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Chapter 7: Problem 12

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. $$ \begin{aligned} &H_{0}: p_{a}=0.2, p_{b}=0.80\\\ &H_{a}: \text { Some } p_{i} \text { is wrong }\\\ &\begin{array}{ll} \mathrm{A} & \mathrm{B} \end{array}\\\ &132 \quad 468 \end{aligned} $$

Short Answer

Expert verified
The expected count for category B is 480, the contribution to the chi-square statistic for category B is 0.3, and the degrees of freedom for the chi-square distribution for this table is 1.

Step by step solution

01

Finding the Expected Count for Category B

The expected count is the total count times the probability under the null hypothesis. The total count here is the sum of the counts for categories A and B which is \( 132 + 468 = 600 \). The given null hypothesis probability for category B is \(0.80\). Therefore, the expected count for category B would be \(0.80 * 600 = 480\)
02

Contribution to the chi-square statistic for category B

The formula for the chi-square statistic is \((observed-expected)^2/expected\). Here, the observed count for category B is 468, so the contribution to the chi-square statistic from category B would be \((468 - 480)^2 / 480 = 0.3\)
03

Degrees of freedom for the chi-square distribution table

The degrees of freedom are typically calculated as the number of categories minus 1. In this case, there are 2 categories (A and B), thus the degrees of freedom would be \(2 - 1 = 1\)

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Key Concepts

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

Chi-square statistic
The chi-square statistic is a crucial component in statistical testing, particularly for good-of-fit tests. It provides a measure of how well the observed data fits the expected distribution under the null hypothesis. The idea behind it is to compare what you observe in your sample data to what you would expect if the null hypothesis is true.
The calculation of the chi-square statistic for each category involves the formula:
  • \((\text{Observed} - \text{Expected})^2 / \text{Expected}\)
This formula is used to find the contribution of each category in your dataset to the overall chi-square statistic. You sum up these contributions across all categories to get the total chi-square statistic.
In essence:
  • The larger the chi-square statistic, the greater the deviation from the expected results, which can suggest that your null hypothesis might not be true.
  • A smaller statistic implies the data fits closely to the expectation, supporting the null hypothesis.
Degrees of freedom
Degrees of freedom is a concept that determines how flexible our statistical model can be when we estimate parameters. In the context of chi-square tests, it corresponds to the number of independent values that can vary in our dataset.
Here鈥檚 a simple way to understand it:
  • For a typical chi-square goodness-of-fit test, the degrees of freedom can be calculated by taking the number of categories you have and subtracting one.
  • Mathematically, it is expressed as:\(\text{Degrees of Freedom} = \text{Number of Categories} - 1\)
Having the right degrees of freedom is vital as it affects the critical value, which determines the threshold for rejecting the null hypothesis. In our given problem, with two categories, the degrees of freedom is 1, which means there is one free value that can vary while the others are constrained by the sum of probabilities.
Expected count
Calculating expected counts is an essential procedure in a goodness-of-fit test. This tells us what distribution we'd expect if the null hypothesis is correct. To find the expected count for any category in a chi-square test, you multiply the total number in the sample by the probability for that category given in the null hypothesis.
Steps involve:
  • First, find the total count by adding up all observed frequencies in your data.
  • Multiply this total count by the probability stated under the null hypothesis for the category in question.
This expected count becomes a benchmark for comparison to see how your observed count deviates in the real dataset.
  • An expected count that is much different from the observed count will lead to a larger contribution to the chi-square statistic, potentially indicating a poor fit.
  • If the expected and observed counts are close, the fit is presumably good, which supports the null hypothesis.

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Most popular questions from this chapter

Which Is More Important: Grades, Sports, or Popularity? 478 middle school (grades 4 to 6 ) students from three school districts in Michigan were asked whether good grades, athletic ability, or popularity was most important to them. \({ }^{18}\) The results are shown below, broken down by gender: $$ \begin{array}{lccc} \hline & \text { Grades } & \text { Sports } & \text { Popular } \\ \hline \text { Boy } & 117 & 60 & 50 \\ \text { Girl } & 130 & 30 & 91 \end{array} $$ (a) Do these data provide evidence that grades, sports, and popularity are not equally valued among middle school students in these school districts? State the null and alternative hypotheses, calculate a test statistic, find a p-value, and answer the question. (b) Do middle school boys and girls have different priorities regarding grades, sports, and popularity? State the null and alternative hypotheses, calculate a test statistic, find a p-value, and answer the question.

Gender and Frequency of "Liking" Content on Facebook Exercise 7.48 introduces a study about users of social networking sites such as Facebook. Table 7.37 shows the frequency of users "liking" content on Facebook, with the data shown by gender. Does the frequency of "liking" depend on the gender of the user? Show all details of the test. $$ \begin{array}{l|rr|r} \hline \downarrow \text { Liking/Gender } \rightarrow & \text { Male } & \text { Female } & \text { Total } \\ \hline \text { Every day } & 77 & 142 & 219 \\ \text { 3-5 days/week } & 39 & 54 & 93 \\ \text { 1-2 days/week } & 62 & 69 & 131 \\ \text { Every few weeks } & 42 & 44 & 86 \\ \text { Less often } & 166 & 182 & 348 \\ \hline \text { Total } & 386 & 491 & 877 \end{array} $$

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. $$ \begin{array}{l} \hline \begin{array}{l} \text { Category } \\ \text { Observed } \\ \text { (Expected) } \end{array} & \begin{array}{c} \mathrm{A} \\ 38(30) \end{array} & \begin{array}{c} \mathrm{B} \\ 55(60) \end{array} & \begin{array}{c} \mathrm{C} \\ 79(90) \end{array} & \begin{array}{c} \mathrm{D} \\ 128(120) \end{array} \\ \hline \end{array} $$

Metal Tags on Penguins In Exercise 6.148 on page 445 we perform a test for the difference in the proportion of penguins who survive over a ten-year period, between penguins tagged with metal tags and those tagged with electronic tags. We are interested in testing whether the type of tag has an effect on penguin survival rate, this time using a chi-square test. In the study, 10 of the 50 metal-tagged penguins survived while 18 of the 50 electronic-tagged penguins survived. (a) Create a two-way table from the information given. (b) State the null and alternative hypotheses. (c) Give a table with the expected counts for each of the four categories. (d) Calculate the chi-square test statistic. (e) Determine the p-value and state the conclusion using a \(5 \%\) significance level.

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. $$ \begin{array}{l|rr|r} \hline & \text { Miscarriage } & \text { No miscarriage } & \text { Total } \\\ \hline \text { NSAIDs } & 18 & 57 & 75 \\ \text { Acetaminophen } & 24 & 148 & 172 \\ \text { No painkiller } & 103 & 659 & 762 \\ \hline \text { Total } & 145 & 864 & 1009 \\ \hline \end{array} $$
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