/*! 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 2 Say as much as you can about the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 2

Say as much as you can about the \(P\)-value for an upper-tailed chi-squared test in each of the following situations: a. \(x^{2}=7.5\), df \(=2\) b. \(x^{2}=13.0\), df \(=6\) c. \(X^{2}=18.0, \mathrm{df}=9\) d. \(\chi^{2}=21.3\), df \(=5\) e. \(x^{2}=5.0, k=4\)

Short Answer

Expert verified
a: P 鈮� 0.023, b: P 鈮� 0.042, c: P 鈮� 0.035, d: P 鈮� 0.0007, e: P 鈮� 0.287.

Step by step solution

01

Understanding Chi-Squared Distribution

The chi-squared distribution is used in hypothesis testing, specifically for categorical data. It is defined by the degrees of freedom (df), which is determined by the number of categories minus one. In an upper-tailed test, we are looking to see if our test statistic falls in the tail of the distribution, indicating it is not aligned with the null hypothesis.
02

Using the Chi-Squared Test Statistic Formula

For each given test statistic value and degrees of freedom, we can use a chi-squared distribution table or software to estimate the P-value. The P-value represents the probability that a chi-squared test statistic as extreme as the one obtained would occur by random chance under the null hypothesis.
03

Case a - Calculate P-Value for \(x^2 = 7.5\), df = 2

Using the chi-squared distribution table or software, look up the value corresponding to the test statistic 7.5 and df = 2. The P-value is the area to the right of 7.5 in this distribution. After computation, we find \( P(x^2 > 7.5) \approx 0.023 \).
04

Case b - Calculate P-Value for \(x^2 = 13.0\), df = 6

Look up the value for x虏 = 13.0 with df = 6 in the chi-squared distribution. The area to the right gives us \( P(x^2 > 13.0) \approx 0.042 \).
05

Case c - Calculate P-Value for \(x^2 = 18.0\), df = 9

For x虏 = 18.0 and df = 9, we check the chi-squared table. The P-value is \( P(x^2 > 18.0) \approx 0.035 \).
06

Case d - Calculate P-Value for \(x^2 = 21.3\), df = 5

Use the table to find the area to the right of 21.3 with df = 5. This results in a P-value \( P(x^2 > 21.3) \approx 0.0007 \).
07

Case e - Calculate P-Value for \(x^2 = 5.0\), df = 4

Check the chi-squared distribution for x虏 = 5.0 and df = 4. The corresponding P-value is \( P(x^2 > 5.0) \approx 0.287 \).

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 91影视!

Key Concepts

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

P-Value
The P-value is a crucial component in statistical hypothesis testing. It represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. In the context of a chi-squared test, a chi-squared distribution table or statistical software can be used to find the P-value associated with a given test statistic and degrees of freedom.

Here鈥檚 a breakdown of what a P-value can reveal in this context:
  • If the P-value is small, typically less than 0.05 or 0.01, it suggests the observed data are less consistent with the null hypothesis. This often leads to the conclusion that there is enough evidence to reject the null hypothesis.
  • A larger P-value indicates that the observed data are more in line with what one would expect under the null hypothesis, leading to the acceptance or retention of the null hypothesis.
This measure is vital for assessing the reliability of results in the chi-squared test by quantifying the evidential weight against the null hypothesis.
Degrees of Freedom
Degrees of Freedom (df) determine the shape of the chi-squared distribution. It is directly influenced by the number of categories or variables being analyzed minus any constraints or parameters set in the analysis.

For example, in a table with observed and expected frequencies:
  • If you have data spread across five categories, the degrees of freedom would typically be 4, because df is calculated as the number of categories minus one.
The degrees of freedom are essential because they directly impact the critical value of the chi-squared statistic needed to evaluate the null hypothesis.

To visualize, as the degrees of freedom increase, the chi-squared distribution shifts, becoming closer to a normal distribution. Understanding this helps in interpreting test outcomes as df changes the threshold for statistical significance.
Upper-Tailed Test
An upper-tailed test in the context of a chi-squared test focuses on the extreme end or the right tail of the distribution. This type of test is used to determine whether there is significant evidence to suggest that the null hypothesis should be rejected.

Here鈥檚 how it works:
  • You calculate the chi-squared test statistic from your data.
  • Compare this statistic against the critical value from the chi-squared distribution corresponding to your chosen significance level and degrees of freedom.
  • If the test statistic exceeds the critical threshold, the test is significant in the upper tail, implying the null hypothesis can be rejected.
Upper-tailed tests are particularly useful in scenarios where you're testing if your observed data shows more variability than expected or if one category is more pronounced than what would be anticipated under the null hypothesis.

