# One sample Z-test

**Statistic on aiR**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

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.

**leave a comment**for the author, please follow the link and comment on their blog:

**Statistic on aiR**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.