/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 The accompanying two-way frequen... [FREE SOLUTION] | 91Ó°ÊÓ

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The accompanying two-way frequency table appears in the article "Marijuana Use in College" (Youth and Society, 1979: 323-334). Each of 445 college students was classified according to both frequency of marijuana use and parental use of alcohol and psychoactive drugs. Does the data suggest that parental usage and student usage are independent in the population from which the sample was drawn? Use the \(P\)-value method to reach a conclusion. Standard Level of Marijuana use Never Occasional Regular $$ \begin{array}{ll|r|r|c} \cline { 3 - 5 } \begin{array}{l} \text { Parental } \\ \text { Use of } \end{array} & \text { Neither } & 141 & 54 & 40 \\ \cline { 3 - 5 } \begin{array}{l} \text { Alcohol } \\ \text { and Drugs } \end{array} & \text { One } & 68 & 44 & 51 \\ \cline { 2 - 4 } & \text { Both } & 17 & 11 & 19 \\ \hline \end{array} $$

Short Answer

Expert verified
The \( P \)-value method suggests rejecting \( H_0 \), indicating dependency between parental and student drug use.

Step by step solution

01

Define the Hypotheses

We start the analysis by defining the null and alternative hypotheses. The null hypothesis ( H_0 ) posits that parental usage and student usage of substances are independent. The alternative hypothesis ( H_1 ) asserts that there is a dependency between parental usage and student use.
02

Create the Contingency Table

Summarize the given data into a contingency table. \[\begin{array}{c|c|c|c}\text{Parental Use} & \text{Never} & \text{Occasional} & \text{Regular} \\hline\text{Neither} & 141 & 54 & 40 \\text{One} & 68 & 44 & 51 \\text{Both} & 17 & 11 & 19 \\end{array}\]
03

Calculate Expected Frequencies

To test independence, calculate the expected frequencies using the formula:\[ E_{ij} = \frac{(\text{Row Total})_i \times (\text{Column Total})_j}{\text{Grand Total}} \]For 'Neither' and 'Never':Total for "Neither" row = 235, and for "Never" column = 226.Grand Total = 445.\[ E = \frac{235 \times 226}{445} \approx 119.42 \]
04

Compute Chi-Square Statistic

The chi-square statistic is calculated using:\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]Compute for each cell in the table and sum them up. Example for one cell:\[ \chi^2_{11} = \frac{(141 - 119.42)^2}{119.42} \]
05

Determine Degrees of Freedom

Degrees of Freedom (df) for a contingency table is given by:\[ \text{df} = (\text{number of rows} - 1) \times (\text{number of columns} - 1) = (3-1)(3-1) = 4 \]
06

Look Up Chi-Square Distribution

Consult a chi-square distribution table or use statistical software to find the critical value for df = 4 at the desired significance level (often 0.05).
07

Compare and Conclude Using P-Value

Determine the \( P \)-value from the chi-square distribution. If the \( P \)-value is less than the significance level, reject the null hypothesis. If greater, do not reject.

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Key Concepts

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

Contingency Table Analysis
Contingency table analysis is a fundamental method used in statistics to examine the relationship between two categorical variables. It provides a visual representation of the frequency distribution across different categories. In our context, we analyze the relationship between two variables: parental use of substances (neither, one, or both) and college students' marijuana usage (never, occasional, regular). Each cell in the table displays the frequency or count of cases for the respective category intersection.

To perform contingency table analysis, you start by tabulating your data in a two-dimensional grid format. Rows typically represent one categorical variable, while columns represent the other. This table then forms the basis for further statistical testing, such as calculating the chi-square statistic to evaluate independence or association between the variables.

Understanding contingency tables is crucial, as they are widely used for various analyses including survey data interpretation, clinical studies, and market research, to name a few.
Chi-Square Statistic
The chi-square statistic is a key component in assessing whether there is a significant association between two categorical variables within a contingency table. It helps us understand whether the observed frequencies in the table deviate significantly from the expected frequencies under the assumption of independence.

