Question

A hockey team is convinced that the coin used to determine the order of play is weighted. The team captain steals this special coin and flips it 14 times to evaluate the hypothesis that the coin is weighted, and it shows up heads 12 times. Test this hypothesis (use alpha=.05).
1. What is the appropriate test?

2. State the null hypothesis:

3. State the alternative hypothesis:

4. Find the critical value:

5. Calculate the obtained statistic:

6. Make a decision:

7. What does your decision mean

Answer 1

Answer:

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

Step-by-step explanation:

Let p be the probability of heads in a single toss of the coin. Then our null hypothesis that the coin is fair will be formulated as

H0 :p 0.5   against   Ha: p ≠ 0.5

The significance level is approximately 0.05

The test statistic to be used is number of heads x.

Critical Region: First we compute the probabilities associated with X the number of heads using the binomial distribution

Heads (x)        Probability (X=x)                        Cumulative     Decumulative

0                        1/16384 (1)             0.000061     0.000061

1                         1/16384  (14)         0.00085             0.000911

2                       1/16384 (91)           0.00555             0.006461

3                       1/16384(364)         0.02222

4                       1/16384(1001)         0.0611

5                       1/16384(2002)       0.122188

6                        1/16384(3003)      0.1833

7                         1/16384(3432)      0.2095

8                        1/16384(3003)       0.1833

9                        1/16384(2002)       0.122188

10                       1/16384(1001)        0.0611

11                       1/16384(364)        0.02222

12                      1/16384(91)            0.00555                             0.006461

13                     1/16384(14)              0.00085                           0.000911

14                       1/16384(1)            0.000061                            0.000061

We use the cumulative and decumulative column as the critical region is composed of two portions of area ( probability) one in each tail of the distribution. If  alpha = 0.05 then alpha by 2 - 0.025 ( area in each tail).

We observe that P (X≤2) =   0.006461 > 0.025

and

P ( X≥12 ) = 0.006461 > 0.025

Therefore true significance level is

∝=  P (X≤0)+P ( X≥14 ) = 0.000061+0.000061= 0.000122

Hence critical region is (X≤0) and ( X≥14)

Computation x= 12

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.