## 5.1           Test Hypotheses

The possible outcomes of the pairwise comparison tests and their relevant hypotheses include:

• Preference towards Multi Microphone Array technique;

o   Test participants may prefer the sound of the Multi Microphone Array recording meaning that despite the post-mix benefits and simplicity of setup associated with the Soundfield microphone, it is not the correct choice in terms of sound quality and listener enjoyment of classical music material. This produces the test hypothesis, HM, which states that the probability of participants choosing the Multi Microphone Array over the Soundfield will be greater than half. Equation 5.1 – Hypothesis M

• Preference towards the Soundfield technique;

o   Test participants may prefer the sound of the Soundfield recording meaning that as well as having operational benefits for the engineer, there are also clear sonic advantages to using this microphone for the recording of classical music. This produces the test hypothesis, HS, which states that the probability of participants choosing the Multi Microphone Array over the Soundfield will be less than half. Equation 5.2 – Hypothesis S

• Equal preference;

o   Overall results may show parity between each technique. This could mean that rather than having a clearly better tool to record with, the engineer can make a decision on array choice without worrying about negative sonic implications. This produces the test hypothesis, HE, which states that the probability of participants choosing the Multi Microphone Array over the Soundfield will be half. Equation 5.3 – Hypothesis E

Given the significant operational differences outlined in Section 1.4, it is unlikely that there will be equal preference, or parity, between a Multi Microphone Array and Soundfield array. Therefore HM, HSand HE are replaced with H0 and HA which are the null and alternative hypotheses respectively as seen in Equation 5.4. These hypotheseswill also be applied to the attribute questions which are detailed in Section 2.3. Equation 5.4 – Null and Alternative Hypotheses

## 5.2           Hypothesis Testing

The null hypothesis, H0, will be tested to establish whether the test question achieves parity. For example, in a situation where a pairwise preference comparison is presented to participants and preference for each is equal, H0 will not be rejected meaning that equal preference, or parity has been established. In the situation where one of the test options is never selected by the participants, H0 will be rejected (Sprent, 1989, pp. 7 – 12).

In a case of null hypothesis rejection, the alternative hypothesiswill be established meaning that parity between the pairwise comparisons options has been discounted as a significant difference between the options has been detected to a certain confidence level.To establish where the preference lies between the two options being tested, a two tailed binomial test is used (Bower, 2014; Elder Laboratory, 2014).

If the null hypothesis can not be rejected, the result will have to reside in the blue shaded region which is between -1.96 and 1.96 as shown in Figure 5.1. If the null hypothesis can be rejected, the numerical result will give a figure less than -1.96 or greater than 1.96, depending on the pairwise option which shows significance. Figure 5.1 – 95% Confidence Interval (e-Discovery Team, 2014)

This numerical region is dependent on the confidence level being used. To determine whether a result is statistically significant, results are commonly compared to a 95% and/or 99% confidence level (Sprent, 1989, p. 8).

To calculate the region, the NORM.S.INV function of Microsoft Excel can be used (Microsoft, 2014). To operate at 99% significance, the critical region is from -2.58 to 2.58. This means that there are more possible answers to the binomial test which can be determined as not rejecting H0 while results which do reject it, can be held to a higher statistical significance.

## 5.3           Analysis Formulae

All calculations were completed using two tailed binomial tests, shown in Equation 5.5 where p is the hypothesised proportion, n is the sample size and x is the number of votes of preference for the first option of the test (Bower, 2014; Elder Laboratory, 2014).

Equation 5.5 – One Proportion Binomial Test

Calculations were made with respect to a confidence level of 95% and where appropriate, 99%. As the distribution criteria outlined in Equation 5.6 were satisfied, the normal approximation of the binomial distribution was used (University, West Chester, 2014).

Equation 5.6 – Distribution Criteria

As stated in Section 4.1 and further discussed in Appendix B.1, a Bonferroni correction of three was applied to the statistical analysis as three comparisons were being made from each set of recording extract stimuli (Goldman, 2014). This adjusts the critical regions which become -2.39 to 2.39 for 95% and -2.94 to 2.94 for 99%.

Required formulae were integrated into the data collection spread sheet of the project. The integrity of the calculations was tested by using the statistical software package, Minitab (Minitab, 2014).

## 5.4           Summary

The possible results of the array comparisons have been discussed and appropriate test hypotheses are highlighted. With consideration given to the project aims, an appropriate hypothesis testing method has been established and described. The one proportion binomial test will be used to test the null and alternative hypotheses have which are H0 and HA respectively.

Next post

6       Results