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Chapter 12: Problem 19

What percentage of the time will a variable that has a \(t\) distribution with the specified degrees of freedom fall in the indicated region? (Hint: See discussion on page 496 ) a. 10 df, between -1.81 and 1.81 b. 24 df, between -2.06 and 2.06 c. 24 df, outside the interval from -2.80 to 2.80 d. 10 df, to the left of -1.81

Short Answer

Expert verified
a. Approximately 90%, b. Approximately 95%, c. 1%, d. 5%

Step by step solution

01

Understanding the Question

Recognition of the problem as one being about the t-distribution is important. The t-distribution is used to calculate probabilities when the sample size is small (generally under 30). The exercise asks for the percentage of time a variable (for example, a test statistic) will fall within a certain range. This percentage can be found by finding the area under the t-distribution curve that corresponds to that range.
02

Using the t-Distribution Table

For each scenario, check the row corresponding to the given degrees of freedom on your t-Distribution table. For a and b, find the number in the row that is closest to 1.81 and 2.06 respectively, and read off the associated probability. This represents the percentage of the time that a randomly selected value will fall between -1.81 and 1.81 or -2.06 and 2.06. For c and d, since the interval is outside a range or to the left of a value, the probabilities need to be subtracted from 1.
03

Apply the Values

a. For 10 df and the value of 1.81, the percentage of time between -1.81 and 1.81 is almost 90%. \n\nb. For 24 df and the value of 2.06, the percentage of time between -2.06 and 2.06 is approximately 95%. \n\nc. For 24 df and the value of 2.80, the percentage of time outside of this interval is 1 - 0.99 = 0.01 or 1%. \n\nd. For 10 df and the value of -1.81, the percentage of the time to the left of -1.81 is 5%.

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Key Concepts

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

Degrees of Freedom
In the context of the t-distribution, the term "degrees of freedom" is a crucial concept to understand. Degrees of freedom (often denoted as df) are a measure of the number of independent pieces of information in a dataset that can vary when estimating statistical parameters.

For example, when calculating the standard deviation, once the mean is calculated, each additional data point contributes one less degree of freedom.
  • In a t-distribution, the degrees of freedom determine the shape of the distribution. More degrees of freedom make the distribution closer in shape to a normal distribution.
  • With fewer degrees of freedom, the tails of the distribution become heavier and the peak becomes lower.
Understanding degrees of freedom helps in reading t-distribution tables, as this value is directly linked to the statistical calculations you perform.
Probability
In statistics, probability is a numerical description of how likely an event is to occur or how likely it is that a proposition is true.

The t-distribution, like other probability distributions, helps in determining the likelihood or probability of a certain range of outcomes.
  • It allows statisticians to make inferences about populations based on sample data, especially when sample sizes are small and the population variance is unknown.
  • In our exercise, probabilities are determined by looking at the area under the curve of the relevant t-distribution, given specific degrees of freedom and t-scores such as -1.81, 1.81, etc.
The t-distribution probability calculations inform us about the percentage of times a test statistic falls within, between, or outside certain t-score values.
Area Under the Curve
The area under the curve (AUC) of a t-distribution plays a vital role in probability calculations. The AUC represents the probability that a random variable, following the t-distribution, falls within a specified range.

In simple terms, the area under the curve is how we quantify probabilities in graphical terms. For the t-distribution:
  • For a two-tailed test between two values (e.g., "between -1.81 and 1.81"), you'd calculate the area under the curve between these points to get the probability or percentage of occurrences.
  • This area can sometimes cover both tails of the distribution, especially for tests looking at extremes (e.g., "outside the interval from -2.80 to 2.80").
  • The total area always equals 1 (or 100%), and segments of this area equate to probabilities of different events occurring.
By understanding areas under the curve, you can grasp the graphical aspects of probability and how statistical tests generate their results.
T-Distribution Table
The t-distribution table is a critical tool in statistics for determining probabilities related to the t-distribution. This table helps find the probability for various values of the t-distribution given specific degrees of freedom.

Here鈥檚 how the table works:
  • Find the "degrees of freedom" that match your sample data. This will correspond to a specific row in the table.
  • Look for the t-score values (such as 1.81, 2.06, etc.) and find probabilities associated with these scores. Typically, these probabilities represent the "tail areas" for those critical values.
  • In two-tailed tests, you may need to look at both sides of the table to find the probability of a value falling within certain range limits.
The table thus serves as a quick reference to convert t-scores into probabilities or vice versa, allowing quick decision-making based on statistical data.

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Most popular questions from this chapter

Because of safety considerations, in May, \(2003,\) the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 pounds as a typical passenger weight (including carry-on luggage) in warm months and 185 pounds as a typical weight in cold months. The Alaska Journal of Commerce (May 25,2003\()\) reported that Frontier Airlines conducted a study to estimate mean passenger plus carry-on weights. They found an mean summer weight of 183 pounds and a winter mean of 190 pounds. Suppose that these estimates were based on random samples of 100 passengers and that the sample standard deviations were 20 pounds for the summer weights and 23 pounds for the winter weights. a. Construct and interpret a \(95 \%\) confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers. b. Construct and interpret a \(95 \%\) confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers. c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during this conversation (San Luis Obispo Telegram-Tribune, August 16, 1995). The mean number of lies per conversation was 0.5 . Suppose that the standard deviation (which was not reported) was 0.4 a. Suppose that this group of 77 is representative of the population of students at this university. Construct a \(95 \%\) confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include \(0 .\) Does this imply that all students lie to their mothers? Explain.

The report "Highest Paying Jobs for 2009-10 Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 was \(\$ 48,722\). Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3,300\). Do the sample data provide strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722 ?\) Test the relevant hypotheses using \(\alpha=0.05 .\)

An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 4 minutes. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage boys and that the sample standard deviation was 1.4 minutes. Is there convincing evidence that the average attention span for teenage boys is less than 5 minutes? Test the relevant hypotheses using \(\alpha=0.01\).

The authors of the paper "Driven to Distraction" (Psychological Science [2001]:462-466) describe a study to evaluate the effect of using a cell phone on reaction time. Subjects were asked to perform a simulated driving task while talking on a cell phone. While performing this task, occasional red and green lights flashed on the computer screen. If a green light flashed, subjects were to continue driving, but if a red light flashed, subjects were to brake as quickly as possible and the reaction time (in msec) was recorded. The following summary statistics were read from a graph that appeared in the paper: $$ n=48 \quad \bar{x}=530 \quad s=70 $$ Construct and interpret a \(95 \%\) confidence interval for \(\mu,\) the mean time to react to a red light while talking on a cell phone. What assumption must be made in order to generalize this confidence interval to the population of all drivers?
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