Difference between revisions of "AP Statistics Curriculum 2007 StudentsT"
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| − | Suppose a researcher wants to examine [http://gateway.nlm.nih.gov/MeetingAbstracts/102282532.html 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% confidence interval for <math>\mu</math>: | + | Suppose a researcher wants to examine [http://gateway.nlm.nih.gov/MeetingAbstracts/102282532.html 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% (<math>\alpha=0.025</math>) confidence interval for <math>\mu</math>: |
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| Variable ||N || N* || Mean ||SE||Mean||StDev ||Minimum || Q1|| Median || Q3 ||Maximum | | Variable ||N || N* || Mean ||SE||Mean||StDev ||Minimum || Q1|| Median || Q3 ||Maximum | ||
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| CD4 || 25|| 0 ||321.4|| 14.8 || 73.8 ||208.0 ||261.5 || 325.0 ||394.0 || 449.0 | | CD4 || 25|| 0 ||321.4|| 14.8 || 73.8 ||208.0 ||261.5 || 325.0 ||394.0 || 449.0 | ||
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: <math>n = 25</math> | : <math>n = 25</math> | ||
| − | : <math>CI(\alpha): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.</math> | + | : <math>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}}.</math> |
: <math>321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}</math> | : <math>321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}</math> | ||
Revision as of 21:49, 3 February 2008
Contents
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 \( \mu\) 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 \(S_n\). 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
- You can see the discretized T-table or
- Use the interactive SOCR T-distribution or
- Use the high precision T-distribution calculator.
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% (\(\alpha=0.025\)) confidence interval for \(\mu\):
| Variable | N | N* | Mean | SE | Mean | StDev | 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
- SOCR Home page: http://www.socr.ucla.edu
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