Chapter 12.1, Problem 29E

Fair die? A gambler rolls a die 600 times to determine whether or not it is fair. Following are the results.

1. Let $p_1$ be the probability that the die comes up 1, let $p_2$ be the probability that the die comes up 2, and so on. Use the chi-square distribution to test the null hypothesis, at the $\alpha =0.05$ level of significance, that the die is fair.
2. The gambler decides to use the test for proportions (discussed in Section 9.4) to test $H_0:p_1=1/6$ for each $p_i$. Find the P-values for each of these tests.
3. Show that the hypothesis $H_0:p_1=1/6$ is rejected at level 0.05.
4. The gambler now reasons as follows: "I reject $H_0:p_4=1/6$, and I conclude that $p_4\ne 1/6$
5. Therefore, I can reject the null hypothesis that the die is fair." Explain why this reasoning is incorrect. (Hint: See Section 11.5 on the multiple testing problem.)
6. Use the Bonferroni correction (Section 11.5) to adjust the P-value for the test of $H_0:p_4=1/6$

Can you now conclude that the die is not fair? Explain.

To determine

The testing of null hypothesis using chi square distribution at $\alpha =0.05$ level of significance.

Answer to Problem 29E

From chi square test, one can clearly say that die is fair

Explanation of Solution

Let us declare the hypothesis,

  ${\text{H}}_0:p_i=\frac{1}{6}:\forall i\in \{1,2,3,4,5,6\}$ V/s

  $H_1:\text{Some of }{p}_i\text{ are different }$

For calculating expected frequencies,

  $E(X_i)=n{p}_i$

  $E(X_i)=100$

Now,

Outcome	Observed Frequency	Expected Frequency
1	113	100
2	101	100
3	106	100
4	81	100
5	108	100
6	91	100

For chi square test, test statistic is given by

  ${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(Observed-Expected\right)}^2}{Expected}}$

Using calculator,

  ${\chi }^2=7.12$ --------------------------------- (i)

Using chi square table, let’s calculate ${\chi }_{0.05,5}^2$ , where 5 is the degree of freedom

  ${\chi }_{0.05,5}^2=\text{11}\text{.1}$ ------------------------------- (ii)

It can be clearly seen that chi square calculated is less than chi square tabulated. Hence, do not have enough evidences to reject null hypothesis.

This implies that die is fair.

To determine

The p-value for the test ${\text{H}}_0:p_i=\frac{1}{6}$ for each $p_i$

Answer to Problem 29E

Hypothesis	p-value
${\text{H}}_0:p_1=\frac{1}{6}$	$0.1415$
${\text{H}}_0:p_2=\frac{1}{6}$	$0.8779$
${\text{H}}_0:p_3=\frac{1}{6}$	$0.4825$
${\text{H}}_0:p_4=\frac{1}{6}$	$0.0413$
${\text{H}}_0:p_5=\frac{1}{6}$	$0.3567$
${\text{H}}_0:p_6=\frac{1}{6}$	$0.3454$

Explanation of Solution

Let’s declare hypothesis

  ${\text{H}}_0:p_1=\frac{1}{6}$ V/s

  $H_1:p_1\ne \frac{1}{6}$

Now,

  $\widehat{p_1}=\frac{113}{600}$

  $\widehat{p_1}=0.1883$

For the test statistic,

  $\begin{array}{l}Z=\frac{\widehat{p_1}-p_1}{\sqrt{\frac{p_1(1-p_1)}{n}}}\\ Z=\frac{0.1883-\frac{1}{6}}{\sqrt{\frac{\frac{1}{6}\left(\frac{5}{6}\right)}{600}}}\end{array}$

Using calculator,

  $Z=1.47$

For $\alpha =0.05$ , p-value one has to find $P(|Z|\le 1.96)$

Using calculator,

  $P(|Z|\le 1.96)=0.1415$

In a similar manner,

Hypothesis	p-value
${\text{H}}_0:p_1=\frac{1}{6}$	$0.1415$
${\text{H}}_0:p_2=\frac{1}{6}$	$0.8779$
${\text{H}}_0:p_3=\frac{1}{6}$	$0.4825$
${\text{H}}_0:p_4=\frac{1}{6}$	$0.0413$
${\text{H}}_0:p_5=\frac{1}{6}$	$0.3567$
${\text{H}}_0:p_6=\frac{1}{6}$	$0.3454$

To determine

The proof that ${\text{H}}_0:p_4=\frac{1}{6}$ is rejected at $\alpha =0.05$ level of significance.

Explanation of Solution

Let’s declare hypothesis,

  ${\text{H}}_0:p_4=\frac{1}{6}$ V/s

  $H_1:p_4\ne \frac{1}{6}$

In the above part, p- value was obtained as $0.0413$ . For $\alpha =0.05$ level of significance, p-value is less than level of significance. Hence null hypothesis is rejected.

To determine

Why the reasoning of gambler is incorrect.

Explanation of Solution

First of all when multiple tests are performed then ‘multiple testing problem’ occurs, which states that as more hypothesis test are performed, small p-values become less meaningful.

Hence, it is preferred to adjust p-value using bonferroni adjustment.

Conclusions has been made without taking bonferroni adjustment into account. Therefore, gambler’s reasoning is incorrect.

To determine

The adjusted p-value using Bonferroni adjustment for the test ${\text{H}}_0:p_4=\frac{1}{6}$

Answer to Problem 29E

Modified p-value $=0.2478$ .

It can be concluded that dice is fair.

Explanation of Solution

Let’s declare hypothesis,

  ${\text{H}}_0:p_4=\frac{1}{6}$ V/s

  $H_1:p_4\ne \frac{1}{6}$

First of all when multiple tests are performed then ‘multiple testing problem’ occurs, which states that as more hypothesis test are performed, small p-values become less meaningful.

Hence, it is preferred to adjust p-value using bonferroni adjustment.In the above part, p- value was obtained as $0.0413$

Modified p-value $=0.2478$

It is clearly seen that p value is greater than $\alpha =0.05$ . Hence, do not have enough evidences to reject null hypothesis.

Now, it can be concluded that dice is fair.