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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. $$ \begin{array}{cccc} \hline \text { Category } & \text { A } & \text { B } & \text { C } \\ \text { Observed } & 61(50) & 35(50) & 54(50) \\ \text { (Expected) } & & & \\ \hline \end{array} $$

Short Answer

Expert verified
The total \(\chi^{2}\) test statistic and the p-value need to be calculated first to determine whether the observed frequencies significantly differ from the expected frequencies. The outcome will depend on the calculated \(\chi^{2}\) test statistic and the corresponding p-value from the \(\chi^{2}\) -distribution table.

Step by step solution

01

- Calculate Chi-Square Test Statistic

Firstly, apply the formula to calculate \(\chi^{2}\) for each category (\(A, B, C\)) using the observed and expected values. Hence, you have\(\chi^{2}_{A}=\frac{{(61-50)^{2}}}{50}\),\(\chi^{2}_{B}=\frac{{(35-50)^{2}}}{50}\),\(\chi^{2}_{C}=\frac{{(54-50)^{2}}}{50}\). After calculation, sum these values to find the total chi-square test statistic.
02

- Calculate Degrees of Freedom

For a chi-square test, the degrees of freedom are calculated as \(df = k - 1\), where \(k\) is the number of categories. In this case, \(k = 3\) (categories A, B, C), therefore, \(df = 3 - 1 = 2\).
03

- Find the p-value

Use a \(\chi^{2}\) -distribution table with 2 degrees of freedom to find the p-value. Locate the chi-square value calculated in step 1 on the table, which corresponds to our calculated chi-square. The p-value is the area to the right of this chi-square value under the chi-square distribution curve.
04

- Interpretation

If the p-value is less than 0.05, we reject the null hypothesis in favor of the alternative hypothesis, meaning there is a statistically significant difference between the observed and expected frequencies. If the p-value is more than 0.05, then we fail to reject the null hypothesis, meaning there is no statistically significant difference between the observed and expected frequencies.

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Key Concepts

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

Categorical Variable
Categorical variables play a crucial role in statistical analysis, particularly when conducting a chi-square test. Unlike numerical variables, which can be measured and ordered, categorical variables represent characteristics that can be sorted into categories but not quantified. For example, categories such as 'eye color' with options like blue, green, and brown are categorical because they describe an attribute rather than a measurement.

Categorical variables can be 'nominal', with no intrinsic ordering to the categories, or 'ordinal', where the categories have a logical order. In the exercise, categories A, B, and C represent nominal categorical variables. The observed counts correspond to the number of occurrences or frequency within each category. Understanding and identifying categorical variables is crucial when interpreting chi-square test results, as these variables define the groups being compared for expected versus observed frequencies.
Null Hypothesis
The null hypothesis, often denoted as H0, is a baseline proposition in statistics claiming that there is no effect or no difference between groups. Essentially, it's a statement of 'no change' or 'no association'. In the context of a chi-square test for categorical data, the null hypothesis states that the observed frequencies in each category are consistent with the expected frequencies.

For instance, in the provided exercise, the null hypothesis would be that there is no significant difference between the observed counts (61, 35, 54) and the expected counts (all 50) across the three categories (A, B, C). Refuting or not refuting this hypothesis becomes the crux of the analysis, directing researchers toward concluding whether their observed data negates the baseline expectation.
P-value
The p-value, or probability value, is a key concept in hypothesis testing and indicates the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct. A low p-value suggests that the observed data are unlikely given the null hypothesis, prompting analysts to consider rejecting H0.

In simple terms, if the p-value is small enough (commonly less than 0.05, representing a 5% significance level), it signifies that what was observed is unusual if we believed the null hypothesis to be true. In the chi-square test from the textbook exercise, the p-value tells us whether the difference between the observed and expected counts is statistically significant or just due to random chance.
Degrees of Freedom
Degrees of freedom (df) is a statistics concept that describes the number of independent values or quantities that can be assigned to a statistical distribution. In the context of the chi-square test, the degrees of freedom are calculated based on the number of categories under examination minus one (df = k - 1).

