# A Method for Ascertaining Statistical Ties

Before proceeding you might wish to read, in less technical terms, what constitutes a statistical tie.

A sample must be is randomly drawn from a population. To use this (approximate) method we must assume that the sample size is less than 10% of the size of the population. In addition, at least 10 polled voters must indicate a preference for each of the candidates being assessed.

## Example

For an example, suppose we randomly select 1359 likely registered voters about the 2000 election, asking "Which of the following candidates do you prefer for President?" At the time of the polling (it is currently February of 2000) there are still a number of candidates. Here are the poll results:

Candidate    # Favoring    % Favoring
Al Gore 315 23.18
George W. Bush   287 21.12
John McCain 185 13.61
Steve Forbes 113 8.31
Alan Keyes 65 4.78
Gary Bauer 27 1.99
Other 38 2.80
Undecided 126 9.27

Let PA and PB be the true (population) results for candidates A and B. Determining a statistical tie is equivalent to testing

H0: PA - PB = 0         HA: PA - PB ¹ 0 .

The poll result is designated a statistical tie if the decision is to not reject H0 at the given significance level a. Almost always, for media polls, a = 5%.

The appropriate test statistic is given by

 or

The first expression -- based on the estimated proportions (the "hatted values") is preferred conceptually because it puts the quantity in the standard form of

 Estimate of difference: divided by the Standard Error of the estimate:

where the standard error shrinks as the sample size grows. The other expression, using the observed counts (rather than proportions) in the two categories -- the O's -- is easier to use and avoids possible round-off errors due to obtaining proportions. The two will give exactly the same value for Z excepting rounding errors.

Obtain this quantity, find the normal tail-area corresponding to it, and double this tail area (both directions of difference are important) to get the P-value.

• If the P-value is less than a--this election is not a statistical tie.
• If the P-value is greater than a--it's a statistical tie.

## Example continued

The question is: Are Al Gore and George Bush in a statistical tie? Use 5% as your error rate in deciding.

Clearly the two are not in an exact tie, either for the poll (Gore has a bit of a lead) or for the population as a whole (it would be extraordinary if two candidates were exactly tied over the entire nation of some 100,000,000 voters).

Let PG be the true proportion of people for Gore; PB the proportion of people for Bush. We test:

H0: PG - PB = 0         HA: PG - PB ¹ 0 .

Note that more than 10 fall into each of the Gore and Bush categories, so this procedure may be used. Using the first expression for Z

 Estimated difference: .2318 - .2112 = .0206. Standard error of estimated difference: .0181.

So, the test statistic is = .0206/.0181 = 1.1412. Tail area: .1269. P-value = 2*.1269 = .2538 or 25.38%. Decision: Do not reject H0 at the 5% significance level. This is equivalent to a statistical tie. At this point Gore and Bush are in a statistical tie.

Care to try some exercises on determining statistical ties?