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significance-testing of share of preference

Hello,

I designed with the choice simulator 3 products based on HB-model and got the following shares of preferences and the std errors:

Shares of preferences:
 1: 27.4%
 2: 23.3%
 3: 22.9%
None-Option: 26.5%

Std. Error:
1: 1,32627%
2: 1,08949%
3: 1,10678%
None: 1,22438%

I would to make sure, that the share of preferences are significant.
Can I test it like this? (similar to 12D in Becoming an Expert in Conjoint Analysis)
→ t (1/2) = 27.1-23.3 / square root(0.0132627²+0.0108949²) = 240.76
→ t (1/3) = 27.1-22.9/ square root(0.0132627²+0.0110678²) = 264.25
→ t (1/4) = 27.1-26.5/ square root(0.0132627²+0.0122438²) = 50.87
→ t (2/3) = 23.3-22.9/ square root(0.0108949²+0.0110678²) = 4.43598
→ t (2/4) = 23.3-26.5/ square root(0.0108949²+0.0122438²) = -35.428
→ t (3/4) = 22.9-26.5/ square root(0.0110678²+0.0122438²) = -218.119

Since all t-values has an absolute magnitude greater than 2.58, it is greater than 99% confident.

Is that right or do I have to test in different?

Thank you an best regards,
Beatrice
asked May 29 by bea007 (255 points)

1 Answer

0 votes
Your calculations are not quite right.  You are off the mark by a factor of 100.

Here's the calculation for product 1 compared to product 2:

t = (27.1-23.3)/SQRT(1.32627^2 + 1.08949^2)
= 2.4

This t-ratio calculation is a quick one that you can do from summary (pooled) statistics of means and standard errors.  But, this test is not quite as strong as the matched sample t-test, as described in Chapter 12 of our "Becoming an Expert..." conjoint book (https://www.sawtoothsoftware.com/169-support/technical-papers/cbc-related-papers/2033-statistical-testing).
answered May 29 by Bryan Orme Platinum Sawtooth Software, Inc. (164,715 points)
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