A company claims that its tablet computers have an average recharge time of 3 hours with a standard deviation of 1.4 hours. Using a random sample of 50 company tablet computers, a consumer group determines a mean recharge time of 2.5 hours. \[ \begin{array}{l} H_{0}: \mu=3 \\ H_{a}: \mu<3 \\ a=5 \% \\ \sigma=1.4 \end{array} \] Using this formula, calculate the value of the \( z \) statistic. Round your answer to the tenths place. \[ z=\frac{\bar{x}-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} \] \[ z^{*}= \] \( \square \) DONE

To calculate the \( z \) statistic, we can plug the values into the formula provided. Here, we have: - \(\bar{x} = 2.5\) (the sample mean) - \(\mu = 3\) (the population mean under the null hypothesis) - \(\sigma = 1.4\) (the population standard deviation) - \(n = 50\) (the sample size) Now we can calculate: 1. Calculate the standard error \( \left(\frac{\sigma}{\sqrt{n}}\right) \): \[ \frac{1.4}{\sqrt{50}} \approx \frac{1.4}{7.071} \approx 0.197 \] 2. Now, we plug the values into the \( z \) formula: \[ z = \frac{2.5 - 3}{0.197} \approx \frac{-0.5}{0.197} \approx -2.537 \] Rounding to the tenths place, we get: \[ z^{*} = -2.5 \] So the final output would be: \( \square \) \(-2.5\)