# Statistical formulas documentation

mTAB Median Calculation from Income Brackets

 Approximate Annual HH Income Median 71,180 Unweighted Sample Total Count 10,811

 Approximate Annual HH Income WeightedResponse (1) Accumulated Response Less than $15,000 11,714 11,714$15,000 - $24,999 46,054 57, 768$25,000 - $34,999 83,965 141,733$35,000 - $44,999 102,093 243,826$45,000 - $59,999 155,721 399,546$60,000 - $74,999 161,435 560,981 <--Median will fall here (3)$75,000 - $99,999 193,540 754,521$100,000 - $124,999 134,706 889,227$125,000 - $149,999 59,748 948,975$150,000 - $199,999 41,971 990,946$200,000 - $249,999 16,391 1,007,337$250,000 or More 32,409 1,039,746 Weighted Subset Total Count 1,039,746 Weighted Sample Total Count 1,255,411

 (1) Calculated Accumulated Weighted Response (2) Divide total (1,039,746) by 2=519,873 519,873 (3) Find first value in Accumulated Response column that is greater than step 2 value The median will fall between the $60,000-$74,999 bracket (4) Step 2 amount (519,873) MINUS preceding break accumulated response 399,546 = 120,327 (5) Acc. Response where Median will fall 560,981 MINUS preceding break 399,546 = 161,435 (6) Step 4 Divided by Step 5 0.74536 (7) Multiply Step 6 by the range 14,999 ($60,000-$75,999) 11180 (8) Add Step 7 to bottom of range $60,000 71,180 mTAB Mean/Weighted Average Calculation from Income Brackets  Approximate Annual HH Income Mean/Weighted Average 83,610 Unweighted Sample Total Count 10,811  (A) (B) (C) Approximate Annual HH Income STAT1 STAT2 Midpoint Less than$15,000 11,714 1 14,999 7,500 87,857,249 $15,000 -$24,999 46,054 15,000 24,999 20,000 921,059,004 $25,000 -$34,999 83,965 25,000 34,999 30,000 2,518,899,346 $35,000 -$44,999 102,093 35,000 44,999 40,000 4,083,654,266 $45,000 -$59,999 155,721 45,000 59,999 52,500 8,175,254,132 $60,000 -$74,999 161,435 60,000 74,999 67,500 10,896,752,251 $75,000 -$99,999 193,540 75,000 99,999 87,500 16,934,669,636 $100,000 -$124,999 134,706 100,000 124,999 112,500 15,154,345,342 $125,000 -$149,999 59,748 125,000 149,999 137,500 8,125,258,359 $150,000 -$199,999 41,971 150,000 199,999 175,000 7,344,910,850 $200,000 -$249,999 16,391 200,000 249,000 225,000 3,688,025,252 \$250,000 or More 32,409 250,000 300,000 275,000 8,912,452,979 Weighted Subset Total Count 1,039,746 86,933,138,666 Weighted Sample Total Count 1,255,411

(1) Find Midpoint of data ranges - Column (B)
(2) Multiply Weighted Counts (A) by Midpoints (B) to generate (C)
(3) Divide the sum of column (C) by the total weighted response at the bottom of column (A)...
86,933,138,666 divided by 1,039,746 = 83,610
You will notice the calculated average matches the mTAB produced average

Standard Deviation Calculation

For categorized questions, each response is assigned 1 or 2 stat weights. If a single weight is assigned, then this is the value used to calculate the standard deviation. If 2 weights are provided, the midpoint is used.

D = Question Mean - Stat Value as described above
SS = Sum of Squares, D*D*Weighted Response Count, for all table responses
Sample = Sum of all Weighted Response Counts for all table responses

Standard Deviation = SQRT(SS/Sample-1));

The calculation is the same for continuous variables except the actual data values are used instead of stat weight.

D = Question Mean - Response Value
SS = Sum of Squares, D*D*Respondent Weight Count for each response
Sample = Sum of all Respondent Weights for each response

Standard Deviation = SQRT(SS/Sample-1));

Comparison of two population means using T-Statistic

When the two populations have equal variances

$$t = \dfrac{m_1-m_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

Where
m1 = Mean of the 1st sample
s1 = Standard Deviation of the 1st sample
n1 = Un-weighted Sample of the 1st sample

m2 = Mean of the 2nd sample
s2 = Standard Deviation of the 2nd sample
n2 = Un-weighted Sample of the 2nd sample

Decision rules:
If |t|<1.65 then the two populations are NOT significantly different at 90%
If |t|≥1.65 then the two populations ARE significantly different at 90%
If |t|<1.95 then the two populations are NOT significantly different at 95%
If |t|≥1.95 then the two populations ARE significantly different at 95%

When the two populations have UNEQUAL variances

$$t = \dfrac{m_1-m_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$$

Where
m1 = Mean of the 1st sample
s1 = Standard Deviation of the 1st sample
n1 = Un-weighted Sample of the 1st sample

m2 = Mean of the 2nd sample
s2 = Standard Deviation of the 2nd sample
n2 = Un-weighted Sample of the 2nd sample

Decision rules:
If |t|<1.65 then the two populations are NOT significantly different at 90%
If |t|≥1.65 then the two populations ARE significantly different at 90%
If |t|<1.95 then the two populations are NOT significantly different at 95%
If |t|≥1.95 then the two populations ARE significantly different at 95%