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Suppose that \(T\) is a \(t\) -distributed random variable. a. If \(T\) has 5 df, use Table 5 , Appendix 3 , to find \(t_{10}\), the value such that \(P\left(T>t_{10}\right)=.10\). Find \(t_{10}\) using the applet Student's \(t\) Probabilities and Quantiles. b. Refer to part (a). What quantile does \(t_{.10}\) correspond to? Which percentile? c. Use the applet Student's t Probabilities and Quantiles to find the value of \(t_{.10}\) for \(t\) distributions with 30, 60, and 120 df. d. When \(Z\) has a standard normal distribution, \(P(Z>1.282)=.10\) and \(z_{.10}=1.282 .\) What property of the \(t\) distribution (when compared to the standard normal distribution) explains the fact that all of the values obtained in part (c) are larger than \(z_{10}=1.282 ?\) e. What do you observe about the relative sizes of the values of \(t_{.10}\) for \(t\) distributions with 30,60 and \(120 \mathrm{df} ?\) Guess what \(t_{.10}\) "converges to" as the number of degrees of freedom gets large. [Hint: Look at the row labeled \(\infty \text { in Table } 5 \text { , Appendix } 3 .\)]

Short Answer

Expert verified
Value \( t_{10} \) for 5 df is approximately 1.476; it's a 90th percentile. For larger df, \( t_{.10} \) values decrease and approach the standard normal value (1.282).

Step by step solution

01

Finding t10 for 5 degrees of freedom

Consult Table 5 in Appendix 3 to find the value of \( t_{10} \) such that \( P(T>t_{10}) = 0.10 \) for 5 degrees of freedom. You'll find that \( t_{10} \approx 1.476 \). This is confirmed using the Student's \( t \) distribution applet.
02

Identifying the Quantile and Percentile

Since \( t_{.10} \) means \( P(T > t_{.10}) = 0.10 \), it corresponds to the 90th percentile because \( P(T \leq t_{.10}) = 0.90 \). Quantiles and percentiles are inversely related in this context.
03

Finding t.10 for Various Degrees of Freedom

Use the Student's \( t \) applet to find \( t_{.10} \) for 30, 60, and 120 degrees of freedom. You'll find values approximately \( 1.310 \), \( 1.296 \), and \( 1.290 \) respectively.
04

Comparing with Standard Normal Distribution

For a standard normal distribution, \( z_{.10} = 1.282 \). Since the values of \( t_{.10} \) for all \( t \) distributions are greater than 1.282, this is explained by the 'tail-heaviness' of the \( t \) distribution, which accounts for this difference especially for smaller degrees of freedom.
05

Observing Convergence of t.10 Values

Notice that as the degrees of freedom increase from 30 to 60 to 120, \( t_{.10} \) values decrease and approach \( z_{.10} = 1.282 \). This indicates \( t_{.10} \) converges to the standard normal distribution value as degrees of freedom get large, aligning with the value for infinity in Table 5 of Appendix 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degrees of Freedom
When working with t-distributions, you'll frequently encounter the term "degrees of freedom." But what exactly does it mean? In simplest terms, degrees of freedom (often denoted as df) refer to the number of independent values that are allowed to vary in an analysis without breaking any constraints. This concept is critical in statistical contexts, especially when estimating population parameters.

For example, if you have a sample of data points, the degrees of freedom can be thought of as the number of values that can change while still computing a particular statistic, like the mean, accurately. In the context of t-distributions, a higher degree of freedom means the distribution will be closer to a standard normal distribution with a bell-shaped curve.
  • Small degrees of freedom result in a more spread out, or 'heavy-tailed', distribution.
  • As the degrees of freedom increase, the t-distribution increasingly resembles a normal distribution.
Understanding degrees of freedom helps in interpreting t-scores and deciding whether they significantly differ from standard normal z-scores, which would inform hypotheses testing or confidence interval estimations.
Percentile
A percentile is a measure that indicates the value below which a given percentage of observations within a dataset fall. For instance, if you are in the 90th percentile in an exam, it means you scored better than 90% of the participants.

In statistical distributions, percentiles are used to communicate how data points relate to the entire set. They are widely used in various contexts, such as education, to rank test scores, or in health statistics, to determine body measurements relative to a population.
  • Percentiles can tell you where a specific score stands in comparison to others.
  • Understanding this measure in the distribution provides insights into data spread and helps identify extremities, such as outliers.
The exercise highlights the 90th percentile, meaning for the t-distribution at a certain degree of freedom, 90% of the data lies below this value, which links directly to the probability concept.
Quantiles
Closely related to percentiles are quantiles, which divide your data into interchangeable parts. Quantiles are points taken at regular intervals from the cumulative distribution function of a random variable, effectively splitting the range of a probability distribution into continuous intervals with equal probabilities.

