Question

A company has developed a hand-held sensor that it claims can detect explosives inside containers. A government agency is interested in buying the devices if they're effective and arranges for an independent test. For the test, experts place explosive material in one of four containers at random. A technician then uses the device to determine which of the four containers holds the explosives. In 60 trials, the technician using the device correctly identifies the explosive container 18 times. Is this convincing evidence that the device is effective in detecting explosives at a rate higher than chance alone?

1. A

   The p-value of the test is 0.25, so the evidence is convincing that that the device works better than guessing and chance alone.

2. B

   The evidence isn't convincing. The p-value of the test is about 0.186. So nearly 19% of the time, you'd expect to find the explosives 18 or more times in 60 trials just by guessing.

   Correct

3. C

   The evidence is inconclusive. There aren't enough successes and failures to do the test.

4. D

   The evidence is convincing. With a p-value of 0.018, you'd reject the null and conclude the device detects explosives at a rate better than chance guessing alone.

   Check your p-value calculation.

I know the answer is B but I can't figure out how to formulate the null and alternate hypothesis with this word problem. The teacher said the P value was 0.186 and I can't figure out how he got that. I kept getting a different value of P. I think it's because I set up the problem wrong in the first place and maybe didn't have my sample size or p hat or something not right. Can you help me? The teacher gave me this correct answer, but I've been at it for hours trying to figure out what I did wrong!

Thanks!

Eileen

Answer 1

here since if it is effective it should be able to tell correctly more than 25% of the times (since if a person guess then he/she will be able to tell 1/4=0.25 times corectly)

null Hypothesis:               Ho:   p	=	0.250
alternate Hypothesis:    Ha: p	>	0.250

sample success x   =	18
sample size          n    =	60
std error   se =√(p*(1-p)/n) =	0.0559
sample proportion p̂ = x/n=	0.3000
test stat z =(p̂-p)/√(p(1-p)/n)=(0.3-0.25)/0.0559 =	0.8944
p value                          =	0.186 (from excel function 1-normsinv(0.8944)