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I struggle to understand the following example...

A new breast cancer screening method is tested to see if it performs better than existing methods in detecting breast cancer. To measure the sensitivity of the test, a total of 10,000 patients known to have various stages of breast cancer are testing using the new method. Of those 10,000 patients, 9,942 are identified by the new method to have breast cancer. Given that the sensitivity of the best current test is 99.3%, is there significant evidence at the α=0.05 level to conclude that the new method has higher sensitivity than existing methods? Hint - H0:p=0.993 and H1:p>0.993

The possible answers below:

No, since the p-value under H0 of no difference is approximately equal to 0.081, which is greater than α=0.05

No, since the p-value under H0 of no difference is approximately equal to 0.063, which is greater than α=0.05

Yes, since the p-value under H0 of no difference is approximately equal to 0.048, which is less than α=0.05

Yes, since the p-value under H0 of no difference is approximately equal to 0.033, which is less than α=0.05


I would be very grateful for an explaination as I stuggle to grasp the concept.

Many thanks! Markus

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  • $\begingroup$ Check stats.stackexchange.com/questions/185817/… $\endgroup$ – Tim Jan 18 '17 at 15:53
  • $\begingroup$ Thanks Tim. I should have explained better. I am specifically struggling with the calculation of the p-value. With the Bayes concept as such I am OK. $\endgroup$ – Markus Knopfler Jan 18 '17 at 15:59
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Don't think Bayesian - asking for p-values is frewquentist statistics. In your case, it is as simple as a binomial test, wether prob=0.993 fits to your data or not. A one sides test ist asked for in the hint:

> binom.test(x=9942, n=10000, p=.993, alternative="greater")

    Exact binomial test

data:  9942 and 10000
number of successes = 9942, number of trials = 10000, p-value = 0.08086
alternative hypothesis: true probability of success is greater than 0.993
95 percent confidence interval:
 0.9927874 1.0000000
sample estimates:
probability of success 
            0.9942 

So with a p-value of 0.081 the first answer is the only one that fits.

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The problem is asking you to calculate the p-value, the probability of observing a result as anomalous or more anomalous than what was seen, under the null hypothesis. Given the alternative hypothesis, anomalous behavior equates to "more patients identified". So, to calculate the p-value, calculate the binomial probability of observing 9942 or more successes in 10,000 trials, when the probability of success is 0.993. This is a bit of a tedious calculation... using an online calculation tool, I find the p-value is 0.081.

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  • $\begingroup$ Many thanks for that both, as you said Bernhard, Bayes got me totally off track! $\endgroup$ – Markus Knopfler Jan 18 '17 at 17:07

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