A company claims that its packages of beads contain, on average, 50 beads with a standard deviation of 5.4 beads. In a hypothesis test of this is the \( z \)-statistic for the sample? Round the claim, \( H_{0} \) is \( \mu=50 \) and \( \mathrm{H}_{\mathrm{a}} \) is \( \mu \neq 50 \), where \( \mu \) is the average number of beads per package. Based on a sample of 20 packages, Celia calculates a mean of 52.5 beads per package. Formula for z-statistic: \( z=\frac{\bar{x}-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} \)

To calculate the \( z \)-statistic for the sample, we can use the formula provided: \[ z = \frac{\bar{x} - \mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} \] Given: - Claim: \( \mu = 50 \) beads - Standard deviation: \( \sigma = 5.4 \) beads - Sample size: \( n = 20 \) packages - Sample mean: \( \bar{x} = 52.5 \) beads Substitute the given values into the formula: \[ z = \frac{52.5 - 50}{\left(\frac{5.4}{\sqrt{20}}\right)} \] Now, we can calculate the \( z \)-statistic. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(52.5-50\right)}{\left(\frac{5.4}{\sqrt{20}}\right)}\) - step1: Remove the parentheses: \(\frac{52.5-50}{\frac{5.4}{\sqrt{20}}}\) - step2: Simplify the root: \(\frac{52.5-50}{\frac{5.4}{2\sqrt{5}}}\) - step3: Divide the terms: \(\frac{52.5-50}{\frac{27}{10\sqrt{5}}}\) - step4: Subtract the numbers: \(\frac{2.5}{\frac{27}{10\sqrt{5}}}\) - step5: Convert the expressions: \(\frac{\frac{5}{2}}{\frac{27}{10\sqrt{5}}}\) - step6: Multiply by the reciprocal: \(\frac{5}{2}\times \frac{10\sqrt{5}}{27}\) - step7: Reduce the numbers: \(5\times \frac{5\sqrt{5}}{27}\) - step8: Multiply: \(\frac{5\times 5\sqrt{5}}{27}\) - step9: Multiply: \(\frac{25\sqrt{5}}{27}\) The \( z \)-statistic for the sample is approximately 2.070433.