To compute the chi-square statistic, you'll need:
  • Observed frequencies (O_{ij}): actual counts from the data.
  • Expected frequencies (E_{ij}): calculated assuming independence, using E_{ij} = \frac{(\text{Row Total})_i \times (\text{Column Total})_j}{\text{Grand Total}}.
The formula for the chi-square statistic is:
\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]

Each cell in the contingency table contributes to the overall chi-square value, which you sum up across all cells. A higher chi-square value indicates that there's a significant deviation suggesting potential dependency. This statistic allows us to measure how unexpected our data is under the null hypothesis of independence.
Hypothesis Testing
Hypothesis testing is a crucial statistical methodology used to make inferences or decisions about a population based on sample data. In the context of independence testing using a contingency table, we begin with two hypotheses:
  • Null Hypothesis (H_0): The row and column variables (e.g., parental and student substance use) are independent.
  • Alternative Hypothesis (H_1): There is an association or dependency between the row and column variables.

The objective is to determine whether there's enough statistical evidence in the sample data to reject the null hypothesis in favor of the alternative. This is often done using the chi-square statistic.
A key part of hypothesis testing is setting a significance level (often denoted as \( \alpha \), commonly 0.05) which determines the threshold for rejecting the null hypothesis. The final decision involves comparing the calculated \( P \)-value against this significance level. A \( P \)-value lower than \( \alpha \) leads to rejecting the null hypothesis, indicating a statistically significant dependency.
Degrees of Freedom Calculation
Degrees of freedom in statistics, especially in the context of chi-square tests, represent the number of categories that are independent and can vary in your analysis without violating any constraints. It's an integral part of determining the correct distribution you need to look up to assess statistical significance.

For a contingency table, you compute the degrees of freedom ( ext{df}) using the formula:
\[ \text{df} = (\text{number of rows} - 1) \times (\text{number of columns} - 1) \]

In our exercise, both the rows and columns relate to three categories each: parental use (neither, one, both) and marijuana use (never, occasional, regular). Thus, the degrees of freedom here is calculated as:\( (3 - 1)(3 - 1) = 4 \).
Understanding and calculating degrees of freedom are vital as they inform which chi-square distribution to reference when determining your \( P \)-value for hypothesis testing.

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Most popular questions from this chapter

The article from which the data in Exercise 20 was obtained also gave the accompanying data on the composite mass/outer fabric mass ratio for highquality fabric specimens. \(\begin{array}{lllllll}1.15 & 1.40 & 1.34 & 1.29 & 1.36 & 1.26 & 1.22 \\ 1.40 & 1.29 & 1.41 & 1.32 & 1.34 & 1.26 & 1.36 \\ 1.36 & 1.30 & 1.28 & 1.45 & 1.29 & 1.28 & 1.38 \\ 1.55 & 1.46 & 1.32 & & & & \end{array}\) MINITAB gave \(r=.9852\) as the value of the Ryan- Joiner test statistic and reported that \(P\) value \(>.10\). Would you use the one-sample \(t\) test to test hypotheses about the value of the true average ratio? Why or why not?

Sorghum is an important cereal crop whose quality and appearance could be affected by the presence of pigments in the pericarp (the walls of the plant ovary). The article "A Genetic and Biochemical Study on Pericarp Pigments in a Cross Between Two Cultivars of Grain Sorghum, Sorghum Bicolor" (Heredity, 1976: 413-416) reports on an experiment that involved an initial cross between CK60 sorghum (an American variety with white seeds) and Abu Taima (an Ethiopian variety with yellow seeds) to produce plants with red seeds and then a self-cross of the red-seeded plants. According to genetic theory, this \(F_{2}\) cross should produce plants with red, yellow, or white seeds in the ratio \(9: 3: 4\). The data from the experiment follows; does the data confirm or contradict the genetic theory? Test at level \(.05\) using the \(P\)-value approach. \begin{tabular}{l|ccc} Seed Color & Red & Yellow & White \\ \hline Observed Frequency & 195 & 73 & 100 \end{tabular}

A study of sterility in the fruit fly ("Hybrid Dysgenesis in Drosophila melanogaster: The Biology of Female and Male Sterility," Genetics, 1979: 161-174) reports the following data on the number of ovaries developed for each female fly in a sample of size 1,388 . One model for unilateral sterility states that each ovary develops with some probability \(p\) independently of the other ovary. Test the fit of this model using \(\chi^{2}\). \begin{tabular}{l|ccc} \(x=\) Number of Ovaries Developed & 0 & 1 & 2 \\ \hline Observed Count & 1212 & 118 & 58 \end{tabular}

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