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Comparison of the means of two independent groups of samples, taken from two populations with known variance.**Statistic on aiR**, and kindly contributed to R-bloggers)Is asked to compare the average heights of two groups. The first group (A) consists of individuals of Italian nationality (the variance of the Italian population is 5); the second group is taken from individuals of German nationality (the variance of German population variance is 8.5). The data are given below:

A: 175, 168, 168, 190, 156, 181, 182, 175, 174, 179

B: 185, 169, 173, 173, 188, 186, 175, 174, 179, 180

B: 185, 169, 173, 173, 188, 186, 175, 174, 179, 180

Since we have the variance of the population, we must proceed with a

**two sample Z-test**. Even in this case is not avalilable in R a function to solve the problem, but we can easily create it ourselves.

$$Z=\frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$$

z.test2sam = function(a, b, var.a, var.b){

n.a = length(a)

n.b = length(b)

zeta = (mean(a) - mean(b)) / (sqrt(var.a/n.a + var.b/n.b))

return(zeta)

}

The function

`z.test2sam`

provides in output the value of zeta, after receiving in input two vectors (`a`

and `b`

), the variance of the first population (`var.a`

) and the variance of the second population (`var.b`

).Using this function we obtain:

a = c(175, 168, 168, 190, 156, 181, 182, 175, 174, 179)

b = c(185, 169, 173, 173, 188, 186, 175, 174, 179, 180)

z.test2sam(a, b, 5, 8.5)

[1] -2.926254

The value of zeta is greater than the value of the critical value zeta tabulated for alpha equal to 0.05 (

**z-tabulated = 1.96**for a two-tailed test): then we reject the null hypothesis in favor of the alternative hypothesis. We conclude that the two means are significantly different.

To

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