Posted by Gayla on Sunday, June 8, 2008 at 7:35pm.
Here might be one way to do this problem:
Null hypothesis is that the coin is fair. Ho: p = .5
Alternate hypothesis is that the coin is unfair. Ha: p not equal to .5
Using the binomial formula: P(x) = (nCx)(p^x)[q^(n - x)]
...where n = number of coin tosses, x = number of times came up heads, p = probability given in the null hypothesis, q = 1 - p.
Using your data:
P(60) = (60C38)(.5^38)(.5^22)
= .0123 (rounded to four decimal places)
If the alternate hypothesis uses "not equal to" then we multiply the results by 2. Therefore, 2 * .0123 = .0246
If the alternate hypothesis would have shown a specific direction, then we could have used .0123 as is.
Reject the null hypothesis if the test statistic above is less than .10 level of significance; otherwise, do not reject null.
The p-value is .0246 (the p-value is the actual level of the test statistic). You should be able to take it from here.
I hope this will help.
Correction (this does not change the outcome):
Using the binomial formula: P(x) = (nCx)(p^x)[q^(n - x)]
...where n = number of coin tosses, x = number of times came up heads, p = probability given in the null hypothesis, q = 1 - p.
Using your data:
P(38) = (60C38)(.5^38)(.5^22)
= .0123 (rounded to four decimal places)
*** Changed P(60) to P(38). Sorry for any confusion!
Thank you for your help.