Loaded Die 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, and 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?

A loaded die is rolled 200 times, and the outcomes on the die are recorded each time.

Assume the sample data is taken from larger population. Also, since the frequency counts are tested and if each expected frequency is at least 5, the requirements of the test are met.

The expected values are computed as,

Let E denote the expected frequencies of the outcomes.

It is expected that the die is fair.

Therefore, the probability of occurrence of each outcome should be equal to:

\(\begin{aligned}{c}p = \frac{1}{6}\\ = 0.16667\end{aligned}\)

Since the die is rolled 200 times,\(n = 200\).

Now, the expected frequency for all the six outcomes is equal to:

\(\begin{aligned}{c}E = np\\ = 200\left( {\frac{1}{6}} \right)\\ = 33.3333\end{aligned}\)

Thus, the chi-square test is applicable.

The null hypothesis for conducting the given test is as follows:

\({H_o}:{p_1} = {p_2} = {p_3} = {p_4} = {p_5} = {p_6}\)

The alternative hypothesis is as follows:

\({H_a}:\)At least one proportion is not equal to the others.

The test is right-tailed.

Let O denote the observed frequencies of the outcomes.

The following values are obtained:

\(\begin{aligned}{l}{O_1} = 27\\{O_2} = 31\\{O_3} = 42\\{O_4} = 40\\{O_5} = 28\\{O_6} = 32\end{aligned}\)

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 1.203333 + 0.163333 + ...... + 0.053333\\ = 5.86\end{aligned}\)

Thus,\({\chi ^2} = 5.86\).

For 6 outcomes, k is equal to 6.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 6 - 1\\ = 5\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 5 degrees of freedom is equal to 11.0705.

The p-value is computed as,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 5.86} \right)\\ = 0.3201\end{aligned}\)

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to be rejected.

There is not enough evidence to conclude that the outcomes are not equally likely.

A die is fair if all the outcomes are equally likely. The result indicates that the loaded die is not different from a fair die.