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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. $$ \begin{aligned} &H_{0}: p_{a}=p_{b}=p_{c}=p_{d}=0.25\\\ &H_{a}:\\\ &\text { Some } p_{i} \neq 0.25\\\ &\begin{array}{lccc|c} \hline \text { A } & \text { B } & \text { C } & \text { D } & \text { Total } \\\ \text { 40 } & 36 & 49 & 35 & 160 \\ \hline \end{array} \end{aligned} $$

Short Answer

Expert verified
The expected count for category B is 40. The contribution to the chi-square statistic for category B is 0.4. The degrees of freedom for the chi-square distribution is 3.

Step by step solution

01

Calculate the Expected Count for Category B

The expected count for each of the four categories (A, B, C, D) under the null hypothesis is given by the total count times the probability for each category. As given by the null hypothesis, the probability for each category is 0.25. Therefore, the expected count for category B is \(0.25 \times 160 = 40\).
02

Calculate the Contribution to the Chi-Square Statistic for Category B

The contribution to the sum of the chi-square statistic from category B is given by \((Observed - Expected)^2 / Expected\). The observed count for category B is 36. Given that the expected count calculated in Step 1 was 40, the contribution from category B to the chi-square statistic is \((36 - 40)^2 / 40 = 0.4\).
03

Calculate the Degrees of Freedom

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

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Key Concepts

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

Expected Count
When conducting a goodness-of-fit test, the concept of expected count plays a pivotal role. It represents the frequency that is anticipated for each category if the null hypothesis holds true. In simple terms, the expected count is the count that would occur if the experimental data perfectly aligned with the probability distribution specified by the null hypothesis.

In our textbook exercise, the null hypothesis suggests that each of the categories A, B, C, and D are equally likely, each with a probability of 0.25. Given a total sample size of 160, the expected count for category B (as well as for the others) is calculated by multiplying the total count by the probability for that category, which is 0.25 in this case. Mathematically, it's expressed as the total count (160) times the probability (0.25), yielding an expected count of 40 for category B.

This count is essential because it provides a benchmark against which the observed count is compared. If the observed count deviates significantly from the expected count, it could indicate that the null hypothesis may not be an accurate representation of the underlying distribution. Thus, calculating the expected count is the first step in evaluating the goodness-of-fit.
Chi-Square Statistic
The chi-square statistic is a measure that quantifies how the observed counts deviate from the expected counts in a categorical data set. It is a crucial element of the goodness-of-fit test, helping statisticians to determine whether there is a statistically significant difference between the expected and observed frequencies in data categories.

To calculate the contribution of each category to the chi-square statistic, we use the formula \( (Observed - Expected)^2 / Expected \). For category B in our exercise, the observed count is 36, while the expected count is 40. Plugging these values into our formula, we obtain \( (36 - 40)^2 / 40 = 0.4 \), suggesting a discrepancy between what was observed in the sample and what we would expect under the null hypothesis.

The chi-square statistic is summed over all categories, and if the overall value is sufficiently large, this could lead to rejection of the null hypothesis, suggesting that the observed data does not fit the expected distribution. Interpretation of the chi-square value always takes into consideration the degrees of freedom and is referenced against a chi-square distribution table to determine statistical significance.
Degrees of Freedom
Degrees of freedom is a concept tied to the reliability of various statistical parameters and distributions, including the chi-square statistic. In the context of a goodness-of-fit test, the degrees of freedom define the number of values that can vary freely for a given statistic, after certain constraints (such as the total sum of observed counts) are taken into account.

For our chi-square test example, the degrees of freedom are determined by the number of categories minus one. Hence, with four categories (A, B, C, and D), the degrees of freedom for the test would be \(4 - 1 = 3\). It's important to note that the ‘minus one’ is to account for the restriction that all the expected frequencies must add up to the total sample size. We can essentially say that the degrees of freedom reflect the amount of ‘wiggle room’ you have in your data.

The degrees of freedom help to identify the correct distribution for evaluating the chi-square statistic and therefore, play a critical role in interpreting the test result. A chi-square distribution varies depending on the degrees of freedom, affecting how the critical value for statistical significance is determined. Therefore, correctly identifying the degrees of freedom is essential to conducting an accurate goodness-of-fit test.

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Most popular questions from this chapter

Painkillers and Miscarriage Exercise A.50 on page 179 describes a study examining the link between miscarriage and the use of painkillers during pregnancy. Scientists interviewed 1009 women soon after they got positive results from pregnancy tests about their use of painkillers around the time of conception or in the early weeks of pregnancy. The researchers then recorded which of the pregnancies were successfully carried to term. The results are in Table \(7.30 .\) (NSAIDs refer to a class of painkillers that includes aspirin and ibuprofen.) Does there appear to be an association between having a miscarriage and the use of painkillers? If so, describe the relationship. If there is an association, can we conclude that the use of painkillers increases the chance of having a miscarriage? 7.44 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. $$ \begin{array}{l|rr|r} \hline & \text { Miscarriage } & \text { No miscarriage } & \text { Total } \\\ \hline \text { NSAIDs } & 18 & 57 & 75 \\ \text { Acetaminophen } & 24 & 148 & 172 \\ \text { No painkiller } & 103 & 659 & 762 \\ \hline \text { Total } & 145 & 864 & 1009 \\ \hline \end{array} $$

Gender and Award Preference Example 2.6 on page 53 contains a two-way table showing preferences for an award (Academy Award, Nobel Prize, Olympic gold medal) by gender for the students sampled in StudentSurvey. The data are reproduced in Table \(7.28 .\) Test whether the data indicate there is some association between gender and preferred award. $$ \begin{array}{l|ccc|c} \hline & \text { Academy } & \text { Nobel } & \text { Olympic } & \text { Total } \\ \hline \text { Female } & 20 & 76 & 73 & 169 \\ \text { Male } & 11 & 73 & 109 & 193 \\ \hline \text { Total } & 31 & 149 & 182 & 362 \\ \hline \end{array} $$

Can People Delay Death? A study indicates that elderly people are able to postpone death for a short time to reach an important occasion. The researchers \({ }^{10}\) studied deaths from natural causes among 1200 elderly people of Chinese descent in California during six months before and after the Harbor Moon Festival. Thirty-three deaths occurred in the week before the Chinese festival, compared with an estimated 50.82 deaths expected in that period. In the week following the festival, 70 deaths occurred, compared with an estimated 52. "The numbers are so significant that it would be unlikely to occur by chance," said one of the researchers. (a) Given the information in the problem, is the \(\chi^{2}\) statistic likely to be relatively large or relatively small? (b) Is the p-value likely to be relatively large or relatively small? (c) In the week before the festival, which is higher: the observed count or the expected count? What does this tell us about the ability of elderly people to delay death? (d) What is the contribution to the \(\chi^{2}\) -statistic for the week before the festival? (e) In the week after the festival, which is higher: the observed count or the expected count? What does this tell us about the ability of elderly people to delay death? (f) What is the contribution to the \(\chi^{2}\) -statistic for the week after the festival? (g) The researchers tell us that in a control group of elderly people in California who are not of Chinese descent, the same effect was not seen. Why did the researchers also include a control group?

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