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You are testing \(H_{0}: \mu=100\) against \(H_{a}: \mu>100\) based on an SRS of 25 observations from a Normal population. The \(t\) statistic is \(t=2.10\). The degrees of freedom for the \(t\) statistic are a. 24 b. 25 c. 26 .

Short Answer

Expert verified
The degrees of freedom for the t statistic is 24 (option a).

Step by step solution

01

Understand the Problem

We need to determine the number of degrees of freedom for the given t-statistic when testing the hypothesis: \(H_0: \mu = 100\) against \(H_a: \mu > 100\). We are given a sample size of 25 observations.
02

Define Degrees of Freedom for t-test

In statistical t-tests, the degrees of freedom ( extit{df}) are calculated as \( n - 1 \), where \( n \) is the sample size. This is a foundational concept for calculating the extit{df} for a t-statistic.
03

Calculate the Degrees of Freedom

Substitute the sample size into the formula to calculate the degrees of freedom. With a sample size of \( n = 25 \), the degrees of freedom is \( 25 - 1 = 24 \).
04

Analyze the Options

Review the given choices: a. 24, b. 25, c. 26. The calculated degrees of freedom, 24, corresponds to option A.

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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 statistical calculations, degrees of freedom (often denoted as df) is a concept that can initially seem abstract, but it is central to understanding the structure of statistical tests such as the t-test. Simply put, degrees of freedom refer to the number of independent values that can vary in an analysis without breaking any constraints. To put it in simpler terms, when calculating a statistic, such as a mean or variance, some quantities have already been used to estimate parameters of our model, thus reducing the freedom of variation we remain with in our data. For a t-test, which is the test used in many hypothesis testing scenarios, the degrees of freedom are usually calculated using the formula: \( n - 1 \), where \( n \) is the sample size. In this scenario, it represents the number of observations that are "free" to vary once a mean is computed, as the mean itself constrains the values. For example, with a sample size of 25, as in the exercise, the degrees of freedom would be 25 minus 1, equaling 24.
Normal Distribution
A Normal distribution is a continuous probability distribution that is symmetrical around its mean, depicting a bell-shaped curve. A key feature of a Normal distribution is its consistent appearance, no matter the data's mean or standard deviation. This consistency is what makes it pivotal in statistics. The Normal distribution, often referred to as the Gaussian distribution, plays a significant role in hypothesis testing because many statistical tests, like the t-test, assume that the data follows a normal distribution.
  • Symmetry around the mean
  • Predictable proportions of data within standard deviation ranges
  • Foundation for advanced statistical techniques
When working with normally distributed data, around 68% of the data will fall within one standard deviation from the mean, 95% within two, and about 99.7% within three standard deviations. This property of normal distribution allows statisticians to make inferences about a population based solely on sample data.
Hypothesis Testing
Hypothesis testing is an essential part of statistical analysis, facilitating the testing of an assumption about a population parameter. In hypothesis testing, two hypotheses are set up: the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). The goal is to determine whether there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.In the context of our exercise, the null hypothesis \( H_0: \mu = 100 \) implies that the mean of the population is assumed to be 100. Conversely, the alternative hypothesis \( H_a: \mu > 100 \) suggests that we believe the mean is greater than 100. The process includes:
  • Formulating the null and alternative hypotheses
  • Choosing a significance level, often denoted by \( \alpha \), commonly set at 0.05
  • Calculating a test statistic, like a t-statistic, to compare against a critical value or to compute a p-value
  • Determining whether the observed data fall in the rejection region of the null hypothesis
If the test statistic exceeds a critical value or the p-value is less than the chosen significance level, we reject \( H_0 \), suggesting that the sample provides enough evidence to support \( H_a \). This is a powerful tool for making educated guesses about broader trends within data sourced from complex real-world conditions.

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Most popular questions from this chapter

Kicking a Helium-Filled Foot ball. Does a football filled with helium travel farther than one filled with ordinary air? To test this, the Columbus Dispatch conducted a study. Two identical footballs, one filled with helium and one filled with ordinary air, were used. A casual observer was unable to detect a difference in the two footballs. A novice kicker was used to punt the footballs. A trial consisted of kicking both footballs in a random order. The kicker did not know which football (the helium-filled or the air-filled football) he was kicking. The distance of each punt was recorded. Then another trial was conducted. A total of 39 trials were run. Here are the data for the 39 trials, in yards that the footballs traveled. The difference (helium minus air) is the response variable. a. Examine the data. Is it reasonable to use the \(t\) procedures? b. If your conclusion in part (a) is Yes, do the data give convincing evidence that the heliumfilled football travels farther than the air-filled football?

Mathematics Scores in Dallas. The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a mathematics test scored from 0 to 500 . A score of 262 is a "basic" mathematics level, and a score of 299 is "proficient." Scores for a random sample of 1100 eighth-graders in Dallas had \(x=264\) with standard error 1.3. 16 a. We don't have the 1100 individual scores, but use of the \(t\) procedures is surely safe. Why? b. Give a \(99 \%\) confidence interval for the mean score of all Dallas eighth- graders. (Be careful: the report gives the standard error of \(x\), not the standard deviation s.) c. Based on your answer in (b), is there good evidence that the mean for all Dallas eighthgraders is different from the basic level? (Just as for z procedures, a level C \(t\) confidence interval for a mean \(\mu\) can be used to test $$ \begin{aligned} &H_{0}: \mu=\mu_{0} \\ &H_{a}: \mu \neq \mu_{0} \end{aligned} $$ at level \(1-C\) by checking whether \(\mu_{0}\) is in the interval.)

Which of these settings does not allow use of a matched pairs \(t\) procedure? a. You interview both the instructor and one of the students in each of 20 introductory statistics classes and ask each how many hours per week homework assignments require. b. You interview a sample of 15 instructors and another sample of 15 students and ask each how many hours per week homework assignments require. c. You interview 40 students in the introductory statistics course at the beginning of the semester and again at the end of the semester and ask how many hours per week homework assignments require.

Because the \(t\) procedures are robust, the most important condition for their safe use is that a. the sample size is at least \(15 .\) b. the population distribution is exactly Normal. c. the data can be regarded as an SRS from the population.

Twenty-five adult citizens of the United States were asked to estimate the average income of all U.S. households. The mean estimate was \(x=\$ 70,000\) and \(s=\$ 15,000\). (Note: The actual average household income at the time of the study was about \(\$ 90,000\).) Assume the 25 adults in the study can be considered an SRS from the population of all adult citizens of the United States. A \(95 \%\) confidence interval for the mean estimate of the average income of all U.S. households is a. \(\$ 67,000\) to \(\$ 73,000\). b. \(\$ 63,808\) to \(\$ 76,192\). c. \(\$ 83,808\) to \(\$ 96,192\).

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