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Chapter 14: Problem 1

Random samples of 200 observations were selected from each of three populations, and each observation was classified according to whether it fell into one of three mutually exclusive categories. Is there sufficient evidence to indicate that the proportions of observations in the three categories depend on the population from which they were drawn? Use the information in the table to answer the questions in Exercises \(1-4 .\) $$ \begin{array}{lrlll} \hline & {\text { Category }} & \\ \text { Population } & 1 & 2 & 3 & \text { Total } \\ \hline 1 & 108 & 52 & 40 & 200 \\ 2 & 87 & 51 & 62 & 200 \\ 3 & 112 & 39 & 49 & 200 \\ \hline \end{array} $$ Give the value of \(\mathrm{X}^{2}\) for the test.

Short Answer

Expert verified
#tag_title#Step 5: Finding the critical value and making a decision#tag_content# To determine whether the null hypothesis should be rejected or not, we need to find the critical value for the \(\chi^2\) distribution. We require the degrees of freedom, which can be calculated as: \(df = (number\_of\_rows - 1) * (number\_of\_columns - 1) = (3 - 1) * (3 - 1) = 4\). Using a significance level of 0.05, we look up the critical value for the \(\chi^2\) distribution with 4 degrees of freedom. The critical value is 9.488. Now we compare the test statistic (\(\chi^2 = 12.819\)) to the critical value (9.488): - If the test statistic is greater than the critical value, we reject the null hypothesis (H0) in favor of the alternative hypothesis (H1). - If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis. Since our test statistic (12.819) is greater than the critical value (9.488), we reject the null hypothesis (H0). There is evidence to suggest that there is an association between the categories and the population. Please write a short answer discussing the implication of our results.

Step by step solution

01

Creating the expected frequencies table

To perform the Chi-square test, we need to create a table with the expected frequencies for each observation. To calculate the expected frequencies, we can use the formula: \(E_{ij} = \frac{(R_i * C_j)}{n}\), where \(E_{ij}\) is the expected frequency in the i-th row and j-th column, \(R_i\) is the total number of observations in the i-th row, \(C_j\) is the total number of observations in the j-th column, and n is the total number of observations.
02

Calculating expected frequencies

Using the formula provided in Step 1, we can calculate the expected frequencies for each observation: $$ \begin{array}{cccc} \hline \text{Population} & \text{Category 1} & \text{Category 2} & \text{Category 3} \\ \hline 1 & \frac{(200)(307)}{600} & \frac{(200)(142)}{600} & \frac{(200)(151)}{600} \\ 2 & \frac{(200)(307)}{600} & \frac{(200)(142)}{600} & \frac{(200)(151)}{600} \\ 3 & \frac{(200)(307)}{600} & \frac{(200)(142)}{600} & \frac{(200)(151)}{600} \\ \hline \end{array} $$
03

Evaluating the expected frequencies

Next, evaluate the expected frequencies: $$ \begin{array}{cccc} \hline \text{Population} & \text{Category 1} & \text{Category 2} & \text{Category 3} \\ \hline 1 & 102.333 & 47.333 & 50.333 \\ 2 & 102.333 & 47.333 & 50.333 \\ 3 & 102.333 & 47.333 & 50.333 \\ \hline \end{array} $$
04

Calculating the Chi-square test statistic

Now, we will calculate the \(\chi^2\) test statistic using the formula: \(\chi^2 = \sum{\frac{(O_{ij} - E_{ij})^2}{E_{ij}}}\), where \(O_{ij}\) and \(E_{ij}\) are the observed and expected frequencies, respectively, for the i-th row and j-th column in the table. $$ \begin{array}{cccc} \hline \text{Population} & \text{Category 1} & \text{Category 2} & \text{Category 3} \\ \hline 1 & \frac{(108 - 102.333)^2}{102.333} & \frac{(52 - 47.333)^2}{47.333} & \frac{(40 - 50.333)^2}{50.333} \\ 2 & \frac{(87 - 102.333)^2}{102.333} & \frac{(51 - 47.333)^2}{47.333} & \frac{(62 - 50.333)^2}{50.333} \\ 3 & \frac{(112 - 102.333)^2}{102.333} & \frac{(39 - 47.333)^2}{47.333} & \frac{(49 - 50.333)^2}{50.333} \\ \hline \end{array} $$ After calculating the \(\chi^2\) values for each cell, sum them up to obtain the test statistic: \(\chi^2 = \sum{\frac{(O_{ij} - E_{ij})^2}{E_{ij}}} = 3.036 + 0.454 + 2.134 + 2.276 + 0.289 + 2.689 + 0.454 + 1.449 + 0.036 = 12.819\). The value of \(\chi^2\) for the test is 12.819.

