Question

A coin was flipped 56 times and came up heads 36 times. At the .10 level of significance, is the coin biased toward heads?

(a-1) H0: formula130.mml ? .50 versus H1: formula130.mml > .50. Choose the appropriate decision rule at the .10 level of significance.

a. Reject H0 if z >1.282 b. Reject H0 if z < 1.282......... a or b

(a-2) Calculate the test statistic. (Carry out all intermediate calculations to at least 4 decimal places. Round your answer to 3 decimal places.) Test statistic

(a-3) The null hypothesis should be rejected. True False

(a-4) The true proportion is greater than .50. True False

(b-1) Find the p-value. (Round your answer to 4 decimal places.) p-value

(b-2) Is the coin biased toward heads? Yes No

Answer 1

The sample proportion here is computed as:

p = 36 / 56 = 0.6429

a-1) As we are testing whether the coin is biased, the null and the alternate hypothesis here are given as:

$\LARGE H_0: p \leq 0.5$

$\LARGE H_1: p >0.5$

a-2) The standard error here is first computed as:

$\LARGE SE = \sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.5(1-0.5)}{56}}$

$\LARGE SE = 0.0668$

Now the test statistic here is computed as:

$\LARGE z^* = \frac{p-P}{SE} = \frac{0.6429 -0.5}{0.0668} = 2.139$

As this is a one tailed test, the p-value here is computed as:

p = P(Z > 2.139 ) = 0.0162

(a-3) As the p-value here is 0.0162 > 0.01 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here.

(a-4) The true proportion is not greater than 0.5 as we cannot reject the null hypothesis. Therefore False

(b-1) The p-value is computed above to be 0.0162

(b-2) The coin is not biased as we were not able to reject the null hypothesis