
   \]

4. **Find the critical value:**
   At a significance level of \( \alpha = 0.05 \) for a right-tailed test with 36 degrees of freedom, we need to find the critical value from the chi-square distribution table.

   The critical value for \( \chi^2_{0.05, 36} \) is approximately 50.998 (based on chi-square distribution tables).

5. **Decision rule:**
   - If \( \chi^2 \) (calculated) > \( \chi^2_{\text{critical}} \), reject the null hypothesis.
   - If \( \chi^2 \) (calculated) ≤ \( \chi^2_{\text{critical}} \), fail to reject the null hypothesis.

6. **Conclusion:**
   Since the calculated \( \chi^2 = 97.7886 \) is greater than the critical value \( 50.998 \), we **reject the null hypothesis**.

### Final conclusion:
There is sufficient evidence at the 0.05 significance level to support the claim that the population standard deviation is greater than 0.01000 g.

---

Do you want more details or have any questions?  
Here are 5 related questions to expand on this concept:

1. How would the hypothesis test change if the sample size were larger?
2. What is the effect of choosing a different significance level (e.g., 0.01) on the test result?
3. How can we compute a confidence interval for the population variance?
4. Why is the chi-square distribution used for variance testing?
5. What are the assumptions behind using the chi-square test for variance?

**Tip:** The chi-square test for variance is sensitive to non-normal data, so it's important to verify the normality assumption of your sample!

A recently collected data set on coin weights gives weights of pennies minted after 1983. Here are the statistics for those weights: sample size equals 37, simple mean equals 2.49910 g, simple standard deviation equals 0.01648 g. Use a 0.05 significance level to test a claim that the sample is from a population of pennies with weights having a standard deviation greater than 0.01000 g.

Learn how to perform a chi-square test to evaluate whether the variance in the weight of pennies (minted after 1983) is greater than 0.01000 g. This problem uses a sample size of 37, a sample standard deviation of 0.01648 g, and a 0.05 significance level, making it ideal for high school students studying hypothesis testing and variance.

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