AP Statistics Curriculum 2007 StudentsT
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* 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!)====  
+  : <math>P(X=6)=</math>? Where <math>X\sim B(12, 0.44)</math>  
: <math>P(X=6)={12\choose 6}p^6(1p)^{6}</math>, with <math>{12\choose 6}=\frac{12!}{6!(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].  : <math>P(X=6)={12\choose 6}p^6(1p)^{6}</math>, with <math>{12\choose 6}=\frac{12!}{6!(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].  
  +  
+  ====Approach II (Approximate)====  
+  : <math>X \sim B(n=12, p=0.44)</math>  
: <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=(np(1p))1/2=1.7]</math>. <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X5.28} \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>  : <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=(np(1p))1/2=1.7]</math>. <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X5.28} \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>  
  +  
+  ====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(1p)/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=0.51/24</math> and <math>p_2=0.5+1/24</math>. Standardize <math>Z = (p0.44)/0.1433</math> 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>  <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>  
Revision as of 04:50, 4 February 2008
Contents 
General AdvancePlacement (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 samplevariance. This requires us to introduce the Tdistribution, which is a oneparameter distribution connecting .
Student's T Distribution
The Student's tdistribution 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 ttests for the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means.
Suppose X_{1}, ..., X_{n} are independent random variables that are Normally distributed with expected value μ and variance σ^{2}. Let
 be the sample mean, and
 be the sample variance. We already discussed the following statistic:
is normally distributed with mean 0 and variance 1, since the sample mean is normally distributed with mean μ and standard deviation .
Gosset studied a related quantity under the pseudonym Student),
which differs from Z in that the (unknown) population standard deviation is replaced by the sample standard deviation S_{n}. Technically, has a Chisquare distribution distribution. Gosset's work showed that T has a specific probability density function, which approaches Normal(0,1) as the degree of freedom (df=samplesize 1) increases.
Computing with Tdistribution
 You can see the discretized Ttable or
 Use the interactive SOCR Tdistribution or
 Use the high precision Tdistribution 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% (α = 0.025) 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?
 s = 73.8
 SE = 14.8
 n = 25
 [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) = 200499 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 roundup. 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
 , with , using SOCR Binomial interactive GUI or calculator.
Approach II (Approximate)

 . , where and X_{1} = 5.5, X_{2} = 6.5. So,
Approach III (Approximate)
 The sample proportion is Thus, , where p_{1} = 0.5 − 1 / 24 and p_{2} = 0.5 + 1 / 24. Standardize Z = (p − 0.44) / 0.1433 to get:
References
 SOCR Home page: http://www.socr.ucla.edu
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