3.4 Confidence Intervals for the Population Mean
As stressed before, we will never estimate the exact value of the population mean of \(Y\) using a random sample. However, we can compute confidence intervals for the population mean. In general, a confidence interval for an unknown parameter is a recipe that, in repeated samples, yields intervals that contain the true parameter with a prespecified probability, the confidence level. Confidence intervals are computed using the information available in the sample. Since this information is the result of a random process, confidence intervals are random variables themselves.
Key Concept 3.7 shows how to compute confidence intervals for the unknown population mean \(E(Y)\).
Key Concept 3.7
Confidence Intervals for the Population Mean
A \(95\%\) confidence interval for \(\mu_Y\) is a random variable that contains the true \(\mu_Y\) in \(95\%\) of all possible random samples. When \(n\) is large we can use the normal approximation. Then, \(99\%\), \(95\%\), \(90\%\) confidence intervals are
\[\begin{align} &99\%\text{ confidence interval for } \mu_Y = \left[ \overline{Y} \pm 2.58 \times SE(\overline{Y}) \right], \\ &95\%\text{ confidence interval for } \mu_Y = \left[\overline{Y} \pm 1.96 \times SE(\overline{Y}) \right], \\ &90\%\text{ confidence interval for } \mu_Y = \left[ \overline{Y} \pm 1.64 \times SE(\overline{Y}) \right]. \end{align}\]These confidence intervals are sets of null hypotheses we cannot reject in a two-sided hypothesis test at the given level of confidence.
Now consider the following statements.
In repeated sampling, the interval \[ \left[ \overline{Y} \pm 1.96 \times SE(\overline{Y}) \right] \] covers the true value of \(\mu_Y\) with a probability of \(95\%\).
We have computed \(\overline{Y} = 5.1\) and \(SE(\overline{Y})=2.5\) so the interval \[ \left[ 5.1 \pm 1.96 \times 2.5 \right] = \left[0.2,10\right] \] covers the true value of \(\mu_Y\) with a probability of \(95\%\).
While 1. is right (this is in line with the definition above), 2. is wrong and none of your lecturers wants to read such a sentence in a term paper, written exam or similar, believe us. The difference is that, while 1. is the definition of a random variable, 2. is one possible outcome of this random variable so there is no meaning in making any probabilistic statement about it. Either the computed interval does cover \(\mu_Y\) or it does not!
In R, testing of hypotheses about the mean of a population on the basis of a random sample is very easy due to functions like t.test() from the stats package. It produces an object of type list. Luckily, one of the most simple ways to use t.test() is when you want to obtain a \(95\%\) confidence interval for some population mean. We start by generating some random data and calling t.test() in conjunction with ls() to obtain a breakdown of the output components.
# set seed
set.seed(1)
# generate some sample data
sampledata <- rnorm(100, 10, 10)
# check the type of the outcome produced by t.test
typeof(t.test(sampledata))## [1] "list"# display the list elements produced by t.test
ls(t.test(sampledata))## [1] "alternative" "conf.int" "data.name" "estimate" "method"
## [6] "null.value" "p.value" "parameter" "statistic"Though we find that many items are reported, at the moment we are only interested in computing a \(95\%\) confidence set for the mean.
t.test(sampledata)$"conf.int"## [1] 9.306651 12.871096
## attr(,"conf.level")
## [1] 0.95This tells us that the \(95\%\) confidence interval is
\[ \left[9.31, 12.87\right]. \]
In this example, the computed interval obviously does cover the true \(\mu_Y\) which we know to be \(10\).
Let us have a look at the whole standard output produced by t.test().
t.test(sampledata)##
## One Sample t-test
##
## data: sampledata
## t = 12.346, df = 99, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 9.306651 12.871096
## sample estimates:
## mean of x
## 11.08887We see that t.test() does not only compute a \(95\%\) confidence interval but automatically conducts a two-sided significance test of the hypothesis \(H_0: \mu_Y = 0\) at the level of \(5\%\) and reports relevant parameters thereof: the alternative hypothesis, the estimated mean, the resulting \(t\)-statistic, the degrees of freedom of the underlying \(t\) distribution (t.test() does use perform the normal approximation) and the corresponding \(p\)-value. This is very convenient!
In this example, we come to the conclusion that the population mean is significantly different from \(0\) (which is correct) at the level of \(5\%\), since \(\mu_Y = 0\) is not an element of the \(95\%\) confidence interval
\[ 0 \in \left[9.31,12.87\right]. \] We come to an equivalent result when using the \(p\)-value rejection rule since
\[ p\text{-value} = 2.2\cdot 10^{-16} \ll 0.05. \]