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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 \(\chi^{2}\) ?

Short Answer

Expert verified
The formula is efficient because it reduces the number of operations needed by removing subtraction and squaring.

Step by step solution

01

Understand the Defining Formula for Chi-Squared

The chi-squared statistic for independence is given by the formula: \[ \chi^{2} = \sum_{i=1}^{I} \sum_{j=1}^{J} \frac{(N_{ij} - E_{ij})^2}{E_{ij}} \] where \(N_{ij}\) is the observed frequency and \(E_{ij}\) is the expected frequency for each cell \((i, j)\). \(n\) is the total sample size.
02

Analyze the Given Formula

The formula given is \( \chi^{2} = \sum_{i=1}^{l} \sum_{j=1}^{J} \left( \frac{N_{ij}^{2}}{\dot{E}_{ij}} \right) - n \). To show that this can be derived from the defining formula, expand \((N_{ij} - E_{ij})^2\) to give \(N_{ij}^2 - 2N_{ij}E_{ij} + E_{ij}^2\).
03

Substitute and Compare

Substitute the expanded form into the defining equation: \[ \chi^{2} = \sum_{i=1}^{I} \sum_{j=1}^{J} \frac{N_{ij}^2 - 2N_{ij}E_{ij} + E_{ij}^2}{E_{ij}} = \sum_{i=1}^{I} \sum_{j=1}^{J} \left( \frac{N_{ij}^2}{E_{ij}} - 2N_{ij} + 1 \right) \] Simplifying further, this becomes \[ \chi^{2} = \sum_{i=1}^{I} \sum_{j=1}^{J} \frac{N_{ij}^2}{E_{ij}} - n \] because \(\sum_{i=1}^{I} \sum_{j=1}^{J} 2N_{ij}\) and \(\sum_{i=1}^{I} \sum_{j=1}^{J} 1\) each sum to \(n\).
04

Computational Efficiency

The formula \( \chi^{2} = \sum_{i=1}^{l} \sum_{j=1}^{J} \left( \frac{N_{ij}^{2}}{\dot{E}_{ij}} \right) - n \) is computationally efficient because it requires fewer operations. Calculating \(N_{ij} - E_{ij}\) involves a subtraction for each term, and squaring adds another operation, which is not needed in the derived form. Instead, by summing \(\frac{N_{ij}^2}{E_{ij}}\) directly, we effectively combine these steps, thus reducing computational effort in iterative contexts.

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Key Concepts

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

Observed Frequency
In the context of the chi-squared test of independence, the term 'observed frequency' refers to the actual counts or occurrences in the different categories of a contingency table. When we perform a chi-squared test, we are essentially investigating whether there is an association or independence between two categorical variables. The observed frequency, denoted as \( N_{ij} \), corresponds to the number of occurrences you observe in the data for each cell of your contingency table. In simpler terms, it's the actual count you would record when you classify data points into categories. Understanding observed frequencies is crucial because it provides the raw data that we compare against our expectations to determine if any significant relationship exists between the variables. It's the starting point in the analysis, and its accuracy is vital because subsequent calculations rely significantly on these numbers.
Expected Frequency
'Expected frequency' refers to what you would expect the frequency count to be in each cell of a contingency table if there were no association between the variables. In the chi-squared test, this is calculated under the assumption of independence between the categories. Knowing how to calculate the expected frequencies \( E_{ij} \) is key, as it represents the hypothesized counts if our categorical variables were independent. The expected frequency for each cell is generally calculated using:\[ E_{ij} = \frac{(Row \, total) \times (Column \, total)}{Overall \, total} \]This formula distributes the total counts into expected counts proportionate to the marginal totals of the rows and columns. Hence, expected frequencies help measure the discrepancy between what we observe and what we hypothesize, which is critical for understanding the strength of the independence or association.
Computational Efficiency
In statistics, computational efficiency deals with how resources, such as time and computing power, are utilized to perform calculations. When conducting a chi-squared test of independence, efficiency can mean the difference between smoothly handling large datasets or being bogged down by computational challenges.The formula \[ \chi^{2} = \sum_{i=1}^{l} \sum_{j=1}^{J} \left( \frac{N_{ij}^{2}}{\dot{E}_{ij}} \right)-n \] for computing the chi-squared statistic is more efficient than the standard formula because it simplifies the calculation process by reducing the number of steps. It omits unnecessary arithmetic operations like subtraction and squaring, which typically add extra computational load, especially with large tables or datasets.By directly computing the ratio \( \frac{N_{ij}^{2}}{E_{ij}} \), we combine operations, minimizing processing time, and making it more efficient. Efficiency is particularly beneficial when tests are run iteratively, where such streamlining can save a significant amount of compute power and time, making it practical for larger or more complex analyses. Simplified calculations not only save resources but reduce potential errors, critical for ensuring accuracy in statistical analyses.

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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 \(\alpha=.01\).

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. \(x^{2}=5.0, k=4\)

The article "The Gap Between Wine Expert Ratings and Consumer Preferences" (Intl. J. of Wine Business Res., 2008: 335-351) studied differences between expert and consumer ratings by considering medal ratings for wines, which could be gold \((G)\), silver (S), or bronze (B). Three categories were then established: 1. Rating is the same \([(\mathrm{G}, \mathrm{G}),(\mathrm{B}, \mathrm{B}),(\mathrm{S}, \mathrm{S})]\); 2. Rating differs by one medal \([(\mathrm{G}, \mathrm{S}),(\mathrm{S}, \mathrm{G}),(\mathrm{S}, \mathrm{B}),(\mathrm{B}, \mathrm{S})]\); and 3. Rating differs by two medals \([(G, B),(B, G)]\). The observed frequencies for these three categories were 69,102 , and 45 , respectively. On the hypothesis of equally likely expert ratings and consumer ratings being assigned completely by chance, each of the nine medal pairs has probability \(1 / 9\). Carry out an appropriate chi-squared test using a significance level of 10 by first obtaining \(P\)-value information.

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.

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