1. Many supermarket chains are working to reduce the amount of arsenic in the chicken theysell. To accomplish this, one chain plans to measure, for each supplier, the amount of arsenicin a random sample of chickens. The chain will cancel its relationship with a supplier if thesample provides sufficient evidence that the average amount of arsenic in chicken providedby that supplier is greater than 80 ppb (parts per billion). A randomization distribution forVSis shown below.testing403020100Left Tail60Two-Tail70Right Tail80null = 8090100samples = 500mean = 80.026std. error = 8.327opriate parameter(s) and state the null and alternative hypotheses forocolate M&M's that are green,ion of milk chocolate M&M's that are green is greater110(a) Suppose the sample of chickens from Supplier A contained 100 ppb of arsenic. Use therandomization distribution to find the p-value. (Hint: There are 500 dots on the graph.What proportion of the dots are at 100 or greater?)(b) Suppose the sample of chickens from Supplier B contained 90 ppb of arsenic. Use therandomization distribution to find the p-value. (Do it in a way similar to Part (a))(c) If the supermarket chain uses a 5% significance level, would it cancel its relationship withSupplier A? Would it cancel its relationship with Supplier B?

The dot plot consists of 500 smaller samples of the same size that are obtained repetitively with replacement. In the dot plot, each dot essentially represents the obtained statistic for the average arsenic amount through a random sample, i.e., a smaller sample obtained from the originally sampled chickens through the resampling procedure with the assumption that the null hypothesis is true. Considering μ as the average amount of arsenic (measured in terms of ppb) at the population level, it is hypothesized that: H0:μ≤80 against H1:μ>80(right-tailed). Part (a) Considering it is a right-tailed test, for the observed test statistic value of 100 ppb, the p-value is inferred as the likelihood of obtaining test results that is at least as extreme as 100 towards the right portion of the randomization distribution. As the randomization distribution consists of 500 samples (or dots), and only 10 dots are at least 100. So the p-value is: p-value=10500=0.02. Part (b) Considering it is a right-tailed test, for the observed test statistic value of 90 ppb, the p-value is inferred as the likelihood of obtaining test results that is at least as extreme as 90 towards the right portion of the randomization distribution. As the randomization distribution consists of 500 samples (or dots), and only 72 dots are at least 90. So the p-value is: p-value=72500=0.144. Part (c) For the given level of significance, α=5%=0.05, the decision rule is: reject H0, if p-value≤α. For Supplier-A: Since p-value =0.02<α =0.05, reject H0. The average amount of arsenic is concluded to be higher than 80 ppb. So the chain may end its relationship with Supplier-A. For Supplier-B: Since p-value =0.144>α =0.05, H0 is retained. The average amount of arsenic is concluded to be not higher than 80 ppb. So the chain would not end its relationship with Supplier B.