Solved (Free): 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

ByDr. Raju Chaudhari

Apr 3, 2021

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

Outcomes 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 Setup the hypothesis

The null and alternative hypothesis are as follows:

$H_0:p_1 =\cdots = p_6 =\frac{1}{6}$

$H_1:$ At least one of the proportion is different from $\frac{1}{6}$.

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 $\alpha =0.05$.

Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.05$. Degrees of freedom $df=k-1=6-1 =5$.

The critical value of $\chi^2$ for $df=5$ and $\alpha=0.05$ level of significance is $\chi^2 =11.0705$.

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

 $$\begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 200*0.1667\\ &=&33.3333. \end{eqnarray*}$$

Outcomes Obs. Freq.$(O)$ Prop. $p_i$ Expe.Freq.$(E)$ $(O-E)^2/E$
1 27 0.1667 33.3333 1.203
2 31 0.1667 33.3333 0.163
3 42 0.1667 33.3333 2.253
4 40 0.1667 33.3333 1.333
5 28 0.1667 33.3333 0.853
6 32 0.1667 33.3333 0.053

The test statistic is

 $$\begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(27-33.3333)^2}{33.3333}+\cdots + \frac{(32-33.33)^2}{33.33}\\ &=& 1.203 +\cdots + 0.053\\ &=& 5.858. \end{eqnarray*}$$

The test statistic is $\chi^2 =5.858$ which falls $outside$ the critical region bounded by the critical value $11.0705$, 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 $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

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