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Question 97:



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We can conduct a 1-sample proportion test. The sample size is large enough that we can use the Normal Approximation to the Binomial. It's usually Ok to use the normal approximation to the binomial when n*p and n*q > 5. We have 130*.32 = 41.6 and 130*.68 = 88.4 so we're fine.

The sample mean is n*p  = 41.6 and the standard deviation is the square root of n*p*q = SQRT (130*.32*.68) = 5.31

We'd find the z-score just like we do if the data were continuous (this data is discrete binary since results were plan or not plan to go into general practice ). To account for the discreteness of the data we have to apply what's called a continuity correction (like making discrete data continuous) by adding .5 to the test mean. The school's claim is that MORE than 28% will go on to be general practitioners. In our sample, 28% is .28*130 = 36.4, and we add .5 to this making it 36.9.

The z-score is then calculated as the sample mean minus the test mean all divided by the standard deviation (41.6-36.9)/5.31 = .885.  To find this area, which will give us the p-value, we can use the z-score to percentile calculator and select 1-sided area, this gets an area of 81.19. In other words, the probability more than 28% of the students would go on to general practice is 81.19% given our sample data.

To check our assumptions, we can use the Binomial probabilities directly. Using the Excel function =BINOMDIST(41.9,130,0.28,TRUE) gets us 84.1%, which is pretty close to the 81.19% approximation.  So there is likely between an 81.19 and 84% probability that more than 28% of students will go on to general practice.

If we set the null hypothesis as less than or equal to 28% of students will go on to general practice, our p-value is .188 (which is just 1-.8119), with the typical alpha of .05 we could not reject the null-hypothesis.

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