In trials of a medical screening test for a particular illness, 23 cases out of

Question

In trials of a medical screening test for a particular illness, 23 cases out of 324 positive results turned out to be false-positives. The screening test is deemed acceptable as long as r, the probability of a positive result being incorrect, is no larger than 10%. (a) Calculate a p-value for the null hypothesis that r greaterthanorequalto 0.1. (b) Use the formula below to construct a 99% upper confidence bound on r: r elementof (0, p + z alpha/n Squareroot x (n - x)/n). (c) Do you think that the screening test is acceptable?

Explanation / Answer

Solution:

We are given

x = 23, n = 324

p-hat = x/n = 23/324 = 0.071

Part a

Here, we have to use z test for population proportion.

H0: r 0.1 versus Ha: r < 0.1

Z = (p-hat – p)/sqrt(pq/n) where q = 1 – p = 1 – 0.1 = 0.9

Z = (0.071 – 0.1)/sqrt(0.1*0.9/324) = -1.74

Test statistic = -1.74

P-value = 0.0409

Part b

Upper confidence bound = p-hat + (Z/n)*sqrt[x(n – x)/n]

Upper confidence bound = 0.071 + (1.6449/324)*sqrt(23*(324 – 23)/324)

Upper confidence bound = 0.071 + 0.023468

Upper confidence bound = 0.094468

r (0, 0.094)

Part c

We do not think that the screening test is acceptable because the hypothesized proportion 0.1 not lies between the confidence interval (0, 0.094).

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