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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

Favorite Skittles Flavor? Exercise 7.13 on page 518 discusses a sample of people choosing their favorite Skittles flavor by color (green, orange, purple, red, or yellow). A separate poll sampled 91 people, again asking them their favorite Skittles flavor, but rather than by color they asked by the actual flavor (lime, orange, grape, strawberry, and lemon, respectively). \(^{19}\) Table 7.32 shows the results from both polls. Does the way people choose their favorite Skittles type, by color or flavor, appear to be related to which type is chosen? (a) State the null and alternative hypotheses. (b) Give a table with the expected counts for each of the 10 cells. (c) Are the expected counts large enough for a chisquare test? (d) How many degrees of freedom do we have for this test? (e) Calculate the chi-square test statistic. (f) Determine the p-value. Do we find evidence that method of choice affects which is chosen? $$ \begin{array}{lcrccc} \hline & \begin{array}{l} \text { Green } \\ \text { (Lime) } \end{array} & \begin{array}{c} \text { Purple } \\ \text { Orange } \end{array} & \begin{array}{c} \text { Red } \\ \text { (Grape) } \end{array} & \begin{array}{c} \text { Yellow } \\ \text { (Strawberry) } \end{array} & \text { (Lemon) } \\ \hline \text { Color } & 18 & 9 & 15 & 13 & 11 \\ \text { Flavor } & 13 & 16 & 19 & 34 & 9 \end{array} $$

{ 7.29 Random } & \text { Digits in } & \text { Students' } & \text { Random }\end{array}\( Numbers? How well can people generate random numbers? A sample of students were asked to write down a "random" four-digit number. Responses from 150 students are stored in the file Digits. The data file has separate variables (RND1, RND2, \)R N D 3,\( and \)R N D 4\( ) containing the digits in each of the four positions. (a) If the numbers are randomly generated, we would expect the last digit to have an equal chance of being any of the 10 digits. Test \)H_{0}\( : \)p_{0}=p_{1}=p_{2}=\cdots=p_{9}=0.10\( using technology and the data in \)R N D 4\(. (b) Since students were asked to produce four-digit numbers, there are only nine possibilities for the first digit (zero is excluded). Use technology to test whether there is evidence in the values of \)R N D 1$ that the first digits are not being chosen by students at random.

Gender and Frequency of Status Updates on Facebook Exercise 7.48 introduces a study about users of social networking sites such as Facebook. Table 7.36 shows the self-reported frequency of status updates on Facebook by gender. Are frequency of status updates and gender related? Show all details of the test. $$ \begin{array}{l|rr|r} \hline \text { IStatus/Gender } \rightarrow & \text { Male } & \text { Female } & \text { Total } \\ \hline \text { Every day } & 42 & 88 & 130 \\ \text { 3-5 days/week } & 46 & 59 & 105 \\ \text { 1-2 days/week } & 70 & 79 & 149 \\ \text { Every few weeks } & 77 & 79 & 156 \\ \text { Less often } & 151 & 186 & 337 \\ \hline \text { Total } & 386 & 491 & 877 \\ \hline \end{array} $$

One True Love by Educational Level In Data 2.1 on page 48 , we introduce a study in which people were asked whether they agreed or disagreed with the statement that there is only one true love for each person. Table 7.29 gives a two-way table showing the answers to this question as well as the education level of the respondents. A person's education is categorized as HS (high school degree or less), Some (some college), or College (college graduate or higher). Is the level of a person's education related to how the person feels about one true love? If there is a significant association between these two variables, describe how they are related. $$ \begin{array}{l|rrr|r} \hline & \text { HS } & \text { Some } & \text { College } & \text { Total } \\\ \hline \text { Agree } & 363 & 176 & 196 & 735 \\ \text { Disagree } & 557 & 466 & 789 & 1812 \\ \text { Don't know } & 20 & 26 & 32 & 78 \\ \hline \text { Total } & 940 & 668 & 1017 & 2625 \\ \hline \end{array} $$

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. \((\) Group \(2,\) No \()\) $$ \begin{array}{l|rr} \hline & \text { Yes } & \text { No } \\ \hline \text { Group 1 } & 720 & 280 \\ \text { Group 2 } & 1180 & 320 \\ \hline \end{array} $$

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