# Solved: The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times.

#### ByDr. Raju Chaudhari

Oct 12, 2020

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 27, 31, 42, 40, 28, 32. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die?

#### Solution

The observed data is

position Obs. Freq.$(O)$ Prop.
1 27 0.1667
2 31 0.1667
3 42 0.1667
4 40 0.1667
5 28 0.1667
6 32 0.1667

#### Step 1 The null and alternative hypothesis are as follows:

$H_0:$ The loaded die does not behave differently than a fair die.

i.e., $H_0: p_1=p_2=p_3=p_4 = p_5=p_6=\frac{1}{6}=0.1666667$

$H_1:$ The loaded die behave differently than a fair die.

#### Step 2 Test statistic

The test statistic for testing above hypothesis is

 $$\begin{equation*} \chi^2= \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)} \end{equation*}$$

#### Step 3 Level of Significance

The level of significance is $\alpha =0.01$.

#### Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.01$. Degrees of freedom $df=k-1=6-1 =5$.

The critical value of $\chi^2$ for $df=5$ and $\alpha=0.01$ level of significance is $\chi^2 =15.0863$.

#### Step 5 Test Statistic

The expected frequencies can be calculated as

 $$\begin{equation*} E_{i} =N*p_i \end{equation*}$$

For example, $E_{1}$ is given by

 $$\begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 200*0.1667\\ &=&33.34. \end{eqnarray*}$$

position Obs. Freq.$(O)$ Prop. $p_i$ Expe.Freq.$(E)$ $(O-E)^2/E$
1 27 0.1667 33.34 1.206
2 31 0.1667 33.34 0.164
3 42 0.1667 33.34 2.249
4 40 0.1667 33.34 1.33
5 28 0.1667 33.34 0.855
6 32 0.1667 33.34 0.054

The test statistic is

 $$\begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(27-33.34)^2}{33.34}+\cdots + \frac{(32-33.34)^2}{33.34}\\ &=& 1.206 +\cdots + 0.054\\ &=& 5.858. \end{eqnarray*}$$

#### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =5.858$ which falls $outside$ the critical region bounded by the critical value $15.0863$, we $\textit{fail to reject}$ the null hypothesis.

OR

#### Step 6 Decision ($p$-value approach)

The p-value is $P(\chi^2_{5}>5.858) =0.32028$.

As the p-value $0.3203$ is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

Thus we conclude that the loaded die does not behaves differently than a fair die.