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The following is the question in particular, I got this question wrong. However nothing I run across explains why it is wrong. I answered C because with a p-value of 0.087 we don't have an "unusual" enough outcome to reject the null though the correct terminology would be "extreme".

Thank you for your time.

Use the following information to answer the question. A janitor at a large office building believes that his supply of light bulbs has too many defective bulbs. The janitor's null hypothesis is that the supply of light bulbs has a defect rate of p = 0.07 (the light bulb manufacturer's stated defect rate). Suppose he does a hypothesis test with a significance level of 0.05. Symbolically, the null and alternative hypothesis are as follows: H0: p = 0.07 and

The janitor calculates a p-value for the hypothesis test of approximately 0.087. Choose the correct interpretation for the p-value.

A) The p-value tells us that the true population rate of defective light bulbs is approximately 0.087.

B) None of these

C) The p-value tells us that if the defect rate is 0.07, then the probability that the janitor will have 27 defective light bulbs out of 300 is approximately 0.087. At a significance level of 0.05, this would not be an unusual outcome.

D) The p-value tells us that the probability of concluding that the defect rate is equal to 0.07, when in fact it is greater than 0.07, is approximately 0.087.

DanBaba
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  • The p-value is the probability of rejecting the null given it is true. In the problem given, the null hypothesis is rejected if the observed number of defects is greater than or equal to some number X. While it's true that you would fail to reject the null because the p-value is greater than the alpha level - that is the observed number of defects is less than the number required for rejecting the null - the p-value tells you nothing about P(X=27). In fact this distribution is likely binomial with n=300 and p=0.07, see that P(X=27) = .03 under the null. – James Bender Apr 22 '15 at 01:15
  • Thank you very much for the reply. Isn't p-value definition you gave actually significance level, or alpha? Or in other words the chance that we reject the null, but in fact it is true?

    Maybe I am having more fundamental problems than this, I think of the p-value as being the location on the normal curve where my test lies so in this case it's z-value is 1.48 which states that 97% of the data is on the left side of the curve?

    Again thank you SO MUCH it helps to talk about it!

     – DanBaba Apr 22 '15 at 01:56
  • The p-value can only be calculated when you actually have observed some data. In this case you reject the null hypothesis given the number of observed defective items (call it X) is greater than or equal to some number c. The number c is the critical value and the alpha level is P(X>=c) given that Ho is true - that is X~binomial(300,0.07). Now suppose we have observed the number of defective items to be 23. Then the p-value is P(X>=23) given the null hypothesis. If this probability is higher than alpha, then 23 must be less than the critical value c. Fail to reject Ho. – James Bender Apr 22 '15 at 02:05
  • Also, there is no normal curve in this situation. The distribution is binomial. You could use a normal approximation to the binomial distribution to get your z-value, but it's the same to just work with the binomial distribution and it's probabilities directly. – James Bender Apr 22 '15 at 02:09

1 Answers1

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You are confusing two different uses of "p" in this problem.

First, p = 0.07 is the proportion of defective light bulbs (the stated defect rate). It simply means that the probability of a randomly selected bulb being defected is 0.07.

Second, the p-value of the hypothesis test is given as 0.087. This means that the janitor has conducted a test. Typically this would be by choosing a certain number of bulbs and seeing how many work. The p-value is as you have described - a measure of how unlikely the observed number of defective bulbs is. In this case you should have the alternative hypothesis $H_1: p>0.07$ so for the p-value is the probability that there are the observed number of faulty light bulbs or more. We aren't told how many he tests or how many faulty bulbs he finds - just the p-value.

So this p-value tells us that the chance of having the observed number of defective bulbs or more is 0.087. Because this is greater than the 5% significance level, we conclude that there is insufficient evidence to support the janitor's claim ($H_1$) and we continue to believe that the defect rate is 0.07.

Answer is "B"

tomi
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    The actual answer on the software we use says it was B, I thought it was C and you made a case for D ugh still confused on this but thank you very much for the input.

    I guess you could argue that D isn't right because it says that you would conclude that the failure rate is a certain percentage when we just failed to reject the null, we don't actually accept anything yet?

     – DanBaba Apr 22 '15 at 20:41
  • I have corrected my answer. – tomi Apr 23 '15 at 12:16