Question: 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

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 H04: 2 80 vs Ha : ,u > 80, where ,u 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 f = 91. Subtracting 11 from the sample data to move the mean down to the null mean off 2 80 results in the following data: 57, 64, 70, 82, 84, 123. Click here to access StatKey. Part 1 (a) Use StotKey or other technology to create the randomization distribution for this test. Find the p-value. Round your answer to three decimal places. Use 1000 samples. p-value = n

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