Help with statistics and probability? Adverse Effect of Chemotherapy 3. Scenario

Question

Help with statistics and probability?

Adverse Effect of Chemotherapy 3. Scenario: When a new chemotherapy is developed, we want to study its toxicity before studying the efficacy. Let T denote the probability of an adverse event. Increasing the strength of chemotherapy is beneficial for killing cancer cells, but it is dangerous because it increases T. To this end, researchers aim to treat cancer patients at a target dose corresponding to 3 (i.e. allowing the probability of an adverse event at .30. Let X 1 for an adverse event and X 0 for no adverse event when a patient is treated at the target dose. Assume n 25 patients are treated at dose 700 mg/kg which is assumed to be the target dose corresponding to 3. Since Bernoulli trials are repeated n 25 times, it is a binomial experiment of size n 25. The research question of interest whether the dose 700 mg corresponds to T 3 or T .3 Data: A researcher recorded a sample (1,0, 0) of size n 25. where "1" indicates an adverse event and "0" indicates no adverse event. The researcher reported the sum Y 13 (and the sample size n 25 The statistic Y E 13 is sufficient to make inference for the parameter T. (We do not need to know the sequence of zeros and ones in the sample. Simulation: In practice, researchers can conduct a binomial experiment of size n 25 once. Using computer simulation, we can observe the randomness in the binomial experiment. Let Y denote the number of patients with an adverse event when n 25 patients are treated at the dose corresponding to 3. Under the assumption, we can simulate Y using the R code below. The code repeats the binomial experiment 10,000 times and displays the probability distribution of Y n-sim 10000 set.seed(123) Y rbinom n.sim, n, p table(Y) n sim barplot( table(Y) n-sim, col "gold", ylab "Probability" xlab Number of Adverse Events", main Simat ation Results sum( Y 13) sum( Y 13)

Explanation / Answer

solution:

a) The simulated probability of Y = 13 is 14

c) dbinom(13,n,p)
[1] 0.01148

d) 18

f)0.0174

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