The concept of degrees of freedom is essential when determining the chi-square test statistic's corresponding p-value. It serves as a parameter for the chi-square distribution, reflecting the complexity of the statistical inference involved. In our exercise example, with 3 categories (k=3), the degrees of freedom would be 2 (df=3-1). This parameter is then used along with the chi-square test statistic to find the p-value from the chi-square distribution table.

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Most popular questions from this chapter

Binge Drinking The American College Health Association - National College Health Assessment survey \(,{ }^{17}\) introduced on page 60 , was administered at 44 colleges and universities in Fall 2011 with more than 27,000 students participating in the survey. Students in the ACHA-NCHA survey were asked "Within the last two weeks, how many times have you had five or more drinks of alcohol at a sitting?" The results are given in Table 7.31 . Is there a significant difference in drinking habits depending on gender? Show all details of the test. If there is an association, use the observed and expected counts to give an informative conclusion in context. $$ \begin{array}{c|rr|r} \hline & \text { Male } & \text { Female } & \text { Total } \\ \hline 0 & 5,402 & 13,310 & 18,712 \\ 1-2 & 2,147 & 3,678 & 5,825 \\ 3-4 & 912 & 966 & 1,878 \\ 5+ & 495 & 358 & 853 \\ \hline \text { Total } & 8,956 & 18,312 & 27,268 \\ \hline \end{array} $$

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; \(\quad n=1200\)

Give a two-way table and specify a particular cell for that table. In each case find the expected count for that cell and the contribution to the chi- square statistic for that cell. \((\mathrm{B}, \mathrm{E})\) cell $$ \begin{array}{l|rrrr|r} \hline & \text { D } & \text { E } & \text { F } & \text { G } & \text { Total } \\ \hline \text { A } & 39 & 34 & 43 & 34 & 150 \\ \text { B } & 78 & 89 & 70 & 93 & 330 \\ \text { C } & 23 & 37 & 27 & 33 & 120 \\ \hline \text { Total } & 140 & 160 & 140 & 160 & 600 \\ \hline \end{array} $$

Another Test for Cocaine Addiction Exercise 7.42 on page 532 describes an experiment on helping cocaine addicts break the cocaine addiction, in which cocaine addicts were randomized to take desipramine, lithium, or a placebo. The results (relapse or no relapse after six weeks) are summarized in Table \(7.38 .\) (a) In Exercise 7.42, we calculate a \(\chi^{2}\) statistic of 10.5 and use a \(\chi^{2}\) distribution to calculate a p-value of 0.005 using these data, but we also could have used a randomization distribution. How would you use cards to generate a randomization sample? What would you write on the cards, how many cards would there be of each type, and what would you do with the cards? (b) If you generated 1000 randomization samples according to your procedure from part (a) and calculated the \(\chi^{2}\) statistic for each, approximately how many of these statistics do you expect would be greater than or equal to the \(\chi^{2}\) statistic of 10.5 found using the original sample? $$ \begin{array}{l|cc|c} \hline & \text { Relapse } & \text { No Relapse } & \text { Total } \\ \hline \text { Desipramine } & 10 & 14 & 24 \\ \text { Lithium } & 18 & 6 & 24 \\ \text { Placebo } & 20 & 4 & 24 \\ \hline \text { Total } & 48 & 24 & 72 \end{array} $$

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. $$ \begin{aligned} &H_{0}: p_{a}=p_{b}=p_{c}=p_{d}=0.25\\\ &: \text { Some }\\\ &H_{a}:\\\ &p_{i} \neq 0.25\\\ &\begin{array}{lccc|c} \hline \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \text { Total } \\ 120 & 148 & 105 & 127 & 500 \\ \hline \end{array} \end{aligned} $$

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