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Key Concepts

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

Random Samples
When investigating statistical questions in probability and statistics, random samples play a crucial role. A random sample is a subset of individuals chosen from a larger population, where each individual is selected by chance. The primary characteristic of a random sample is that it should be representative of the whole population, meaning every member of the population has an equal probability of being included in the sample.

This ensures that the inferences and conclusions drawn from the sample are likely to be valid for the entire population. The exercise, which involves examining whether the proportions of observations in categories depend on the population, is fundamentally based on the principle of random sampling. The sampling should prevent any bias and reflect the true distribution of categories across different populations.
Expected Frequencies
Expected frequencies are a fundamental concept in conducting a Chi-square test. They represent the hypothesized or theoretical distribution of observations across different categories under the assumption that no association exists between the variables being studied. In other words, if the observed frequencies perfectly match the expected frequencies, one would conclude that there is no relationship between the population and categories.

The expected frequency for each cell in a contingency table is calculated based on the row and column totals and the overall sample size. It indicates how many observations are statistically predicted to fall into each category by chance alone. Calculating expected frequencies provides a benchmark against which the actual, or observed, frequencies can be compared.
Observed Frequencies
Contrasting expected frequencies, observed frequencies are the actual counts collected from the data or experiment. They directly reflect the number of occurrences in each category for the sample that was taken. The difference between observed and expected frequencies is critical for the Chi-square test. The larger the discrepancy, the stronger the evidence against the null hypothesis, which usually suggests no effect or no association.

In the exercise provided, the observed frequencies are given for three populations across three categories. By comparing these to the expected frequencies, we determine whether there's evidence that the population influences categorization, evaluating if the observed distributions are significantly different from what randomness would predict.
Probability and Statistics
The fields of probability and statistics are deeply interconnected. Probability deals with predicting the likelihood of future events, while statistics involves analyzing the frequency of past events. The Chi-square test, used in the exercise, is a statistical method to assess whether observed frequencies differ from expected frequencies. It is grounded in probability theory, as it uses the concepts of random samples and distributions to form expectations.

The p-value obtained from the Chi-square test statistic helps determine the probability of observing a statistic at least as extreme as the one calculated, under the assumption that the null hypothesis is true. This is essential for making informed decisions on whether or not to reject the null hypothesis. The exercise's resolution through the Chi-square test encapsulates the application of both probability and statistics to draw conclusions from data.

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Most popular questions from this chapter

Give the rejection region for a chi-square test of independence if the contingency table involves \(r\) rows and \(c\) columns. $$r=2, c=4, \alpha=.05$$

According to Americans, access to healthcare and the cost of healthcare remain the most urgent health problems. However, a recent Gallup poll \(^{11}\) shows that concern about substance abuse jumped from \(3 \%\) to \(14 \%\) in 2017 . Based on samples of size 200 for each year, the data that follow reflect the results of that poll. $$ \begin{array}{lrr} \hline \text { Concern } & 2016 & 2017 \\ \hline \text { Access } & 40 & 48 \\ \text { Cost } & 54 & 32 \\ \text { Substance abuse } & 6 & 28 \\ \text { Cancer } & 24 & 22 \\ \text { Obesity } & 16 & 14 \\ \text { Other } & 60 & 56 \\ \hline \text { Total } & 200 & 200 \\ \hline \end{array} $$ a. Calculate the proportions in each of the categories for 2016 and 2017 . Test for a significant change in proportions for the healthcare concerns listed from 2016 to 2017 using \(\alpha=.05\) b. How would you summarize the results of the analysis in part a? Can you conclude that the change in the proportion of adults whose concern was substance abuse is significant? Why or why not?

Give the rejection region for a chi-square test of independence if the contingency table involves \(r\) rows and \(c\) columns. $$r=3, c=5, \alpha=.01$$

A response can fall into one of \(k=4\) categories with hypothesized cell probabilities given If 250 responses are recorded, what are the four expected cell counts for the chi-square test? $$ p_{1}=.25, p_{2}=.15, p_{3}=.10, p_{4}=? $$

Find the appropriate degrees of freedom for the chisquare test of independence. $$\text { five rows and four columns }$$
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