Is it possible to find standard deviation from sample size, population mean?

Question

When the distribution is assumed normal.

Let's say 24/100 of one population votes for a candidate, and 32/100 of another population votes for the same candidate. If I want to find the observed test statistic to see if there's a difference between the groups and distribution of said test statistic, I need the standard deviation. Is it possible to calculate this? (assume significance level = 5%)

Answer 1

You are comparing the proportions between the two populations:

Z = (p1 - p2)/ sqrt ( p(1-p)(1/n1 + 1/n2))

where p = (x1+x2)/(d1+d2) for numerators x and d <--- the POOLED proportion

p = (24+32)/(100+100) = 56/200 = 28/100 = 7/25

(24/100 - 32/100) = -8/100 is the numerator

sqrt ( (7/25)(18/25)(1/100+ 1/100))

= sqrt ( (126/625)(2/100))

= sqrt ( 252/62500)

252 = 2 * 126 = 2 * 2 * 63 = 2*2*7*9

so sqrt(252) = 2*3*sqrt(7) = 6 * sqrt(7)

the denominator is 6*sqrt(7) / 250

= 3 * sqrt(7)/125

dividing: -8/100 divided by 3 * sqrt(7) / 125

= -8/100 * 125/(3 sqrt(7))

= -1000/ (300 * sqrt(7))

= (-10/21)*sqrt(7)

The test stat is just under -1.26