Testing a Drug A drug manufacturer claims that a drug cures a ...

Question

Testing a Drug A drug manufacturer claims that a drug cures a rare skin disease 75% of the time. The claim is checked by testing the drug on 100 patients. If at least 70 patients are cured, then this claim will be accepted. Use this information in Exercises 31 and 32.

Find the probability that the claim will be accepted, assuming that the actual probability that the drug cures the skin disease is 65%.

Find the probability that the claim will be accepted, assuming that the actual probability that the drug cures the skin disease is 65%.

Accepted Answer

All right, hello, everyone. So this question says, a new allergy medication is claimed to be effective for 68% of patients. In a trial with 90 patients, the claim is accepted if at least 60 patients show improvement. If the true effectiveness is 60%, what is the probability that the claim will be accepted? And here we've got 4 different answer choices labeled A through D. So first and foremost, what do we actually know? Well, first, we know the claimed effectiveness. To be 68%. And we also know the true effectiveness is 60%. That is the actual probability of success for this context for this experiment is 60%. Now the sample size n is equal to 90. And the claim will be accepted if at least 60 patients show improvement. So we want to know the probability that at least 60 improve. So we're looking for the probability. Of P equal to 0.60. Now, this is a binomial probability problem, but because the sample size is relatively large, we can use the normal approximation. So for the normal approximation. X follows the binomial distribution. And in addition, N is equal to 90 and P is equal to 0.60. So rewriting the question, therefore, Means that we are looking for PX being greater than or equal to 60. So let's begin here by finding the mean. Now the mean or mew. Is equal to and multiplied by P. So that's 90, multiplied by 0.60, which equals 54. So now you can find the standard deviation. So sigma is equal to the square root of NP multiplied by one subtracted by P. This equals the square root of 90, multiplied by 0.60. Multiplied by 1 subtracted by 0.60, which equals 0.40. After rounding, sigma is equal to approximately 4.647. So now to apply the continuity correction. What we're solving for is P of X greater than or equal to 60. And so after applying the continuity correction, our new expression is P of Y greater than or equal to 59.5. So now, we have to find the Z score that corresponds to 59.5. So Z Is equal to 59.5. Subtracted by me, which is 54. Divided by sigma, which is 4.647. So Z is approximately equal to 1.183. And now on to finding the probability. Now, what you'll notice is that our Z score of 1.183 is between two values in the standard normal tape. Specifically, it's between 1.18 and 1.19. So from the sin normal table, you would now find phi. For both 1.18. And 1.19. So phi of 1.18 is equal to 0.8810. Whereas for 1.19, it's 0.8830. And so to find The probability that we're solving for, we're going to have to interpolate between 1.18 and 1.19. So 5 of. 1.183. Is going to be equal to approximately. By 1.18. Added to 1.183. Subtracted by 1.18. And multiplied by By 1.19. Subtracted by 5 of 1.18. Divided by 1.19 subtracted by 1.18. And so after plugging in the information that we have, 5 of 1.183, Is ultimately equal to. 0.8816 But this doesn't mean that that's your final answer. Because recall that what we're solving for. Is in actuality P of X greater than or equal to 60? Which can be rewritten as P of Z greater than or equal to 1.183. And to solve for this You're going to have to subtract 5 of 1.183 from 1. So that's one subtracted by. By a 1.183. Which equals 1 subtracted by 0.8816. This in actuality leads you now to your final answer. Which is 0.1184. And now Scrolling up to see the answer choices. Our final answer of 0.1184 matches with option C in the multiple choice. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Binomial Distribution

Hypothesis Testing

Cumulative Probability

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