b. Can it be said that the sample is above average? 2. The owner of a factory that sells a particular bottled fruit juice claims that the average capacity of their product is 250 ml . to test the claim, a consumer group gets a sample of 100 such bottles, calculates of each bottle, and then finds the mean capacity to be 248 ml . The standard deviation \( s \) is 15 ml . Is the claim true? 3. In a plant nursery, the owner thinks that the lengths of the seedlings in a box sprayed with 3. \( \begin{array}{l}\text { ander andard }\end{array} \) days and standard

Let's break down the problem step by step. ### Known Conditions 1. **Factory Juice Claim**: - Claimed average capacity: \( \mu_0 = 250 \) ml - Sample size: \( n = 100 \) - Sample mean capacity: \( \bar{x} = 248 \) ml - Sample standard deviation: \( s = 15 \) ml ### Step 1: Hypothesis Testing To determine if the claim is true, we will perform a hypothesis test. - **Null Hypothesis (\( H_0 \))**: The average capacity of the bottles is 250 ml (\( \mu = 250 \)). - **Alternative Hypothesis (\( H_a \))**: The average capacity of the bottles is not 250 ml (\( \mu \neq 250 \)). ### Step 2: Calculate the Test Statistic We will use the formula for the test statistic \( z \): \[ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Substituting the known values: - \( \bar{x} = 248 \) - \( \mu_0 = 250 \) - \( s = 15 \) - \( n = 100 \) Let's calculate \( z \). ### Step 3: Perform the Calculation Now, I will calculate the value of \( z \). \[ z = \frac{248 - 250}{15 / \sqrt{100}} \] This simplifies to: \[ z = \frac{-2}{15 / 10} = \frac{-2}{1.5} \] Now, let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{-2}{\left(\frac{15}{10}\right)}\) - step1: Remove the parentheses: \(\frac{-2}{\frac{15}{10}}\) - step2: Reduce the fraction: \(\frac{-2}{\frac{3}{2}}\) - step3: Multiply by the reciprocal: \(-2\times \frac{2}{3}\) - step4: Multiply: \(-\frac{2\times 2}{3}\) - step5: Multiply: \(-\frac{4}{3}\) The calculated value of the test statistic \( z \) is approximately: \[ z = -\frac{4}{3} \approx -1.33 \] ### Step 4: Determine the Critical Value For a two-tailed test at a significance level (commonly \( \alpha = 0.05 \)), we need to find the critical z-values. The critical z-values for a two-tailed test at \( \alpha = 0.05 \) are approximately \( \pm 1.96 \). ### Step 5: Compare the Test Statistic to the Critical Value - If \( z < -1.96 \) or \( z > 1.96 \), we reject the null hypothesis. - If \( -1.96 < z < 1.96 \), we fail to reject the null hypothesis. Since \( z \approx -1.33 \) falls within the range \( -1.96 < z < 1.96 \), we fail to reject the null hypothesis. ### Conclusion Based on the sample data, there is not enough evidence to reject the factory owner's claim that the average capacity of the bottled fruit juice is 250 ml. Therefore, we conclude that the claim is likely true. ### Additional Question Regarding the question about the seedlings in a plant nursery, it seems that the information is incomplete. If you provide the details about the seedlings, I can assist you further with that part of the problem.