AP Statistics Curriculum 2007 StudentsT

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(New page: == General Advance-Placement (AP) Statistics Curriculum - Student's T Distribution== Very frequently in practive we do now know the population variance ...)
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==[[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==
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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 <math>Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}</math>.
+
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 <math>Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}</math>.
[[Image:SOCR_Distribution_StudentT_density.JPG|200px|thumbnail|right]]
[[Image:SOCR_Distribution_StudentT_density.JPG|200px|thumbnail|right]]
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is normally distributed with mean 0 and variance 1, since the sample mean <math>\scriptstyle \overline{X}_n </math> is normally distributed with mean <math> \mu</math> and standard deviation <math>\scriptstyle\sigma/\sqrt{n}</math>.
is normally distributed with mean 0 and variance 1, since the sample mean <math>\scriptstyle \overline{X}_n </math> is normally distributed with mean <math> \mu</math> and standard deviation <math>\scriptstyle\sigma/\sqrt{n}</math>.
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Gosset studied a related quantity under the pseudonym ''Student''),
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Gosset studied a related quantity under the pseudonym ''Student'',
:<math>T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},</math>
:<math>T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},</math>
which differs from ''Z'' in that the (unknown) population standard deviation <math>\scriptstyle \sigma</math> is replaced by the sample standard deviation <math>S_n</math>.  Technically, <math>\scriptstyle(n-1)S_n^2/\sigma^2</math> has a [http://en.wikipedia.org/wiki/Chi-square_distribution Chi-square distribution <math>\scriptstyle\chi_{n-1}^2</math> 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 <math>\scriptstyle \sigma</math> is replaced by the sample standard deviation <math>S_n</math>.  Technically, <math>\scriptstyle(n-1)S_n^2/\sigma^2</math> has a [http://en.wikipedia.org/wiki/Chi-square_distribution Chi-square distribution <math>\scriptstyle\chi_{n-1}^2</math> 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.
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===Computing with T-distribution===
===Computing with T-distribution===
* You can see the discretized [http://socr.ucla.edu/Applets.dir/T-table.html T-table] or  
* You can see the discretized [http://socr.ucla.edu/Applets.dir/T-table.html T-table] or  
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* Use the [ http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive SOCR T-distribution] or
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* Use the [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive SOCR T-distribution] or
* Use the [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm high precision T-distribution calculator].
* Use the [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm high precision T-distribution calculator].
===Example===
===Example===
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To parallel the example in the [[AP_Statistics_Curriculum_2007#Estimating_a_Population_Mean:_Large_Samples |large sample case]], we consider again the ''number of sentences per advertisement'' as a measure of readability for magazine advertisementsA 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 <math>\mu</math> (sample-mean=22.1 and sample-variance=737.88).
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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 clinicShe 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=0.05</math>) confidence interval for <math>\mu</math>:
 +
 
