# Two sample Student’s t-test #1

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t-Test to compare the means of two groups under the assumption that both samples are random, independent, and come from normally distributed population with unknow but equal variances**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.

Here I will use the same data just seen in a previous post. 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

To solve this problem we must use to a Student’s t-test with two samples, assuming that the two samples are taken from populations that follow a Gaussian distribution (if we cannot assume that, we must solve this problem using the non-parametric test called

**Wilcoxon-Mann-Whitney test**; we will see this test in a future post). Before proceeding with the

*t-test*, it is necessary to evaluate the sample variances of the two groups, using a

**Fisher’s F-test**to verify the

*homoskedasticity*(

*homogeneity of variances*). In R you can do this in this way:

a = c(175, 168, 168, 190, 156, 181, 182, 175, 174, 179) b = c(185, 169, 173, 173, 188, 186, 175, 174, 179, 180) var.test(a,b) F test to compare two variances data: a and b F = 2.1028, num df = 9, denom df = 9, p-value = 0.2834 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.5223017 8.4657950 sample estimates: ratio of variances 2.102784

We obtained p-value greater than 0.05, then we can assume that the two variances are homogeneous. Indeed we can compare the value of F obtained with the tabulated value of F for alpha = 0.05, degrees of freedom of numerator = 9, and degrees of freedom of denominator = 9, using the function

`qf(p, df.num, df.den)`

:qf(0.95, 9, 9) [1] 3.178893

Note that the value of F computed is less than the tabulated value of F, which leads us to accept the null hypothesis of homogeneity of variances.

**NOTE:**The F distribution has only one tail, so with a confidence level of 95%,

`p = 0.95`

. Conversely, the *t-distribution*has two tails, and in the R’s function

`qt(p, df)`

we insert a value `p = 0975`

when you’re testing a two-tailed alternative hypothesis.Then call the function t.test for homogeneous variances (

`var.equal = TRUE`

) and independent samples (`paired = FALSE`

: you can omit this because the function works on independent samples by default) in this way:t.test(a,b, var.equal=TRUE, paired=FALSE) Two Sample t-test data: a and b t = -0.9474, df = 18, p-value = 0.356 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -10.93994 4.13994 sample estimates: mean of x mean of y 174.8 178.2

We obtained p-value greater than 0.05, then we can conclude that the averages of two groups are significantly similar. Indeed the value of t-computed is less than the tabulated t-value for 18 degrees of freedom, which in R we can calculate:

qt(0.975, 18) [1] 2.100922

This confirms that we can accept the null hypothesis H0 of equality of the means.

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