# AP Statistics Curriculum 2007 StudentsT

(Difference between revisions)
 Revision as of 04:39, 4 February 2008 (view source)IvoDinov (Talk | contribs)← Older edit Current revision as of 00:18, 11 March 2011 (view source)IvoDinov (Talk | contribs) m (→Approach III (Approximate)) (15 intermediate revisions not shown) Line 1: Line 1: ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Student's T Distribution== ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Student's T Distribution== - Very frequently in practice 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 $Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}$. + Very frequently in practice, we do not know the population variance. Therefore we need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting $Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}$. [[Image:SOCR_Distribution_StudentT_density.JPG|200px|thumbnail|right]] [[Image:SOCR_Distribution_StudentT_density.JPG|200px|thumbnail|right]] Line 16: Line 16: 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 $\scriptstyle\sigma/\sqrt{n}$. 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''), + Gosset studied a related quantity under the pseudonym ''Student'', :$T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},$ :$T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},$ 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 [http://en.wikipedia.org/wiki/Chi-square_distribution Chi-square distribution $\scriptstyle\chi_{n-1}^2$ distribution].  Gosset's work showed that ''T'' has a specific [http://en.wikipedia.org/wiki/Student%27s_t_distribution probability density function], which approaches Normal(0,1) as the degree of freedom (df=sample-size -1) increases. 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 [http://en.wikipedia.org/wiki/Chi-square_distribution Chi-square distribution $\scriptstyle\chi_{n-1}^2$ distribution].  Gosset's work showed that ''T'' has a specific [http://en.wikipedia.org/wiki/Student%27s_t_distribution probability density function], which approaches Normal(0,1) as the degree of freedom (df=sample-size -1) increases. Line 26: Line 26: ===Example=== ===Example=== - 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% ($\alpha=0.025$) confidence interval for $\mu$: + 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% ($\alpha=0.05$) confidence interval for $\mu$:
Line 43: Line 43: : $n = 25$ : $n = 25$ - : $CI(\alpha)=CI(0.025): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.$ + : $CI(\alpha)=CI(0.05): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.$ : $321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}$ : $321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}$ Line 64: Line 64: *A biologist obtained body weights of male reindeer from a herd during the seasonal round-up.  He measured the weight of a random sample of 102 reindeer in the herd, and found the sample mean and standard deviation to be 54.78 kg and 8.83 kg, respectively.  Suppose these data come from a normal distribution. Calculate a 99% confidence interval. *A biologist obtained body weights of male reindeer from a herd during the seasonal round-up.  He measured the weight of a random sample of 102 reindeer in the herd, and found the sample mean and standard deviation to be 54.78 kg and 8.83 kg, respectively.  Suppose these data come from a normal distribution. Calculate a 99% confidence interval. - * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types what is the probability that exactly 6 subjects have type O blood type in the sample? + * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types. What is the probability that exactly 6 subjects have type O blood type in the sample? + ====Approach I (exact)==== + : $P(X=6)=$? Where $X\sim B(12, 0.44)$ + : $P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068$, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator]. - **Approach I (exact!) : $P(X=6)=$? Where $X\sim B(12, 0.44)$ + ====Approach II (Approximate)==== - : $P(X=6)={12\choose 6}p^6(1-p)^{6}$, with ${12\choose 6}=\frac{12!}{6!(6)!}=0.2068$, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator]. + : $X \sim B(n=12, p=0.44)$. + : $X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7]$. $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where $Z = {{X-5.28} \over {1.7}}$ and $X_1=5.5$, $X_2=6.5. So, [itex]P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211. - **Approach II (Approximate): [itex]X \sim B(n=12, p=0.44)$ + ====Approach III (Approximate)==== - : $X (approx.) \sim N [m = n p = 5.28; (np(1-p))1/2=1.7]$ $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where $Z = (X – 5.28)/1.7$ and $X_1=5.5$, $X_2=6.5$. So, $P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.$ + : $X \sim B(n=12, p=0.44)$. The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$. Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where $p_1=0.5-1/24$ and $p_2=0.5+1/24$. Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits ($p_1$ and $p_2$), using the $Z = (p-0.44)/0.1433$ transformation, to get - + :$P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$ - ** Approach III (Approximate): $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where $p_1=0.5 – 1/24$ and $p_2=0.5+1/24$. Standardize $Z = (p-0.44)/0.1433$ to get: + - $P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$ +

- ===References=== + ===[[EBook_Problems_StudentsT|Problems]]===

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

Very frequently in practice, we do not know the population variance. Therefore we 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

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 = 0.05) confidence interval for μ:

 Variable N N* Mean SE of 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
$CI(\alpha)=CI(0.05): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.$
$321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}$
$321.4 \pm 2.064\times 14.8$
[290.85,351.95]

#### CI Interpretation

Still, does this CI (290.85, 351.95) mean anything to us? Consider the following information: The U.S. Government classification of AIDS has three official categories of CD4 counts – asymptomatic = greater than or equal to 500 cells/uL

• AIDS related complex (ARC) = 200-499 cells/uL
• AIDS = less than 200 cells/uL
• Now how can we interpret our CI?

### SOCR CI Experiments

The SOCR Confidence Interval Experiment provides empirical evidence that the definition and the construction protocol for Confidence intervals are consistent.

### Activities

• A biologist obtained body weights of male reindeer from a herd during the seasonal round-up. He measured the weight of a random sample of 102 reindeer in the herd, and found the sample mean and standard deviation to be 54.78 kg and 8.83 kg, respectively. Suppose these data come from a normal distribution. Calculate a 99% confidence interval.
• Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types. What is the probability that exactly 6 subjects have type O blood type in the sample?

#### Approach I (exact)

P(X = 6) = ? Where $X\sim B(12, 0.44)$
$P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068$, using SOCR Binomial interactive GUI or calculator.

#### Approach II (Approximate)

$X \sim B(n=12, p=0.44)$.
$X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7]$. $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where $Z = {{X-5.28} \over {1.7}}$ and X1 = 5.5, X2 = 6.5. So, $P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.$

#### Approach III (Approximate)

$X \sim B(n=12, p=0.44)$. The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$. Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where p1 = 0.5 − 1 / 24 and p2 = 0.5 + 1 / 24. Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits (p1 and p2), using the Z = (p − 0.44) / 0.1433 transformation, to get
$P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$