<center>
<center>
{| class="wikitable" style="text-align:center; width:75%" border="1"
{| class="wikitable" style="text-align:center; width:75%" border="1"
|-
|-
-
| 16 || 9 || 14 ||  11||  17 ||  12|| 99 || 18 || 13|| 12
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| 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
|}
|}
</center>
</center>
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A ''confidence interval estimate'' of <math>\mu</math> 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. 
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What do we know from the background information?
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: <math>\overline{y}= 321.4</math>
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Before you find an interval estimate, you should first determine how confident you want to be that your interval estimate contains the population mean. 
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: <math>s = 73.8</math>
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: <math>SE = 14.8</math>
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* [[AP_Statistics_Curriculum_2007_Normal_Critical | Recall these critical values for Standard Normal distribution]]:
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: <math>n = 25</math>
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:80% confidence (0.80), <math>\alpha=0.1</math>, [http://socr.ucla.edu/Applets.dir/T-table.html t<sub>(df=9)</sub> = 1.383]
+
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:90% confidence (0.90), <math>\alpha=0.05</math>, [http://socr.ucla.edu/Applets.dir/T-table.html t<sub>(df=9)</sub> = 1.833]
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:95% confidence (0.95), <math>\alpha=0.025</math>, [http://socr.ucla.edu/Applets.dir/T-table.html t<sub>(df=9)</sub> = 2.262]
+
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:99% confidence (0.99), <math>\alpha=0.005</math>, [http://socr.ucla.edu/Applets.dir/T-table.html t<sub>(df=9)</sub> = 3.250]
+
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Notice that for a fixed <math>\alpha</math>, the ''t-critical values'' (for any degree-of-freedom) exceeds the [[AP_Statistics_Curriculum_2007#Estimating_a_Population_Mean:_Large_Samples |corresponding normal z-critical values]], which are used int he large-sample interval estimation.
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: <math>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}}.</math>
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====Known Variance====
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: <math>321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}</math>
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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 <math>\sigma=16</math>).
+
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* For <math>\alpha=0.1</math>, the <math>80% CI(\mu)</math> is constructed by:
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: <math>321.4 \pm 2.064\times 14.8</math>
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<center> <math>\overline{x}\pm 1.383{16\over \sqrt{10}}=14.77 \pm 1.28{16\over \sqrt{10}}=[15.10 ; 29.10]</math></center>
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* For <math>\alpha=0.05</math>, the <math>90% CI(\mu)</math> is constructed by:
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: <math>[290.85, 351.95]</math>
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<center> <math>\overline{x}\pm 1.833{16\over \sqrt{10}}=14.77 \pm 1.833{16\over \sqrt{10}}=[12.83 ; 31.37]</math></center>
+
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* For <math>\alpha=0.005</math>, the <math>99% CI(\mu)</math> is constructed by:
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====CI Interpretation====
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<center> <math>\overline{x}\pm 3.250{16\over \sqrt{10}}=14.77 \pm 3.250{16\over \sqrt{10}}=[5.66 ; 38.54]</math></center>
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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?
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Notice the increase of the CI's (directly related to the decrease of <math>\alpha</math>) reflecting our choice for higher confidence.
+
===SOCR CI Experiments===
 +
The [http://socr.ucla.edu/htmls/SOCR_Experiments.html SOCR Confidence Interval Experiment] provides empirical evidence that the definition and the construction protocol for Confidence intervals are consistent.  
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====Unknown variance====
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===Activities===
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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 <math>s=\hat{\sigma}=27.16390579</math>).
+
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* For <math>\alpha=0.1</math>, the <math>80% CI(\mu)</math> is constructed by:
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*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.
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<center> <math>\overline{x}\pm 1.383{27.16390579\over \sqrt{10}}=14.77 \pm 1.383{16\over \sqrt{10}}=[10.22 ; 33.98]</math></center>
+
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* For <math>\alpha=0.05</math>, the <math>90% CI(\mu)</math> is constructed by:
+
* 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?
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<center> <math>\overline{x}\pm 1.833{27.16390579\over \sqrt{10}}=14.77 \pm 1.833{16\over \sqrt{10}}=[6.35 ; 37.85]</math></center>
+
====Approach I (exact)====
 +
: <math>P(X=6)=</math>? Where <math>X\sim B(12, 0.44)</math>
 +
: <math>P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068</math>, 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].
-
* For <math>\alpha=0.005</math>, the <math>99% CI(\mu)</math> is constructed by:
+
====Approach II (Approximate)====
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<center> <math>\overline{x}\pm 3.250{27.16390579 \over \sqrt{10}}=14.77 \pm 3.250{16\over \sqrt{10}}=[-5.82 ; 50.02]</math></center>
+
: <math>X \sim B(n=12, p=0.44)</math>.
 +
: <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7]</math>. <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X-5.28} \over {1.7}}</math> and <math>X_1=5.5</math>, <math>X_2=6.5</math>. So, <math>P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.</math>
-
Notice the increase of the CI's (directly related to the decrease of <math>\alpha</math>) reflecting our choice for higher confidence.
+
====Approach III (Approximate)====
 +
: <math>X \sim B(n=12, p=0.44)</math>. The sample proportion is <math>\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]</math>. Thus, <math>P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)</math>, where <math>p_1=0.5-1/24</math> and <math>p_2=0.5+1/24</math>. 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 (<math>p_1</math> and <math>p_2</math>), using the <math>Z = (p-0.44)/0.1433</math> transformation, to get
 +
:<math>P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.</math>
-
===Hands-on activities===
 
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*See the [[SOCR_EduMaterials_Activities_CoinfidenceIntervalExperiment | SOCR Confidence Interval Experiment]].
 
-
* Sample statistics, like the sample-mean and the sample-variance, may be easily obtained using [http://socr.ucla.edu/htmls/SOCR_Charts.html 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, [[AP_Statistics_Curriculum_2007_Estim_L_Mean#Example | reported above]].
 
-
<center>[[Image:SOCR_EBook_Dinov_Estimates_L_Mean_020208_Fig3.jpg|400px]]
 
-
[[Image:SOCR_EBook_Dinov_Estimates_L_Mean_020208_Fig4.jpg|400px]]</center>
 
<hr>
<hr>
-
===References===
+
===[[EBook_Problems_StudentsT|Problems]]===
<hr>
<hr>

Current revision as of 00:16, 11 March 2011

Contents

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.

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 = 0.05) confidence interval for μ:

Variable N N* Mean SE of MeanStDev 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.



Problems




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