Difference between revisions of "AP Statistics Curriculum 2007 StudentsT"

From SOCR
Jump to: navigation, search
(New page: == General Advance-Placement (AP) Statistics Curriculum - Student's T Distribution== Very frequently in practive we do now know the population variance ...)
 
m (Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/")
 
(29 intermediate revisions by 4 users 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 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]]
Line 16: Line 16:
 
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>.
  
−
Gosset studied a related quantity under the pseudonym ''Student''),
+
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.
Line 22: Line 22:
 
===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  
−
* Use the [ http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive SOCR T-distribution] or
+
* 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===
−
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 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 <math>\mu</math> (sample-mean=22.1 and sample-variance=737.88).
+
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=0.05</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
+
| 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>
  
−
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. 
+
What do we know from the background information?
−
 
+
: <math>\overline{y}= 321.4</math>
−
Before you find an interval estimate, you should first determine how confident you want to be that your interval estimate contains the population mean. 
+
: <math>s = 73.8</math>
−
 
+
: <math>SE = 14.8</math>
−
* [[AP_Statistics_Curriculum_2007_Normal_Critical | Recall these critical values for Standard Normal distribution]]:
+
: <math>n = 25</math>
−
: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]
 
−
: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]
 
−
: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]
 
−
: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]
 
−
 
 
−
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.
 
  
−
====Known Variance====
+
: <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>
−
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>).
 
  
−
* For <math>\alpha=0.1</math>, the <math>80% CI(\mu)</math> is constructed by:
+
: <math>321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}</math>
−
<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>
 
  
−
* For <math>\alpha=0.05</math>, the <math>90% CI(\mu)</math> is constructed by:
+
: <math>321.4 \pm 2.064\times 14.8</math>
−
<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>
 
  
−
* For <math>\alpha=0.005</math>, the <math>99% CI(\mu)</math> is constructed by:
+
: <math>[290.85, 351.95]</math>
−
<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>
 
  
−
Notice the increase of the CI's (directly related to the decrease of <math>\alpha</math>) reflecting our choice for higher confidence.
+
====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?
  
−
====Unknown variance====
+
===SOCR CI Experiments===
−
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>).  
+
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.  
  
−
* For <math>\alpha=0.1</math>, the <math>80% CI(\mu)</math> is constructed by:
+
===Activities===
−
<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>
 
  
−
* For <math>\alpha=0.05</math>, the <math>90% CI(\mu)</math> is constructed by:
+
*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.
−
<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>
 
  
−
* For <math>\alpha=0.005</math>, the <math>99% 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?
−
<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>
+
====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].
  
−
Notice the increase of the CI's (directly related to the decrease of <math>\alpha</math>) reflecting our choice for higher confidence.
+
====Approach II (Approximate)====
 +
: <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>
  
−
===Hands-on activities===
+
====Approach III (Approximate)====
−
*See the [[SOCR_EduMaterials_Activities_CoinfidenceIntervalExperiment | SOCR Confidence Interval Experiment]].
+
: <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
−
* 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]].
+
:<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>
−
<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>
 
* SOCR Home page: http://www.socr.ucla.edu
 
* SOCR Home page: http://www.socr.ucla.edu
  
−
{{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_StudentsT}}
+
"{{translate|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_StudentsT}}

Latest revision as of 14:28, 3 March 2020

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 \(Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}\).

SOCR Distribution StudentT density.JPG

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

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% (\(\alpha=0.05\)) confidence interval for \(\mu\):

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 \(X_1=5.5,\) \(X_2=6.5.\) 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 \(p_1=0.5-1/24\) 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 \[P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.\]

Problems

"-----

Translate this page:

(default)
Uk flag.gif

Deutsch
De flag.gif

Español
Es flag.gif

Français
Fr flag.gif

Italiano
It flag.gif

Português
Pt flag.gif

日本語
Jp flag.gif

България
Bg flag.gif

الامارات العربية المتحدة
Ae flag.gif

Suomi
Fi flag.gif

इस भाषा में
In flag.gif

Norge
No flag.png

한국어
Kr flag.gif

中文
Cn flag.gif

繁体中文
Cn flag.gif

Русский
Ru flag.gif

Nederlands
Nl flag.gif

Ελληνικά
Gr flag.gif

Hrvatska
Hr flag.gif

Česká republika
Cz flag.gif

Danmark
Dk flag.gif

Polska
Pl flag.png

România
Ro flag.png

Sverige
Se flag.gif

Retrieved from "https://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_StudentsT&oldid=16918"