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

Short Answer

Expert verified
The expected count for category B is 125. The contribution to the sum of the chi-square statistic for category B is 4.416. The degrees of freedom for the chi-square distribution for the table is 3.

Step by step solution

01

Calculation of Expected Count for Category B

According to the null hypothesis, \(p_b = 0.25\). We will multiply this probability with the total count to get the expected count for Category B. Hence, Expected count \(E_B = 0.25 \times Total Count = 0.25 \times 500 = 125.\)
02

Calculation of Chi-square Statistic Contribution for Category B

To find the contribution of B to the chi-square statistic, we subtract the expected from the observed, square it, and then divide by the expected: \((Observed - Expected)^2 / Expected = (148 - 125)^2 / 125 = 4.416.\)
03

Calculation of Degrees of Freedom

Degrees of freedom for the chi-square distribution is the total number of categories minus one. In this case, there are four categories (A, B, C, D), hence degrees of freedom = 4 - 1 = 3.

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Key Concepts

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

Expected Count Calculation
Expected count calculation is a fundamental step in performing a goodness-of-fit test. It involves determining the theoretical frequencies of each category within a dataset based on a specific hypothesis. In the context of a chi-square goodness-of-fit test, these expected counts are crucial for measuring how well the observed data fit the expected distribution.

For example, if you have a null hypothesis that four categories A, B, C, and D are equally likely, and you have a total sample size of 500, then the expected frequency for each category is 500 (total count) multiplied by 0.25 (the expected probability of each category assuming they are equally likely).

Understanding the Calculation

For the category B, with the null hypothesis stating that the expected probability, represented as \(p_b\), equals 0.25, the expected count (\(E_B\)) is calculated by multiplying this probability by the total sample size. The formula looks like this:\[E_B = p_b \times \text{Total Count} = 0.25 \times 500\]Following this approach would yield an expected count of 125 for the category labeled B. This step is fundamental as it sets the stage for comparing what we expect to see if the null hypothesis holds true with the actual observed data.
Chi-square Statistic
The chi-square statistic is a measure calculated in goodness-of-fit tests that quantifies how observed values diverge from what was expected. After establishing the expected counts, this statistic helps to determine if the discrepancies between observed and expected counts are due to random chance or if they signify a meaningful discrepancy, suggesting that the null hypothesis may not accurately reflect reality.

Calculating Contribution to the Chi-square Statistic

For any given category, such as B in our example, the contribution to the chi-square statistic is computed using the following formula:\[(\text{Observed} - \text{Expected})^2 / \text{Expected} = (148 - 125)^2 / 125\]In this instance, the calculation yields a result of 4.416. This number contributes to the sum of the chi-square statistic, which will eventually be compared against a critical value from the chi-square distribution table to assess the goodness-of-fit. The larger this statistic, the less likely it is that the observed difference is due to random variation, indicating a potential rejection of the null hypothesis.
Degrees of Freedom
Degrees of freedom in a chi-square test serve as a way to account for the amount of independent information in your calculated statistic. It is associated with the number of categories from which the expected counts were derived, minus any parameters estimated from the data, typically one for each probability in the null hypothesis calculated from the data.The degrees of freedom are essential to determine the critical value of the chi-square distribution that corresponds to a desired level of significance. In our textbook example, the degrees of freedom (\(df\)) are calculated by subtracting one from the total number of categories since the category probabilities are all pre-specified in the null hypothesis and not estimated from the data:\[df = \text{Number of categories} - 1 = 4 - 1 = 3\]For a chi-square distribution, the degrees of freedom directly impact the shape of the distribution. As degrees of freedom increase, the distribution becomes more symmetric and bell-shaped, akin to the normal distribution. In hypothesis testing, it's crucial to use the correct degrees of freedom to accurately assess the likelihood of the observed data given the null hypothesis.

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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. $$ \begin{aligned} &\text { 7.4 } H_{0}: p_{1}=0.7, p_{2}=0.1, p_{3}=0.1, p_{4}=0.1 ;\\\ &n=400 \end{aligned} $$

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? $$ \begin{array}{l|cc|c} \hline & \text { Relapse } & \text { No Relapse } & \text { Total } \\ \hline \text { Desipramine } & 10 & 14 & 24 \\ \text { Lithium } & 18 & 6 & 24 \\ \text { Placebo } & 20 & 4 & 24 \\ \hline \text { Total } & 48 & 24 & 72 \end{array} $$

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; \(\quad n=1200\)

Handedness and Occupation Is the career someone chooses associated with being left- or right-handed? In one study \(^{20}\) a sample of Americans from a variety of professions were asked if they consider themselves left-handed, right-handed, or ambidextrous (equally skilled with the left and right hand). The results for five professions are shown in Table \(7.33 .\) (a) In this sample, what profession had the greatest proportion of left-handed people? What profession had the greatest proportion of right-handed people? (b) Test for an association between handedness and career for these five professions. State the null and alternative hypotheses, calculate the test statistic, and find the p-value. (c) What do you conclude at the \(5 \%\) significance level? What do you conclude at the \(1 \%\) significance level? $$ \begin{array}{l|rrr|r} \hline & \text { Right } & \text { Left } & \text { Ambidextrous } & \text { Total } \\ \hline \text { Psychiatrist } & 101 & 10 & 7 & 118 \\ \text { Architect } & 115 & 26 & 7 & 148 \\ \text { Orthopedic surgeon } & 121 & 5 & 6 & 132 \\ \text { Lawyer } & 83 & 16 & 6 & 105 \\ \text { Dentist } & 116 & 10 & 6 & 132 \\ \hline \text { Total } & 536 & 67 & 32 & 635 \\ \hline \end{array} $$

Gender and ACTN3 Genotype We see in the previous two exercises that sprinters are more likely to have allele \(R\) and genotype \(R R\) versions of the ACTN3 gene, which makes these versions associated with fast-twitch muscles. Is there an association between genotype and gender? Computer output is shown for this chi-square test, using the control group in the study. In each cell, the top number is the observed count, the middle number is the expected count, and the bottom number is the contribution to the chi-square statistic. What is the p-value? What is the conclusion of the test? Is gender associated with the likelihood of having a "sprinting gene"? \(\begin{array}{lrrrr} & \text { RR } & \text { RX } & \text { XX } & \text { Total } \\ \text { Male } & 40 & 73 & 21 & 134 \\ & 40.26 & 69.20 & 24.54 & \\\ & 0.002 & 0.208 & 0.509 & \\ \text { Female } & 88 & 147 & 57 & 292 \\ & 87.74 & 150.80 & 53.46 & \\ & 0.001 & 0.096 & 0.234 & \\ \text { Total } & 128 & 220 & 78 & 426\end{array}\) \(\mathrm{Chi}-\mathrm{Sq}=1.050, \mathrm{DF}=2, \mathrm{P}\) -Value \(=0.592\)

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