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Question 429:



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  1. Q429_1_MegaStatExcelInstructionsQ429.xls

First we need to see if we can use the normal approximation to the binomial. If np >5 and nq >5 we can, n is the sample size and p is the proportion and q is 1-p. We get p = 38/60 = .6333 and np = 38 and nq = 22 so we're fine.

  1. The Null Hypothesis is that coin is fair Ho: π = .5
  2. The Alternative Hypothesis is that the coin is not fair π  .5
  3. The alpha value for the 2-tailed test is .05 or .10 for the 1-tailed test (meaning in favor of heads).
  4. The test statistic z* is found as the difference between the observed proportion and test proportion = .633-.50 = .1333, divided by the standard error of the proportion (SE).
  5. The SE for a proportion is found as the SQRT(pq/n), where p and q were the proportion and 1-proportion defined above (.633 and .3667). We get an SE value of .06455
  6. This makes the test statistic z* = .133/.06455 = 2.0656
  7. We need to lookup the 1-sided area for this z-score (1-sided since we're only interested in whether the coin is biased in favor of heads).
  8. Using the z-score to percentile calculator gives us a one-sided probability of  .0194 or 1.94%, which is less than the .10 alpha level.  We can also use the excel function =1-NORMSDIST(2.0656). The 2-tailed probability is twice this amount or .0389, which is also less than α =.10.
  9. We would reject the NULL hypothesis and conclude this coin is not a fair coin and biased in favor of heads.
See the attached file for Instructions and results using MegaStat

See also question 319

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