Statistical Significance On Variance

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maudise

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Nov 15, 2013
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Hi,

It has been many years since I last did this sort of thing and it's late on a Friday!

I've got two independent samples and I want to know whether they are statistically different (appreciate this depends on what my confidence interval is). I am testing for the existance of whether a person has a blue car from a survey, and the amount they spent on it.

Sample A were questionned under duress whilst Sample B were questionned under relaxed conditions.

Sample A had 1,025 respondents; 91 (8.88%) had a blue car and their average cost of the car was £4,304.

Sample B had 1,036 respondents; 67 (6.47%) had a blue car and their average cost was £3,345.

Together 7.67% had a blue car and the average spend was £3,898

If I want to say Sample B is statistically significant to Sample A on either the value or the propensity to have a blue car (or combined) how would I go about this?

Thanks!

Maudise
 
maudise said:
Hi,

It has been many years since I last did this sort of thing and it's late on a Friday!

I've got two independent samples and I want to know whether they are statistically different (appreciate this depends on what my confidence interval is). I am testing for the existance of whether a person has a blue car from a survey, and the amount they spent on it.

Sample A were questionned under duress whilst Sample B were questionned under relaxed conditions.

Sample A had 1,025 respondents; 91 (8.88%) had a blue car and their average cost of the car was £4,304.

Sample B had 1,036 respondents; 67 (6.47%) had a blue car and their average cost was £3,345.

Together 7.67% had a blue car and the average spend was £3,898

If I want to say Sample B is statistically significant to Sample A on either the value or the propensity to have a blue car (or combined) how would I go about this?

Thanks!

Maudise
Click to expand...
The propensity question is a binomial distribution - either has or has not a blue car. [Perhaps "admits to have" or "claims to have" a blue car would fit the scenario better?] Do you remember the formulas for mean and standard deviation of a binomial distribution?

......\(\displaystyle \mu = n\ p,\;\;\;\;\;\;\;\;\sigma = \sqrt{n\ p\ (1-p)}\)

The numbers are large enough that you should approximate the two binomial distributions as normal distributions with the same \(\displaystyle \mu,\ \sigma\). You can ask how many standard deviations apart the two means are, or there are fancier tests fot the equivalence of two distributions.

You havn't reported enough data to say anything about comparing costs. You have to know the standard deviations from each sample, as well as the means. That requires going back to the individual data.
 
Hi,

Thanks for the prompt reply!

The standard Deviation for Sample A: 0.57689 and for Sample B: 25.07381 (this seems high?).

Also for your formulas N = Number of observations, what does P stand for?

I'm heading home shortly so will be without access to the raw data, will do what I can from there.

Maudise
 
maudise said:
Hi,

Thanks for the prompt reply!

The standard Deviation for Sample A: 0.57689 and for Sample B: 25.07381 (this seems high?).

Also for your formulas N = Number of observations, what does P stand for?

I'm heading home shortly so will be without access to the raw data, will do what I can from there.

Maudise
Click to expand...
In a binomial distribution, p is the (universal) probability of "success" on a single trial, and n is the number of trials. You estimate p to be the observed percentage.

To test for A=B, the null hypothesis is A - B = 0. If A and B are statistically independent - which they are because they are different random sets of people - then the Variances of A and B add, so that the standard deviation of the difference is the square root of the sum of the squares:

\(\displaystyle \displaystyle \sigma_{A-B} = \left(\sigma_A^2 + \sigma_B^2\right)^{1/2}\)

Use that standard deviation and a table of the normal distribution to find the confidence level that A-B=0.
 
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