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Question 350:



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We'd conduct a 1-sample proportion test and since the sample is reasonably large we can use the normal approximation to the binomial. We are testing whether the observed proportion 27/300 = .09 is greater than .05 (or .05*300=15) while taking into account chance fluctuations.

  1. The standard rule of thumb is that you're usually ok using the normal approximation if n*p > 5 and n*q > 5, where q is just 1-p. For this example that would be n*p = 300*.05 = 15 and for q  300*.95 = 285.
  2. We need to get a z-score out of the observed proportion. For continuous data we usually have the mean and standard deviation. We only have the mean (27) and no standard deviation. Fortunately, for discrete distributions, the standard deviation is equal to the square root of n*p*q.  For this data that would be SQRT ( 300*.05*.95 ) = a standard deviation of 3.77.
  3. We now need to make a continuity correction since we have a discrete distribution and want to get as close as we can to a continuous one. We add .5 to our value of 15 that we are testing for to get 15.5.
  4. Now we compute the z-score with the new data (27 - 15.5)/3.77 = a z of 3.05. Using the z-score to percentile calculator gives us a one-sided probability of  .00147 or .147%, less than the 1% alpha level set.
  5. In other words, the probability 27 defects out of 300 actually comes from a population less than or equal to .05 is about .147%--very remote.
  6. So we'd conclude the robot needs repair.

If we were to use the binomial probabilities instead of approximating, we'd have a probability of .00127 versus the normal-approximation of .00147, which is pretty close.

See also question 319

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