One of the most common quantiles is the quartile, dividing data into four equal parts, but there are others like deciles (ten parts) or percentiles (hundred parts). The quantile corresponding to a particular percentile provides the cutoff value of that percentile. In the exercise above, when looking at t- or z-distributions, quantiles help us find these cutoff values for certain probabilities in hypothesis testing and confidence interval estimation.
  • Quantiles give a statistical breakdown of a dataset.
  • They aid in understanding data dispersion and inequality.
Quantiles are integral in analyzing how your data or test statistic compares over a distribution and effectively communicate probability concepts between observed and theoretical data.
Standard Normal Distribution
The standard normal distribution is a very special and well-known type of normal distribution. It is a continuous probability distribution that is symmetric around its mean, with a mean (\(\mu\)) of 0 and a standard deviation (\(\sigma\)) of 1.

This distribution plays a crucial role in statistics, primarily because it provides a baseline for z-scores, which are used to identify how far away a particular data point is from the mean in terms of standard deviations.
  • It informs much of inferential statistics, as many statistic tests (like z-tests) codify assumptions based on this distribution.
  • "Z-scores" help standardize different datasets to aid in comparison.
Comparing t-distributions to standard normal distributions highlights key differences, chiefly how t-distributions are 'heavier' in the tails, affecting probabilities especially when degrees of freedom are low. As we learned in the exercise, as the degrees of freedom increase, the t-distribution approximates the standard normal distribution more closely.

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Most popular questions from this chapter

The quality of computer disks is measured by the number of missing pulses. Brand X is such that \(80 \%\) of the disks have no missing pulses. If 100 disks of brand \(X\) are inspected, what is the probability that 15 or more contain missing pulses?

Shear strength measurements for spot welds have been found to have standard deviation 10 pounds per square inch (psi). If 100 test welds are to be measured, what is the approximate probability that the sample mean will be within 1 psi of the true population mean?

Access the applet Normal Approximation to Binomial Distribution (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html). When the applet is started, it displays the details in Example 7.11 and Figure 7.9 . Initially, the display contains only the binomial histogram and the exact value (calculated using the binomial probability function) for \(p(8)=P(Y=8) .\) Scroll down a little and click the button "Toggle Normal Approximation" to overlay the normal density with mean 10 and standard deviation \(\sqrt{6}=2.449\) the same mean and standard deviation as the binomial random variable \(Y\). You will get a graph superior to the one in Figure \(7.9 .\) a. How many probability mass or density functions are displayed? b. Enter 0 in the box labeled "Begin" and press the enter key. What probabilities do you obtain? c. Refer to part (b). On the line where the approximating normal probability is displayed, you see the expression Normal: $$ P(-0.5<=k<=8.5)=0.2701 $$ Why are the .5s in this expression?

Suppose that \(Y\) has a binomial distribution with \(n=5\) and \(p=.10\) a. Use the Normal Approximation to Binomial Distribution applet to find exact and approximate values for \(P(Y \leq 1)\) b. The normal approximation is not particularly good. Why?

Suppose that \(Z\) has a standard normal distribution and that \(Y\) is an independent \(\chi^{2}\) -distributed random variable with \(\nu\) df. Then, according to Definition 7.2, $$T=\frac{Z}{\sqrt{Y / \nu}}$$ has a \(t\) distribution with \(\nu\) df. \(^{*}\) a. If \(Z\) has a standard normal distribution, give \(E(Z)\) and \(E\left(Z^{2}\right)\). [Hint: For any random variable, \(\left.E\left(Z^{2}\right)=V(Z)+(E(Z))^{2} \cdot\right]\) b. According to the result derived in Exercise \(4.112(\mathrm{a}),\) if \(Y\) has a \(\chi^{2}\) distribution with \(\nu\) df, then $$E\left(Y^{a}\right)=\frac{\Gamma([\nu / 2]+a)}{\Gamma(\nu / 2)} 2^{a}, \quad \text { if } \nu>-2 a$$. Use this result, the result from part (a), and the structure of \(T\) to show the following. [Hint: Recall the independence of \(Z \text { and } Y \)] i. \(E(T)=0,\) if \(\nu>1\) ii. \(V(T)=\nu /(\nu-2),\) if \(\nu>2\)

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