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Chapter 13: Problem 427

Find the probability that a t-distribution has a t score (a) greater than \(1.740\) when d.o.f. \(=17\) (b) less than \(-1.323\) when d.o.f. \(=21\).

Short Answer

Expert verified
The probability of a t-score greater than 1.740 with 17 degrees of freedom is 0.95, and the probability of a t-score less than -1.323 with 21 degrees of freedom is 0.10.

Step by step solution

01

(a) Finding probability for t-score greater than 1.740 with d.o.f. = 17

: 1. First, identify the degrees of freedom (d.o.f.), which is 17 in this case. 2. Using a t-distribution table, look for the row corresponding to the given d.o.f. 3. Locate the t-score (1.740) in this row. If the exact value is not available, choose the closest value. 4. Look at the column header for this t-score, which will provide you with the one-tailed probability. 5. Since we need the probability for the t-score being greater than 1.740, subtract this one-tailed probability from 1. Now using a t-distribution table, the closest t-score to 1.740 with 17 d.o.f. is 1.740 itself, which corresponds to a one-tailed probability of 0.05. Hence, the probability of a t-score greater than 1.740 is: \(1 - 0.05 = 0.95\)
02

(a) Result

: The probability of a t-score greater than 1.740 with 17 degrees of freedom is 0.95.
03

(b) Finding probability for t-score less than -1.323 with d.o.f. = 21

: 1. First, identify the degrees of freedom (d.o.f.), which is 21 in this case. 2. Using a t-distribution table, look for the row corresponding to the given d.o.f. 3. Locate the t-score (1.323) in this row. Note: we will be using the positive value of -1.323 as the t-distribution table has only positive t-scores, which doesn't affect our results because t-distribution is symmetric. 4. If the exact value is not available, choose the closest value. 5. Look at the column header for this t-score, which will provide you with the one-tailed probability. 6. Since we need the probability for the t-score being less than -1.323, the one-tailed probability is the value we need. From the t-distribution table, the closest t-score to 1.323 with 21 d.o.f. is 1.321, which corresponds to a one-tailed probability of 0.10.
04

(b) Result

: The probability of a t-score less than -1.323 with 21 degrees of freedom is 0.10.

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Key Concepts

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

Understanding Degrees of Freedom
When dealing with t-distributions, a critical concept to understand is "degrees of freedom" (d.o.f.). In essence, degrees of freedom refer to the number of independent values or quantities which can vary in a statistical calculation.

Degrees of freedom impact the shape of the t-distribution:
  • More degrees of freedom make the t-distribution closer to a normal distribution.
  • Fewer degrees of freedom result in a distribution with fatter tails, indicating more variability.
In our exercise, the degrees of freedom are 17 and 21 for the two scenarios, affecting how closely the distribution of values resembles a normal curve. Understanding this helps you gauge how likely a particular t-score is under these different conditions.
Calculating the t-score
A t-score is a standardized value that indicates how far a specific value deviates from the mean in terms of the standard error. It is used to determine the position of a sample mean in a t-distribution. The formula for a t-score generally looks like this:

\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

Where:
  • \( \bar{x} \) is the sample mean.
  • \( \mu \) is the population mean.
  • \( s \) is the sample standard deviation.
  • \( n \) is the sample size.
When using a t-distribution table, you use these scores to find the probability that a statistic is observed to a certain degree of extremity, given the degrees of freedom. For example, in our exercise, we are interested in a t-score greater than 1.740 and another less than -1.323, which we find by referencing this theoretical distribution.
Probability Calculation in t-Distribution
Finding probability in a t-distribution involves using tables or technology to determine the likelihood of obtaining a t-score more extreme than your observed value. Let's break down the steps through examples from the exercise:

  • Identify your degrees of freedom, as it determines the row in your t-distribution table. For instance, 17 and 21 are used in our exercise.
  • Compare your t-score against the value in the table. If needed, find the closest t-score available in the table for each of the degrees of freedom.
  • Look at the associated probability in the table. This probability represents the t-score's extremeness (either greater or less depending on the direction you are checking).
In our exercise:
  • For a score greater than 1.740, with a 17 d.o.f, the probability suggests the likelihood is 0.05, so we subtract from 1, yielding 0.95.
  • For a score less than -1.323, with a 21 d.o.f, the probability is found as 0.10 directly from the table.
These calculations help assess how unusual or rare a specific observation is, guiding decisions about statistical significance.

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Most popular questions from this chapter

Suppose that light bulbs made by a standard process have an average life of 2000 hours, with a standard deviation of 250 hours, and suppose that it is worthwhile to change the process if the mean life can be increased by at least 10 percent. An engineer wishes to test a proposed new process, and he is willing to assume that the standard deviation of the distribution of lives is about the same as for the standard process. How large a sample should he examine if he wishes the probability to be about \(.01\) that he will fail to adopt the new process if in fact it produces bulbs with a mean life of 2250 hours?

How large a sample must be taken in order that you are 99 percent certain that \(\underline{X}_{n}\) is within \(.5 \sigma\) of \(\mu\) ?

The chi-square density function is the special case of a gamma density with parameters \(\alpha=\mathrm{K} / 2\) and \(\lambda=1 / 2\). Find the mean, variance and moment-generating function of a chi-square random variable.

Let \(U\) be a chi-square random variable with \(\mathrm{m}\) degrees of freedom. Let \(\mathrm{V}\) be a chi-square random variable with \(\mathrm{n}\) degrees of freedom. Consider the quantity \(\mathrm{X}=[(\mathrm{U} / \mathrm{m}) /(\mathrm{V} / \mathrm{n})]\), where \(\mathrm{U}\) and \(\mathrm{V}\) are independent. Using the change of variable technique, find the probability density of \(\mathrm{X}\).

Briefly discuss the Central Limit Theorem.
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