/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Find the critical value(s) and r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the critical value(s) and rejection region(s) for the type of t-test with level of significance \(\alpha\) and sample size \(n\). Two-tailed test, \(\alpha=0.05, n=18\)

Short Answer

Expert verified
Critical values are \(t=\pm2.110\). Reject the null if \(t<-2.110\) or \(t>2.110\).

Step by step solution

01

Understanding the Problem

We need to determine the critical values and rejection regions for a two-tailed t-test with \(n=18\) and \(\alpha=0.05\). This requires finding the t-values on both tails of the t-distribution since it's a two-tailed test.
02

Calculating Degrees of Freedom

In a t-test, the degrees of freedom (df) is calculated as \(n-1\). For \(n=18\), the degrees of freedom is \(18-1 = 17\). This tells us which t-distribution to reference.
03

Finding Critical Values

For a two-tailed test and \(\alpha=0.05\), we split \(0.05\) into two tails: \(0.025\) in each tail. Using a t-distribution table or calculator with \(df=17\), we find the critical t-values corresponding to the tails. These values are approximately \(t=\pm2.110\).
04

Defining Rejection Regions

The rejection region for a two-tailed test at \(\alpha=0.05\) and \(df=17\) are all t-values less than \(-2.110\) or greater than \(2.110\). These are the ranges where we would reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Critical Values in T-Tests
In the realm of statistical testing, critical values play a significant role. They are the thresholds that indicate the boundary between region where the null hypothesis is retained and rejected. In a two-tailed t-test, as in our example, we are particularly interested in two critical values - one for each tail.

The critical values are determined by the chosen significance level, denoted as \(\alpha\). For a significance level of \(\alpha=0.05\) in a two-tailed test, we split this into two equal parts for each tail, resulting in 0.025 in each tail. Using a t-distribution table or appropriate software, and knowing the degrees of freedom, we find the approximate critical values. For our case with \(df=17\), these values are \(t=\pm2.110\).

Accurately identifying critical values is essential because it helps define the rejection regions, the areas where the null hypothesis is not sustainable. Without these critical markers, one can't systematically decide to reject or fail to reject the null hypothesis.
Recognizing Rejection Regions
Rejection regions are crucial for hypothesis testing, as they delineate where we consider the null hypothesis to be unlikely enough to reject it. In our two-tailed t-test example, the rejection region is composed of two distinct parts on the t-distribution graph.

Simply put, these regions are located at the far-left and far-right extremes of the distribution. For our test, much like the critical values dictate, they begin at \(t < -2.110\) and \(t > 2.110\) for \(df=17\). If the test statistic falls into these extremes, it suggests stronger evidence against the null hypothesis, leading us to reject it.

These rejection regions ensure that only the most extreme and unlikely sample results lead to rejection of the null hypothesis. It prevents us from hastily dismissing the null hypothesis based on a sample result that isn't sufficiently rare under the assumed null data distribution. This safeguarding against errors in decision-making underscores the practical utility of well-defined rejection regions.
Calculating Degrees of Freedom
Degrees of freedom (df) are a fundamental concept in statistical tests, and they are essentially the number of values in the final calculation of a statistic that are free to vary. In a t-test, degrees of freedom are calculated slightly differently depending on the context, but often it's the sample size minus one.

For our case, with \(n=18\), the degrees of freedom is \(17\) (calculated as \(n-1\)). This is significant because it impacts which t-distribution you should use to find critical values. The shape of the t-distribution becomes more like the normal distribution as the degrees of freedom increase, and thus the critical t-values change.

Understanding degrees of freedom helps in interpreting the spread and variability in the data and ensures the correct application of the t-test. In hypothesis testing, selecting the proper degrees of freedom allows for more accurate determination of critical values and rejection regions, ultimately leading to more reliable test results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Test the claim about the population variance \(\sigma^{2}\) or standard deviation \(\sigma\) at the level of significance \(\alpha\). Assume the population is normally distributed. Claim: \(\sigma^{2} \leq 60 ; \alpha=0.025\). Sample statistics: \(s^{2}=72.7, n=15\)

(a) identify the claim and state \(H_{0}\) and \(H_{a}\), (b) use technology to find the \(P\)-value, (c) decide whether to reject or fail to reject the null hypothesis, and (d) interpret the decision in the context of the original claim. Assume the population is normally distributed. An education publication claims that the mean score for grade 12 students on a science achievement test is more than 145 . You want to test this claim. You randomly select 36 grade 12 test scores. The results are listed below. At \(\alpha=0.1\), can you support the publication's claim? (Adapted from National Center for Education Statistics) $$\begin{array}{rrrrrrrrrrrr}188 & 80 & 175 & 195 & 201 & 143 & 119 & 81 & 118 & 119 & 165 & 222 \\ 109 & 134 & 200 & 110 & 199 & 181 & 79 & 135 & 124 & 205 & 90 & 120 \\ 216 & 167 & 198 & 183 & 173 & 187 & 143 & 166 & 147 & 219 & 206 & 97\end{array}$$

(a) identify the claim and state \(H_{0}\) and \(H_{a}\), (b) use technology to find the \(P\)-value, (c) decide whether to reject or fail to reject the null hypothesis, and (d) interpret the decision in the context of the original claim. Assume the population is normally distributed. An education researcher claims that the overall average score of 15 -year-old students on an international mathematics literacy test is 494. You want to test this claim. You randomly select the average scores of 33 countries. The results are listed below. At \(\alpha=0.05\), do you have enough evidence to reject the researcher's claim? (Source: National Center for Education Statistics) $$\begin{array}{lllllllllll}561 & 554 & 536 & 531 & 523 & 518 & 515 & 511 & 506 & 500 & 499 \\ 493 & 490 & 489 & 485 & 482 & 482 & 479 & 477 & 466 & 453 & 448 \\ 439 & 432 & 423 & 421 & 413 & 407 & 394 & 388 & 386 & 376 & 368\end{array}$$

(a) identify the claim and state \(H_{0}\) and \(H_{a}\), (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic \(z\), (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. An environmental researcher claims that the mean amount of sulfur dioxide in the air in U.S. cities is \(1.15\) parts per billion. In a random sample of 134 U.S. cities, the mean amount of sulfur dioxide in the air is \(0.93\) parts per billion. Assume the population standard deviation is \(2.62\) parts per billion. At \(\alpha=0.01\), is there enough evidence to reject the claim? (Source: U.S. Environmental Protection Agency)

Test the claim about the population mean \(\mu\) at the level of significance \(\alpha\). Assume the population is normally distributed. Claim: \(\mu<8.25 ; \alpha=0.01 ; \sigma=0.017\). Sample statistics: \(\bar{x}=8.246, n=40\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.