Question 467:

Asked on October 7, 2008

Tags: Normal Approximation , 1-Proportion Test

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Answer:

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1. Q467_1_MegaStatInstructionsQ467.doc

To answer this question we can conduct a 1-sample proportion test which would use the normal approximation to the Binomial. We are testing whether the observed proportion 485/500= .97 is greater than .95 while taking into account chance fluctuations.

We can use the normal approximation to the binomial if np>5 and nq >5 where n is the sample size, p is the proportion and q is 1-p. We have np = 500*.97 = 485 and  nq =500*.03=15, so we're fine using this approach.

1. The test statistic z* is found as the difference between the observed proportion and test proportion = .97-.95 = .02, divided by the standard error of the proportion (SE).
2. The SE for a proportion is found as the SQRT(pq/n), where p and q were the proportion and 1-proportion defined above (.97 and .03). We get an SE value of .00975.
3. This makes the test statistic z* = .02/.00975 = 2.0519567. 
4. We need to lookup the 1-sided area for this z-score (1-sided since we're only interested in whether they exceeded the goal.
   
5. Using the z-score to percentile calculator gives us a one-sided probability of  .0201 or 2.01%, which is just less than the 2.5% alpha level.
6. In other words, the probability 485 out of 500 orders actually comes from a population less than or equal to .95 is about 2.01%--which is rather unlikely.
7. So we'd conclude this exceeds their goal.

See the attached file for Instructions and results using MegaStat

Seee also question 319