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Chapter 7: Problem 5

In Exercises 7.5 to 7.8 , the categories of a categorical variable are given along with the observed counts from a sample. The expected counts from a null hypothesis are given in parentheses. Compute the \(\chi^{2}\) -test statistic, and use the \(\chi^{2}\) -distribution to find the p-value of the test. $$ \begin{array}{lccc} \hline \text { Category } & \text { A } & \text { B } & \text { C } \\ \text { Observed } & 35(40) & 32(40) & 53(40) \\ \text { (Expected) } & & & \\ \hline \end{array} $$

Short Answer

Expert verified
This answer varies depending on the way the calculations from Step 2 and the p-value from Step 4 are done, but you should be able to calculate the \(\chi^{2}\) statistic correctly using the formula, then understand the process of finding a p-value using this \(\chi^{2}\) statistic and the degrees of freedom.

Step by step solution

01

Calculate the Chi-Square Statistic

Calculate the \(\chi^{2}\) statistic using the formula: \(\chi^{2}\) = \(\Sigma\)\(\frac{(O-E)^{2}}{E}\), where \(O\) is observed frequency and \(E\) is expected frequency. After substituting the given values, the individual components of the formula become: \((35-40)^{2}/40\), \((32-40)^{2}/40\), \((53-40)^{2}/40\)
02

Calculate the Sum

Add the individual components of the \(\chi^{2}\) statistic: \((35-40)^{2}/40 + (32-40)^{2}/40 + (53-40)^{2}/40\) to get the \(\chi^{2}\) statistic.
03

Find the Degrees of Freedom

The degrees of freedom for the \(\chi^{2}\)-test is the number of categories minus 1. This means there are \(3-1 = 2\) degree(s) of freedom.
04

Find the P-value

In order to find the p-value, utilize a Chi-square distribution table with the \(\chi^{2}\) statistic and degree(s) of freedom calculated in previous steps. Since tables vary, the p-value is often found by interpolating between the closest tabulated values or it can be found using statistical software. The closer to zero the p-value, the stronger the evidence against the null hypothesis.

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Key Concepts

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

Categorical Variable
In statistics, a categorical variable is a variable that can take on one of a limited, and usually fixed, number of possible values, assigning each individual or other unit of observation to a particular group or nominal category on the basis of some qualitative property.

For instance, in the given exercise, the categorical variable could be types of a product with categories A, B, and C. Each observed count represents the number in a sample that falls into the respective category. Understanding the nature of categorical variables is crucial when performing chi-square tests since these tests are specifically designed to analyze categorical data.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It's a statement that there is no effect or no difference, and it is the presumption that any kind of difference or significance you see in a set of data is due to chance.

In the context of a Chi-Square test, the null hypothesis typically states that the observed frequencies for a categorical variable match the expected frequencies. The expected frequencies are usually derived from a theoretical distribution or population. If the observed data diverge too much from what the null hypothesis predicts, you might have enough evidence to reject it in favor of the alternative hypothesis.
Chi-Square Distribution
The Chi-Square distribution is a theoretical distribution that is used in statistical hypothesis testing. It is applicable when you're looking at the distributions of categorical data and is a fundamental aspect of the Chi-Square test. When you compute a Chi-Square test statistic, it is compared against the Chi-Square distribution to determine the probability of observing your data if the null hypothesis is true.

The shape of the Chi-Square distribution depends on the degrees of freedom in your test. As the degrees of freedom increase, the distribution becomes more symmetrical. Understanding the distribution is key to interpreting the results of a Chi-Square test, as it allows us to determine the p-value associated with our test statistic.
P-value
The p-value is a fundamental concept in the field of statistical hypothesis testing. It鈥檚 the probability of observing a test statistic as extreme as, or more extreme than, the statistic you observed, assuming that the null hypothesis is true.

In the Chi-Square test context, a low p-value indicates that the observed data are unlikely under the null hypothesis. It is a tool for deciding whether to reject the null hypothesis. Common thresholds for the p-value are 0.05, 0.01, or 0.001, with a p-value below such thresholds suggesting that there is sufficient evidence to reject the null hypothesis in favor of the alternative.
Degrees of Freedom
The term degrees of freedom in statistics refers to the number of values in the final calculation of a statistic that are free to vary. When it comes to the Chi-Square test, the degrees of freedom are typically calculated as the number of categories minus one.

The degrees of freedom are critical in determining the correct Chi-Square distribution to use when finding the p-value. In our exercise example, there are three categories, so we have 2 degrees of freedom (\(3 - 1 = 2\)). This figure helps us decide which Chi-Square distribution to refer to when determining the significance of our Chi-Square test statistic.

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Most popular questions from this chapter

In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. $$ \begin{aligned} &\text { 7.4 } H_{0}: p_{1}=0.7, p_{2}=0.1, p_{3}=0.1, p_{4}=0.1 ;\\\ &n=400 \end{aligned} $$

Examining Genetic Alleles in Fast-Twitch Muscles Exercise 7.24 discusses a study investigating the \(A C T N 3\) genotypes \(R R, R X,\) and \(X X .\) The same study also examines the \(A C T N 3\) genetic alleles \(R\) and \(X,\) also associated with fast-twitch muscles. Of the 436 people in this sample, 244 were classified \(R\) and 192 were classified \(X .\) Does the sample provide evidence that the two options are not equally likely? (a) Conduct the test using a chi-square goodnessof-fit test. Include all details of the test. (b) Conduct the test using a test for a proportion, using \(H_{0}: p=0.5\) where \(p\) represents the proportion of the population classified \(R .\) Include all details of the test. (c) Compare the p-values and conclusions of the two methods.

In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. \(H_{0}:\) All three categories \(A, B, C\) are equally likely; \(\quad n=1200\)

Exercises 7.9 to 7.12 give a null hypothesis for a goodness-of-fit test and a frequency table from a sample. For each table, find: (a) The expected count for the category labeled B. (b) The contribution to the sum of the chi-square statistic for the category labeled \(\mathrm{B}\). (c) The degrees of freedom for the chi-square distribution for that table. $$ \begin{aligned} &H_{0}: p_{a}=0.2, p_{b}=0.80\\\ &H_{a}: \text { Some } p_{i} \text { is wrong }\\\ &\begin{array}{ll} \mathrm{A} & \mathrm{B} \end{array}\\\ &132 \quad 468 \end{aligned} $$

Treatment for Cocaine Addiction Cocaine addiction is very hard to break. Even among addicts trying hard to break the addiction, relapse is common. (A relapse is when a person trying to break out of the addiction fails and uses cocaine again.) Data 4.7 on page 323 introduces a study investigating the effectiveness of two drugs, desipramine and lithium, in the treatment of cocaine addiction. The subjects in the six-week study were cocaine addicts seeking treatment. The 72 subjects were randomly assigned to one of three groups (desipramine, lithium, or a placebo, with 24 subjects in each group) and the study was double-blind. In Example 4.34 we test lithium vs placebo, and in Exercise 4.181 we test desipramine vs placebo. Now we are able to consider all three groups together and test whether relapse rate differs by drug. Ten of the subjects taking desipramine relapsed, 18 of those taking lithium relapsed, and 20 of those taking the placebo relapsed. (a) Create a two-way table of the data. (b) Find the expected counts. Is it appropriate to analyze the data with a chi-square test? (c) If it is appropriate to use a chi-square test, complete the test. Include hypotheses, and give the chi-square statistic, the p-value, and an informative conclusion. (d) If the results are significant, which drug is most effective? Can we conclude that the choice of treatment drug causes a change in the likelihood of a relapse?
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