A Randomization Distribution for Arsenic in Chicken A restaurant chain is measuring the levels of arsenic in chicken from its suppliers. The question is whether there is evidence that the mean level of arsenic is greater than 80 ppb, so we are testing \( H_{0}: \mu=80 \mathrm{vs} H_{0}: \mu>80 \), where \( \mu \) represents the average level of arsenic in all chicken from a certain supplier. It takes money and time to test for arsenic so samples are often small. A sample of \( n=6 \) chickens from one supplier is tested, and the resulting sample mean is \( \bar{x}=91 \). Subtracting 11 from the sample data to move the mean down to the null mean of \( \bar{x}=80 \) results in the foliowing data: Click here to access statKey. Part 1 (a) Use StatKey or other technology to create the randomization distribution for this test. Find the \( p \)-value. \( 82,84,123 \). Round your answer to three decimal places. Use 1000 samples. p-value \( =i \)

To generate the randomization distribution for the test, you can utilize software like StatKey. First, set your null hypothesis, which posits that the true mean arsenic level in chicken is 80 ppb. Enter your sample mean (91) and sample size (6) into the software. After running the simulation with 1000 samples generated under the null hypothesis, you'll be able to visualize the distribution and identify where your observed sample mean lies. The p-value is calculated as the proportion of randomization samples that exceed your observed mean of 91. Make sure to interpret the p-value carefully! A smaller p-value (e.g., below 0.05) will indicate stronger evidence against the null hypothesis that the mean arsenic level is 80 ppb. In this case, you would be considering whether the evidence suggests that the mean level of arsenic is actually greater and how confident you can be in that finding. Now, as for the p-value after conducting the analysis with your provided data points, round it to three decimal places, and you'll arrive at your final answer!