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Three samples given are showing the number of years the US presidents, popes, and British monarchs lived after their inauguration, election or coronation respectively.

As the number of samples is three and the equality of medians of the populations of the samples need to be tested, the Kruskal Wallis test is required to be conducted.

The null hypothesis for testing the equality of medians is as follows:

Thethree samples come from populations with the same median.

The alternative hypothesis is as follows:

Thethree samples do not come from populations with the same median.

Combine the three samples and write the A for presidents, B for pope, and C for monarchs as the sample name.

Denote a rank of 1 to the smallest observation, 2 to the next smallest observation until all the observations are assigned ranks.

If some values are equal, assign the mean value of the ranks to all the similar values.

The following table shows the ranks of all the values:

Let\({n_1}\)denote the sample size corresponding to presidents鈥� ages.

Let\({n_2}\)denote the sample size corresponding to popes鈥� ages.

Let\({n_3}\)denote the sample size corresponding to monarchs鈥� ages.

Thus,

\(\begin{array}{l}{n_1} = 38\\{n_2} = 24\\{n_3} = 14\end{array}\)

The value of N is equal to

\(\begin{array}{c}N = 38 + 24 + 14\\ = 76\end{array}\)

The sum of the ranks corresponding to presidents is computed below:

\(\begin{array}{c}{R_1} = 26.5 + 69 + ... + 70.5\\ = 1485.5\end{array}\)

The sum of the ranks corresponding to popes is computed below:

\(\begin{array}{c}{R_2} = 6 + 22.5 + .... + 65.5\\ = 806.5\end{array}\)

The sum of the ranks corresponding to popes is computed below:

\(\begin{array}{c}{R_3} = 46.5 + 15.5 + .... + 39.5\\ = 634\end{array}\)

Thus, the value of the test statistic is computed as follows:

\(\begin{array}{c}H = \frac{{12}}{{N\left( {N + 1} \right)}}\left( {\frac{{{R_1}^2}}{{{n_1}}} + \frac{{{R_2}^2}}{{{n_2}}} + \frac{{{R_3}^2}}{{{n_3}}}} \right) - 3\left( {N + 1} \right)\\ = \frac{{12}}{{76\left( {76 + 1} \right)}}\left( {\frac{{{{1485.5}^2}}}{{38}} + \frac{{{{806.5}^2}}}{{24}} + \frac{{{{634}^2}}}{{14}}} \right) - 3(76 + 1)\\ = 2.5288\end{array}\)

Thus, the value of H comes out to be equal to 2.5288.

Let k be the number of samples.

Thus, k=3.

The degrees of freedom are computed as follows:

\(\begin{array}{c}df = k - 1\\ = 3 - 1\\ = 2\end{array}\)

The critical value of\({\chi ^2}\)with 2 degrees of freedom at\(\alpha = 0.05\)is equal to 5.9915.

Since the value of H is less than the critical value, the decision is fail to reject the null hypothesis.

There is not enough evidence to warrant rejection of the claim that the three samples come from the populations with the same median.