AP Statistics Curriculum 2007 StudentsT

From Socr

Jump to: navigation, search


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:

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


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

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.


  • 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.


Translate this page:









الامارات العربية المتحدة


इस भाषा में









Česká republika





Personal tools