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Chapter 11: Problem 58

Which of the following explains why one of the conditions for performing the chi-square test is met in this case? (a) The sample is large, 4877 teenagers in all. (b) The sample is random. (c) All the observed counts are greater than 100. (d) We used software to do the calculations. (e) Both variables are categorical.

Short Answer

Expert verified
(c) All the observed counts are greater than 100.

Step by step solution

01

Understanding the Chi-Square Test

The chi-square test is used to test the relationship between categorical variables. One of the conditions that must be met for the test to be valid is that the sample must contain expected counts that are sufficiently large to ensure the accuracy of the chi-square approximation.
02

Identify Which Conditions Apply

Review the given options to determine which explains why the chi-square test can be performed: (a) Sample size is large, (b) Sample is random, (c) All observed counts > 100, (d) Software used for calculations, (e) Variables are categorical.
03

Assessing the Options

Analyze each option in terms of conditions required for the chi-square test: (a) A large sample size ensures the expected count requirement is satisfied even if they were not specified. (b) Random sampling ensures the results are representative but is not directly a requirement. (c) Observed counts being > 100 suggests expected counts are likely large enough. (d) Using software is irrelevant to meeting conditions. (e) Both variables being categorical is necessary for the test.
04

Choosing the Correct Explanation

The key condition related to the chi-square test validity concerned with expected counts is best met by option (c), which ensures sufficiency of expected counts as they are derived from observed counts being naturally high.

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Key Concepts

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

Categorical Variables
In a chi-square test, one of the most fundamental requirements is the presence of categorical variables. This means that the variables being studied can be divided into distinct groups or categories. For instance, species of plants, types of cuisine, yes/no responses, and colors are examples of categories.
These variables are not numerical per se and are used to categorize data based on characteristics.
  • Categorical variables make it possible to classify observations into different groups.
  • The chi-square test is specifically designed to analyze frequencies of categorical data.
  • Each category's frequency or count forms the basis of the test.
Without categorical variables, the chi-square test cannot be properly applied as it depends on comparing the distribution of counts across these groups.
Expected Counts
Expected counts are the theoretical frequencies we expect in each category if there is no association between the variables.
They are derived based on the assumption of independence and provide a benchmark for comparing observed data.
In the chi-square test, it's important that these expected counts are large enough to ensure the test's validity.
  • Expected counts, not to be confused with observed counts, help identify deviations caused by random error or actual effect.
  • If expected counts are too low, it may affect the reliability of the chi-square approximation.
  • The rule of thumb is that each expected cell frequency is at least 5.
High expected counts demonstrate that observed differences are more likely due to actual differences rather than random chance.
Statistical Conditions
Several statistical conditions must be met when conducting a chi-square test to ensure the results are valid and reliable.
These conditions involve aspects like sample selection and count size.
Meeting these statistical prerequisites is crucial in transforming raw data into meaningful conclusions. To perform a valid chi-square test, consider the following conditions:
  • Categorical Variables: Both examined variables must be categorical.
  • Random Sampling: Although not a direct condition, random sampling helps reduce biases, thus promoting representativeness.
  • Expected Count Size: Each expected frequency should ideally be 5 or more.
Not meeting these conditions may lead to erroneous conclusions or render the test invalid.
Large Sample Size
Having a large sample size is often advisable in statistical analyses, particularly in chi-square tests. A large sample size can help ensure that expected counts, and subsequently the test results, are accurate and reliable.
  • Larger samples usually bring about larger expected counts due to more varied data spread.
  • This adequacy satisfies the chi-square test's needs for robustness and reliability.
  • While smaller samples can sometimes suffice, larger sizes help minimize errors related to sampling variability.
Large sample sizes inherently help fulfill certain chi-square test conditions, like sufficiently large expected counts, thereby amplifying the validity of conclusions drawn from the test.

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Most popular questions from this chapter

Birds in the trees Researchers studied the behavior of birds that were searching for seeds and insects in an Oregon forest. In this forest, 54% of the trees were Douglas firs, 40% were ponderosa pines, and 6% were other types of trees. At a randomly selected time during the day, the researchers observed 156 red-breasted nuthatches: 70 were seen in Douglas firs, 79 in ponderosa pines, and 7 in other types of trees.2 Do these data suggest that nuthatches prefer particular types of trees when they鈥檙e searching for seeds and insects? Carry out a chi-square goodness-of-fit test to help answer this question.

The chi-square statistic is (a) \(\frac{(18-25)^{2}}{25}+\frac{(22-25)^{2}}{25}+\frac{(39-25)^{2}}{25}+\frac{(21-25)^{2}}{25}\) (b) \(\frac{(25-18)^{2}}{18}+\frac{(25-22)^{2}}{22}+\frac{(25-39)^{2}}{39}+\frac{(25-21)^{2}}{21}\) (c) \(\frac{(18-25)}{25}+\frac{(22-25)}{25}+\frac{(39-25)}{25}+\frac{(21-25)}{25}\) (d) \(\frac{(18-25)^{2}}{100}+\frac{(22-25)^{2}}{100}+\frac{(39-25)^{2}}{100}+\frac{(21-25)^{2}}{100}\) (e) \(\frac{(0.18-0.25)^{2}}{0.25}+\frac{(0.22-0.25)^{2}}{0.25}+\frac{(0.39-0.25)^{2}}{0.25}\) \(+\frac{(0.21-0.25)^{2}}{0.25}\)

The P-value for a chi-square goodness-of-fit test is 0.0129. The correct conclusion is (a) reject H0 at A 0.05; there is strong evidence that the trees are randomly distributed. (b) reject H0 at A 0.05; there is not strong evidence that the trees are randomly distributed. (c) reject H0 at A 0.05; there is strong evidence that the trees are not randomly distributed. (d) fail to reject H0 at A 0.05; there is not strong evidence that the trees are randomly distributed. (e) fail to reject H0 at A 0.05; there is strong evidence that the trees are randomly distributed.

Roulette Casinos are required to verify that their games operate as advertised. American roulette wheels have 38 slots鈥�18 red, 18 black, and 2 green. In one casino, managers record data from a random sample of 200 spins of one of their American roulette wheels. The one-way table below displays the results. \(\begin{array}{llll}{\text { Color: }} & {\text { Red }} & {\text { Black }} & {\text { Green }} \\ \hline \text { Count: } & {85} & {99} & {16} \\\ \hline\end{array}\) (a) State appropriate hypotheses for testing whether these data give convincing evidence that the distribution of outcomes on this wheel is not what it should be. (b) Calculate the expected counts for each color. Show your work.

Opinions about the death penalty The General Social Survey asked a random sample of adults, 鈥淒o you favor or oppose the death penalty for persons convicted of murder?鈥� The following table gives the responses of people whose highest education was a high school degree and of people with a bachelor鈥檚 degree: \(\begin{array}{lll}{\text { High school }} & {1010} & {369} \\ {\text { Bachelor's }} & {319} & {185} \\ \hline\end{array}\) We can test the hypothesis of 鈥渘o difference鈥� in support for the death penalty among people in these educational categories in two ways: with a two-sample z test or with a chi-square test. (a) Minitab output for a chi-square test is shown below. State appropriate hypotheses and interpret the P-value in context. What conclusion would you draw? (b) Minitab output for a two-sample z test is shown below. Explain how these results are consistent with the test in part (a).
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