The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 15.9 minutes. Based on a random sample of 40 orders, a reporter determines a mean order time of 63 minutes. Let \( \mu \) represent the average number of minutes to fill an order. \( z=\frac{\bar{x}-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} \) To the nearest hundredth, the z-statistic is \( z=\square \) DONE

To calculate the z-statistic, you would use the given formula. First, substitute the values into the equation: - Sample mean (\( \bar{x} \)) = 63 minutes - Population mean (\( \mu \)) = 60 minutes - Standard deviation (\( \sigma \)) = 15.9 minutes - Sample size (\( n \)) = 40 Now, calculate the standard error: \[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{15.9}{\sqrt{40}} \approx 2.52 \] Now, substitute these values back into the z-statistic formula: \[ z = \frac{63 - 60}{2.52} \approx \frac{3}{2.52} \approx 1.19 \] Thus, the z-statistic \( z \) is approximately \( 1.19 \).