Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 18 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.11 oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.99 oz and 12.51 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.13 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.13 oz. Use a 0.01 significance level. Complete parts (a) through (d) below.

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Find the P-Value

(Round to Four Decimal places)

### Statistical Analysis of Soda Can Volumes: An Educational Overview

Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 18 cans of the soda drink. These volumes have a mean of 12.19 oz and a standard deviation of 0.11 oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.99 oz and 12.51 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.13 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.13 oz. Use a 0.01 significance level. Complete parts (a) through (d) below.

**a. State the hypotheses:**
   - \( H_0 \): \( \sigma = 0.13 \) oz
   - \( H_1 \): \( \sigma \neq 0.13 \) oz

**b. Compute the test statistic:**

The test statistic for the chi-square test can be calculated using the formula:

\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2_0} \]

where:
  - \( n \) is the sample size (18),
  - \( s \) is the sample standard deviation (0.11 oz),
  - \( \sigma_0 \) is the population standard deviation under the null hypothesis (0.13 oz).

Using this formula, the computed test statistic is:

\[ \chi^2 = 12.172 \]

(Round to three decimal places as needed.)

**c. Find the P-value:**
   - The P-value needs to be calculated using chi-square distribution tables or statistical software. It represents the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct.
   
(Round to four decimal places as needed.)

**d. Conclusion:**
   - Compare the P-value with the significance level (0.01) to determine whether to reject the null hypothesis. If the P-value is less than 0.01, we reject the null hypothesis, indicating that the standard deviation is not equal to 0.13 oz. If the P-value is greater than 0.01,
Transcribed Image Text:### Statistical Analysis of Soda Can Volumes: An Educational Overview Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 18 cans of the soda drink. These volumes have a mean of 12.19 oz and a standard deviation of 0.11 oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.99 oz and 12.51 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.13 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.13 oz. Use a 0.01 significance level. Complete parts (a) through (d) below. **a. State the hypotheses:** - \( H_0 \): \( \sigma = 0.13 \) oz - \( H_1 \): \( \sigma \neq 0.13 \) oz **b. Compute the test statistic:** The test statistic for the chi-square test can be calculated using the formula: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2_0} \] where: - \( n \) is the sample size (18), - \( s \) is the sample standard deviation (0.11 oz), - \( \sigma_0 \) is the population standard deviation under the null hypothesis (0.13 oz). Using this formula, the computed test statistic is: \[ \chi^2 = 12.172 \] (Round to three decimal places as needed.) **c. Find the P-value:** - The P-value needs to be calculated using chi-square distribution tables or statistical software. It represents the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct. (Round to four decimal places as needed.) **d. Conclusion:** - Compare the P-value with the significance level (0.01) to determine whether to reject the null hypothesis. If the P-value is less than 0.01, we reject the null hypothesis, indicating that the standard deviation is not equal to 0.13 oz. If the P-value is greater than 0.01,
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