# AP Statistics Curriculum 2007 StudentsT

## General Advance-Placement (AP) Statistics Curriculum - Student's T Distribution

Very frequently in practive we do now know the population variance and therefore need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting $Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}$.

### Student's T Distribution

The Student's t-distribution arises in the problem of estimating the mean of a normally distributed population when the sample size is small and the population variance is unknown. It is the basis of the popular Student's t-tests for the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means.

Suppose X1, ..., Xn are independent random variables that are Normally distributed with expected value μ and variance σ2. Let

$\overline{X}_n = {X_1+X_2+\cdots+X_n \over n}$ be the sample mean, and
${S_n}^2=\frac{1}{n-1}\sum_{i=1}^n\left(X_i-\overline{X}_n\right)^2$ be the sample variance. We already discussed the following statistic:
$Z=\frac{\overline{X}_n-\mu}{\sigma/\sqrt{n}}$

is normally distributed with mean 0 and variance 1, since the sample mean $\scriptstyle \overline{X}_n$ is normally distributed with mean μ and standard deviation $\scriptstyle\sigma/\sqrt{n}$.

Gosset studied a related quantity under the pseudonym Student),

$T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},$

which differs from Z in that the (unknown) population standard deviation $\scriptstyle \sigma$ is replaced by the sample standard deviation Sn. Technically, $\scriptstyle(n-1)S_n^2/\sigma^2$ has a Chi-square distribution $\scriptstyle\chi_{n-1}^2$ distribution. Gosset's work showed that T has a specific probability density function, which approaches Normal(0,1) as the degree of freedom (df=sample-size -1) increases.

### Example

To parallel the example in the large sample case, we consider again the number of sentences per advertisement as a measure of readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed. Use this sample to find point estimate for the population mean μ (sample-mean=22.1 and sample-variance=737.88).

 16 9 14 11 17 12 99 18 13 12

A confidence interval estimate of μ is a range of values used to estimate a population parameter (interval estimates are normally used more than point estimates because it is very unlikely that the sample mean would match exactly with the population mean) The interval estimate uses a margin of error about the point estimate.

Before you find an interval estimate, you should first determine how confident you want to be that your interval estimate contains the population mean.

80% confidence (0.80), α = 0.1, t(df=9) = 1.383
90% confidence (0.90), α = 0.05, t(df=9) = 1.833
95% confidence (0.95), α = 0.025, t(df=9) = 2.262
99% confidence (0.99), α = 0.005, t(df=9) = 3.250

Notice that for a fixed α, the t-critical values (for any degree-of-freedom) exceeds the corresponding normal z-critical values, which are used int he large-sample interval estimation.

#### Known Variance

Suppose that we know the variance for the number of sentences per advertisement example above is known to be 256 (so the population standard deviation is σ = 16).

• For α = 0.1, the 80%CI(μ) is constructed by:
$\overline{x}\pm 1.383{16\over \sqrt{10}}=14.77 \pm 1.28{16\over \sqrt{10}}=[15.10 ; 29.10]$
• For α = 0.05, the 90%CI(μ) is constructed by:
$\overline{x}\pm 1.833{16\over \sqrt{10}}=14.77 \pm 1.833{16\over \sqrt{10}}=[12.83 ; 31.37]$
• For α = 0.005, the 99%CI(μ) is constructed by:
$\overline{x}\pm 3.250{16\over \sqrt{10}}=14.77 \pm 3.250{16\over \sqrt{10}}=[5.66 ; 38.54]$

Notice the increase of the CI's (directly related to the decrease of α) reflecting our choice for higher confidence.

#### Unknown variance

Suppose that we do not know the variance for the number of sentences per advertisement but use the sample variance 737.88 as an estimate (so the sample standard deviation is $s=\hat{\sigma}=27.16390579$).

• For α = 0.1, the 80%CI(μ) is constructed by:
$\overline{x}\pm 1.383{27.16390579\over \sqrt{10}}=14.77 \pm 1.383{16\over \sqrt{10}}=[10.22 ; 33.98]$
• For α = 0.05, the 90%CI(μ) is constructed by:
$\overline{x}\pm 1.833{27.16390579\over \sqrt{10}}=14.77 \pm 1.833{16\over \sqrt{10}}=[6.35 ; 37.85]$
• For α = 0.005, the 99%CI(μ) is constructed by:
$\overline{x}\pm 3.250{27.16390579 \over \sqrt{10}}=14.77 \pm 3.250{16\over \sqrt{10}}=[-5.82 ; 50.02]$

Notice the increase of the CI's (directly related to the decrease of α) reflecting our choice for higher confidence.

### Hands-on activities

• See the SOCR Confidence Interval Experiment.
• Sample statistics, like the sample-mean and the sample-variance, may be easily obtained using SOCR Charts. The images below illustrate this functionality (based on the Bar-Chart and Index-Chart) using the 30 observations of the number of sentences per advertisement, reported above.