Comparison of the sample mean with know population mean and standard deviation.

Suppose that 10 volunteers have done an intelligence test; here are the results obtained. The mean obtained at the same test, from the entire population is 75. You want to check if there is a statistically significant difference (with a significance level of 95%) between the means of the sample and the population, assuming that the sample variance is known and equal to 18.


To solve this problem it is necessary to develop a one sample Z-test. In R there isn’t a similar function, so we can create our function.
Recalling the formula for calculating the value of z, we will write this function:

$$Z=\frac{\overline{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}$$

z.test = function(a, mu, var){
   zeta = (mean(a) - mu) / (sqrt(var / length(a)))
   return(zeta)
}

We have built so the function z.test; it receives in input a vector of values (a), the mean of the population to perform the comparison (mu), and the population variance (var); it returns the value of zeta. Now apply the function to our problem.

a = c(65, 78, 88, 55, 48, 95, 66, 57, 79, 81)

z.test(a, 75, 18)
[1] -2.832353

The value of zeta is equal to -2.83, which is higher than the critical value Zcv = 1.96, with alpha = 0.05 (2-tailed test). We conclude therefore that the mean of our sample is significantly different from the mean of the population.