Let's solve the problem step by step: ### **Step 1: Identify the Given Information** - Population mean (\(\mu\)) = 80 - Sample size (\(n\)) = 6 - Sample mean after treatment (\(M\)) = 72 **Explanation:** We are given the population mean, sample size, and the sample mean after treatment. We are likely being asked to determine how unusual this sample mean is, possibly by calculating a z-score or evaluating the likelihood of getting such a result. --- ### **Step 2: State the Formula to Use** To evaluate how unusual the sample mean is, we use the z-score formula for sample means: \[ z = \frac{M - \mu}{\sigma_M} \] where \(\sigma_M\) is the standard error of the mean. If population standard deviation (\(\sigma\)) is not given, we cannot compute a numerical answer. Let's assume for now that it is not given in the problem. **Explanation:** The z-score formula tells us how many standard errors the sample mean is away from the population mean. Since the standard deviation is not given, we can only show the formula setup. --- ### **Step 3: Set Up the Calculation** Since the population standard deviation (\(\sigma\)) is not provided, the formula for standard error is: \[ \sigma_M = \frac{\sigma}{\sqrt{n}} \] So, the z-score formula becomes: \[ z = \frac{72 - 80}{\sigma/\sqrt{6}} \] **Explanation:** We have set up the correct formula and plugged in the known values. The exact answer depends on the missing value of \(\sigma\). --- ## **Final Answer** **Without the population standard deviation (\(\sigma\)), we cannot calculate a numerical z-score.** But the formula to evaluate the sample mean is: \[ z = \frac{72 - 80}{\sigma/\sqrt{6}} \] If \(\sigma\) is given, plug it in to compute the z-score and determine how unusual the sample mean is.