Consider the exercise we have just seen before.
It was made an intelligence test in 10 subjects, and here are the results obtained. The average result of the population whici received the same test, is equal to 75. You want to check if the sample mean is significantly similar (when the significance level is 95%) to the average population, assuming that the variance of the population is not known.
65, 78, 88, 55, 48, 95, 66, 57, 79, 81
Contrary to the one sample Z-test, the Student's t-test for a single sample have a pre-set function in R we can apply immediately.
It is the
t.test (a, mu), we can see below applied.
a = c(65, 78, 88, 55, 48, 95, 66, 57, 79, 81)
t.test (a, mu=75)
One Sample t-test
t = -0.783, df = 9, p-value = 0.4537
alternative hypothesis: true mean is not equal to 75
95 percent confidence interval:
mean of x
t.teston one sample provides in output the value of t calculated; also gives us degrees of freedom, the confidence interval and the average (mean of x).
In order to take your statistic decision, you can proceed in two ways. We can compare the value of t with the value of the tabulated student t with 9 degrees of freedom. If we do not have tables, we can calculate the value t-tabulated in the following way:
qt (p, df)returns the value of t computed considering the significance level (we chose a significance level equal to 95%, which means that each tail is the 2.5% which corresponds to the value of p = 1 - 0.025), and the degrees of freedom. By comparing the value of t-tabulated with t-computed, t-computed appears smaller, which means that we accept the null hypothesis of equality of the averages: our sample mean is significantly similar to the mean of the population.
Alternatively we could consider the p-value. With a significance level of 95%, remember this rule: If p-value is greater than 0.05 then we accept the null hypothesis H0; if p-value is less than 0.05 then we reject the null hypothesis H0 in favor of the alternative hypothesis H1.