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

Each headlight on an automobile undergoing an annual vehicle inspection can be focused either too high \((H)\), too low \((L)\), or properly \((N)\). Checking the two headlights simultaneously (and not distinguishing between left and right) results in the six possible outcomes \(H H, L L, N N\), \(H L, H N\), and \(L N\). If the probabilities (population proportions) for the single headlight focus direction are \(P(H)=\theta_{1}, P(L)=\theta_{2}\), and \(P(N)=1-\theta_{1}-\theta_{2}\) and the two headlights are focused independently of one another, the probabilities of the six outcomes for a randomly selected car are the following: $$ \begin{aligned} &p_{1}=\theta_{1}^{2} \quad p_{2}=\theta_{2}^{2} \quad p_{3}=\left(1-\theta_{1}-\theta_{2}\right)^{2} \\ &p_{4}=2 \theta_{1} \theta_{2} \quad p_{5}=2 \theta_{1}\left(1-\theta_{1}-\theta_{2}\right) \\ &p_{6}=2 \theta_{2}\left(1-\theta_{1}-\theta_{2}\right) \end{aligned} $$ Use the accompanying data to test the null hypothesis $$ H_{0}: p_{1}=\pi_{1}\left(\theta_{1}, \theta_{2}\right), \ldots, p_{6}=\pi_{6}\left(\theta_{1}, \theta_{2}\right) $$ where the \(\pi_{1}\left(\theta_{1}, \theta_{2}\right)\) s are given previously. $$ \begin{array}{l|cccccc} \text { Outcome } & H H & L L & N N & H L & H N & L N \\ \hline \text { Frequency } & 49 & 26 & 14 & 20 & 53 & 38 \end{array} $$ [Hint: Write the likelihood as a function of \(\theta_{1}\) and \(\theta_{2}\), take the natural \(\log\), then compute \(\partial / \partial \theta_{1}\) and \(\partial / \partial \theta_{2}\), equate them to 0 , and solve for \(\hat{\theta}_{1}, \hat{\theta}_{2}\).

Qualifications of male and female head and assistant college athletic coaches were compared in the article "Sex Bias and the Validity of Believed Differences Between Male and Female Interscholastic Athletic Coaches" (Research Quarterly for Exercise and Sport, 1990: 259-267). Each person in random samples of 2225 male coaches and 1141 female coaches was classified according to number of years of coaching experience to obtain the accompanying twoway table. Is there enough evidence to conclude that the proportions falling into the experience categories are different for men and women? Use \(\alpha=.01\).

Suppose that in a particular state consisting of four distinct regions, a random sample of \(n_{k}\) voters is obtained from the \(k\) th region for \(k=1,2,3,4\). Each voter is then classified according to which candidate \((1,2\), or 3 ) he or she prefers and according to voter registration \((1=\) Dem., \(2=\) Rep., 3 = Indep.). Let \(p_{\text {igk }}\) denote the proportion of voters in region \(k\) who belong in candidate category \(i\) and registration category \(j\). The null hypothesis of homogeneous regions is \(H_{0}: p_{i j 1}=p_{i j 2}=p_{i j 3}=p_{i j 4}\) for all \(i, j\) (i.e., the proportion within each candidate/registration combination is the same for all four regions). Assuming that \(H_{0}\) is true, determine \(\hat{p}_{i j k}\) and \(\hat{e}_{i j k}\) as functions of the observed \(n_{i j k}\) 's, and use the general rule of thumb to obtain the number of degrees of freedom for the chi-squared test.

An information-retrieval system has ten storage locations. Information has been stored with the expectation that the long-run proportion of requests for location \(t\) is given by \(p_{i}=(5.5-|i-5.5|) / 30\). A sample of 200 retrieval requests gave the following frequencies for locations \(1-10\), respectively: \(4,15,23,25,38,31,32,14,10\), and 8 . Use a chi-squared test at significance level . 10 to decide whether the data is consistent with the a priori proportions (use the P-value approach).

Show that the chi-squared statistic for the test of independence can be written in the form $$ \chi^{2}=\sum_{i=1}^{l} \sum_{j=1}^{J}\left(\frac{N_{i j}^{2}}{\dot{E}_{i j}}\right)-n $$ Why is this formula more efficient computationally than the defining formula for \(\chi^{2}\) ?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation
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.