## Question 429:

1## Answer:

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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.

- The Null Hypothesis is that coin is fair Ho: π = .5
- The Alternative Hypothesis is that the coin is not fair π ≠ .5
- The alpha value for the 2-tailed test is .05 or .10 for the 1-tailed test (meaning in favor of heads).
- 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).
- 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
- This makes the test statistic z* = .133/.06455 = 2.0656
- 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).
- 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.
- We would reject the NULL hypothesis and conclude this coin is not a fair coin and biased in favor of heads.