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July 31, 2014

July 31, 2014

Posted by **Gayla** on Sunday, June 8, 2008 at 7:35pm.

- Statistics -
**MathGuru**, Tuesday, June 10, 2008 at 10:05amHere 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.

- Statistics -
**MathGuru**, Tuesday, June 10, 2008 at 10:31amCorrection (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!

- Statistics -
**Gayla**, Wednesday, June 11, 2008 at 5:12amThank you for your help.

- Statistics -

- Statistics -

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