AP Statistics Curriculum 2007 StudentsT

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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.

Computing with T-distribution

Example

Suppose a researcher wants to examine CD4 counts for HIV(+) patients seen at his clinic. She randomly selects a sample of n = 25 HIV(+) patients and measures their CD4 levels (cells/uL). Suppose she obtains the following results and we are interested in calculating a 95% (α = 0.025) confidence interval for μ:

Variable N N* Mean SEMeanStDev Minimum Q1 Median Q3 Maximum
CD4 25 0 321.4 14.8 73.8 208.0 261.5 325.0 394.0 449.0

What do we know from the background information?

\overline{y}= 321.4
s = 73.8
SE = 14.8
n = 25
CI(\alpha)=CI(0.025): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.
321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}
321.4 \pm 2.064\times 14.8
[290.85,351.95]

References




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