Question

American Idol Contestants on the TV show American Idol competed to win a singing contest. At one point, the website WhatNotToSing.com listed the actual numbers of eliminations for different orders of singing, and the expected number of eliminations was also listed. The results are in the table below. Use a 0.05 significance level to test the claim that the actual eliminations agree with the expected numbers. Does there appear to be support for the claim that the leadoff singers appear to be at a disadvantage?

Singing Order	1	2	3	4	5	6	7–12
Actual Eliminations	20	12	9	8	6	5	9
Expected Eliminations	12.9	12.9	9.9	7.9	6.4	5.5	13.5

Short answer

There is not enough evidence to conclude thatthe actual number of eliminations is not the same as the expected number of eliminations.

Since the actual number of eliminations for the lead-off singers is 20 and the expected eliminations should have been 12.9, it can be said that the lead-off singers appear to be at a disadvantage.

Step-by-step solution

Given information

The actual number and the expected number of eliminations are tabulated for different orders for singing.

Hypotheses

The null hypothesis is as follows:

The actual number of eliminations is the same as the expected number of eliminations.

The alternative hypothesis is as follows:

The actual number of eliminations is not the same as the expected number of eliminations.

It is considered a right-tailed test.

If the test statistic value is greater than the critical value, then the null hypothesis is rejected, otherwise not.

Observed and expected frequencies

Observed Frequency:

The frequencies provided in the table are the observed frequencies. They are denoted by\(O\).

The following table shows the observed frequency for each order:

\({O_1}\)=20

\({O_2}\)=12

\({O_3}\)=9

\({O_4}\)=8

\({O_5}\)=6

\({O_6}\)=5

\({O_{7 - 12}}\)=9

Expected Frequency:

The expected frequencies are also tabulated in the problem. The values are written below:

\({E_1}\)=12.9

\({E_2}\)=12.9

\({E_3}\)=9.9

\({E_4}\)=7.9

\({E_5}\)=6.4

\({E_6}\)=5.5

\({E_{7 - 12}}\)=13.5

Test statistic

The test statistic is computed as shown below:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}\;\;\; \sim } {\chi ^2}_{\left( {k - 1} \right)}\\ = \frac{{{{\left( {20 - 12.9} \right)}^2}}}{{12.9}} + \frac{{{{\left( {12 - 12.9} \right)}^2}}}{{12.9}} + ....... + \frac{{{{\left( {9 - 13.5} \right)}^2}}}{{13.5}}\\ = 5.624\end{aligned}\]

Thus, \({\chi ^2} = 5.624\).

Obtain critical value, p-value and determine the conclusion of the test

The degrees of freedom are computed below:

\(\begin{aligned}{c}df = \left( {k - 1} \right)\\ = \left( {7 - 1} \right)\\ = 6\end{aligned}\)

The critical value of\({\chi ^2}\)for 6 degrees of freedom at 0.05 level of significance for a right-tailed test is equal to 12.5916.

The corresponding p-value is approximately equal to 0.466598.

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

There is not enough evidence to conclude thatthe actual number of eliminations is not the same as the expected number of eliminations.

Since the actual number of eliminations for the lead-off singers is 20 and the expected eliminations should have been 12.9, it can be said that the lead off singers appear to be at a disadvantage.