The following exercise provides us with an excellent example of the ways in which basic statistical analyses can be used - and indeed are used - in everyday transactions. In the problem before us, we have information about the credit card balances that are currently held by male and female customers of the Capital Bank. Its marketing department wishes to know if men or women typically keep a higher balance on their credit cards and, if one group does indeed do so, if the other group can be lured into increasing their balances.
Following are the calculations of mean and variance. The mean is the arithmetic average. In calculating the mean, one adds each of the given numbers in a category (i.e. male or female customers) and then divides that sum by total count of objects in the class. The total count is equal to how many given rational numbers were added together.
male $ 746.51 $86,743.26 $ 294.52
female $ 778.13 $87,152.24 $ 295.22
Variance is calculated as follows: s = [nSx¦-(Sx)¦]/n(n-1)
1. I would guess the difference in means is significant, but the difference in variances is not.
2. Null hypothesis: If the entire population was considered, there would be no difference in the two means. The apparent difference is due to sampling error.
Alternative hypothesis: The observed difference in means reflects a real difference in the population.
We use a t-test to estimate the probability that the means would differ by this amount under the null hypothesis and find:
T-test assuming equal variances: 0.437119
T-test assuming unequal variances: 0.438774
At the .05 level, we easily conclude the differences are not statistically significant. My guess was wrong. The bank should conclude there is no reason to market differently to men and women.
3. Null hypothesis: If the entire population was considered, there would be no difference in the two variances. The observed difference
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