Question

a die is tested for fairness using Ha : p does not equal 1/6 where p is the proportions of the sixes that occur . Would ubserving 12 sixes in 114 rolls of the die be statistically significant evidence that the true probability of a six is different from 1/6? use 10% significance level.

Answer 1

Solution:

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: A die is fair.

Alternative hypothesis: Ha: A die is not fair.

H0: p = 1/6 = 0.1667 versus Ha: p ≠ 0.1667

This is a two tailed test.

We are given

Level of significance = α = 0.10

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 12      

n = sample size = 114

p̂ = x/n = 12/114 = 0.105263158

p = 0.1667

q = 1 - p = 0.8333

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.105263158 - 0.1667)/sqrt(0.1667*0.8333/114)

Z = -1.7600

Test statistic = -1.76

P-value = 0.0784

(by using z-table)

P-value < α = 0.10

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the die is fair.