Question 1040525: This is a 2 parter using the same table. I do not understand the concept, can you explain this process and the solution. Thanks
Suppose that for a certain illness, the probability is 55% that a given patient will improve without treatment. The probability that at least n out of 30 patients will improve without treatment is given in the following table.
N 19 20 21 22 23 24 24
Probability that at least n improve 0.23 0.14 0.069 0.031 0.012 0.0040
We give an experimental drug to 30 patients who have this illness. Suppose we count the test significant if the p-value is 0.06 or less. How many patients must show improvement in order to make the test statistically significant?
19.
21.
23.
22.
10. Suppose that for a certain illness, the probability is 55% that a given patient will improve without treatment. The probability that at least n out of 30 patients will improve without treatment is given in the following table.
An experimental drug is given to 30 patients who have this illness. If 20 people show improvement, what is the p-value?
0.23.
0.069.
0.012.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! This problem requires the application of the binomial probability distribution
:
recheck the probabilities you have listed, they do not match with the values of n
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there are 7 values of n but only 6 probabilities but the 0.23 probability is suspect
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you need to sum the probabilities and calculate the mean and standard deviation
:
standard error(SE) = sample standard deviation / square root(30)
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test statistic = (probability from list - sample mean) / SE
you will have 6 or 7 test statistics
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consult z-tables for the p-value corresponding to each test statistic
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for the second part, using a binomial calculator
P ( X = 20 ) = 0.0656374
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the probability in the choices is 0.069
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