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.
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.