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Chapter 10: Problem 6

The contingency table shows the results of a random sample of individuals by gender and type of vehicle owned. At \(\alpha=0.01\), can you conclude that gender is related to the type of vehicle owned? $$ \begin{array}{|l|r|c|c|c|} \cline { 2 - 5 } \multicolumn{1}{c|}{} & \multicolumn{4}{c|}{\text { Type of vehicle owned }} \\ \hline \text { Gender } & \text { Car } & \text { Truck } & \text { SUV } & \text { Van } \\ \hline \text { Male } & 85 & 95 & 44 & 8 \\ \text { Female } & 110 & 73 & 61 & 4 \\ \hline \end{array} $$

Short Answer

Expert verified
Yes, gender is related to the type of vehicle owned if the chi-square statistic is greater than the critical value.

Step by step solution

01

Understand the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) posits that gender and the type of vehicle owned are independent, meaning there is no relationship between them. The alternative hypothesis (\(H_1\)) suggests that there is a relationship between gender and the type of vehicle owned.
02

Set the Significance Level

The significance level \(\alpha\) is given as 0.01. This means we will reject the null hypothesis if the p-value from our test is less than 0.01.
03

Create the Observed Frequency Table

The observed frequency table is already provided: \[\begin{array}{|l|r|c|c|c|} \hline \text { Gender } & \text { Car } & \text { Truck } & \text { SUV } & \text { Van } \\hline \text { Male } & 85 & 95 & 44 & 8 \\text { Female } & 110 & 73 & 61 & 4 \\hline \end{array} \]
04

Calculate Expected Frequencies

Using the formula for expected frequency \(E = \frac{( ext{Row Total} \times ext{Column Total})}{ ext{Grand Total}}\), calculate the expected frequency for each cell in the contingency table. Calculate the row, column, and grand totals first. The grand total is 480.For example, the expected frequency for Male Car: \(E = \frac{232 \times 195}{480} = 94.25\) and so on for each cell.
05

Compute Chi-Square Statistic

For each cell, calculate \((O - E)^2 / E\), where \(O\) is the observed frequency and \(E\) is the expected frequency calculated in the previous step. Sum these values across all cells to get the Chi-square statistic (\(\chi^2\)).
06

Determine Degrees of Freedom

The degrees of freedom for this test are calculated as \((r - 1) \times (c - 1)\), where \(r\) is the number of rows, and \(c\) is the number of columns. Here, \(df = (2 - 1) \times (4 - 1) = 3\).
07

Find the Critical Value and P-Value

Using the Chi-square distribution table, find the critical value for \(df = 3\) at \(\alpha = 0.01\). The critical value is approximately 11.345. Compare the \(\chi^2\) statistic to this critical value or find the p-value for the chi-square statistic.
08

Make a Decision

If the \(\chi^2\) statistic is greater than the critical value or the p-value is less than 0.01, reject the null hypothesis. If not, do not reject it.

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Key Concepts

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

Contingency Table
In statistical analysis, a contingency table is like a detailed data map. It effectively organizes and represents categorical data. It's a simple grid format that displays the frequency distribution of variables and shows how different variables relate to each other. In our exercise, the contingency table categorizes individuals by gender and the type of vehicle they own.

Here's what each part means:
  • Rows: Represent different categories of one variable, in this case, 'Gender' (Male and Female).
  • Columns: Represent different categories of another variable, 'Type of Vehicle Owned' (Car, Truck, SUV, Van).
  • Cells: Each cell shows the count or frequency of occurrences for the intersection of a particular row and column.
Using this structured format allows researchers to clearly understand and analyze the relationships between different data points.
Null Hypothesis
The null hypothesis ( H_0 ) is a fundamental concept in statistical testing. It proposes that there is no relationship or difference between the variables being studied. For our chi-square exercise, the null hypothesis indicates that gender is not related to the type of vehicle owned.

Here鈥檚 why the null hypothesis is important:
  • Starting Point: It provides a baseline assumption so that any deviation can be statistically tested.
  • Objectivity: Ensures that analysis does not start with bias or preconceptions.
  • Testing Ground: Acts as the hypothesis that researchers try to challenge with evidence.
  • Decision Making: Helps determine if findings are due to chance or a true effect.
In statistical tests like a chi-square test, the null hypothesis is either rejected or not rejected based on the data.
Alternative Hypothesis
The alternative hypothesis ( H_1 ) is the counterpart to the null hypothesis. While the null hypothesis assumes no effect or relationship, the alternative hypothesis suggests that there is indeed a significant relationship or effect to be discovered. In our exercise, the alternative hypothesis posits that there is a relationship between gender and the type of vehicle a person owns.

Key aspects of the alternative hypothesis include:
  • Contradiction to Null: It challenges the assumption of the null hypothesis.
  • Evidence-Based: Supersedes the null hypothesis only if strong empirical evidence exists.
  • Research Objective: Represents what the researcher aims to prove.
  • Outcome Determinant: If accepted, leads to new understanding or insights.
Accepting the alternative hypothesis implies that the data provides enough evidence of a real association between the studied variables.
Degrees of Freedom
Degrees of freedom (\text{df}) are crucial in statistical tests and determine how many values in a calculation are free to vary. They are derived from the number of categories minus the constraints necessarily applied to get meaningful analysis. For our contingency table with 2 genders and 4 types of vehicles:

The formula is:
\[\text{df} = (r - 1) \times (c - 1)\]
Where:
  • r: is the number of rows (2 gender categories).
  • c: is the number of columns (4 vehicle types).
This results in:
\[\text{df} = (2 - 1) \times (4 - 1) = 3\]

Degrees of freedom influence how the chi-square statistic is interpreted against the critical value. They essentially provide the test with the needed "flexibility" to fairly assess evidence and make conclusions about the null hypothesis.

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Most popular questions from this chapter

A quality technician claims that the variance of the insert diameters produced by a plastic company's new injection mold for automobile dashboard inserts is less than the variance of the insert diameters produced by the company's current mold. The table shows samples of insert diameters (in centimeters) for both the current and new molds. At \(\alpha=0.05\), can you support the technician's claim? \begin{aligned} &\begin{array}{|l|l|l|l|l|l|l|} \hline \text { New } & 9.611 & 9.618 & 9.594 & 9.580 & 9.611 & 9.597 \\ \hline \text { Current } & 9.571 & 9.642 & 9.650 & 9.651 & 9.596 & 9.636 \\ \hline \end{array}\\\ &\begin{array}{|l|l|l|l|l|l|l|} \hline \text { New } & 9.638 & 9.568 & 9.605 & 9.603 & 9.647 & 9.590 \\ \hline \text { Current } & 9.570 & 9.537 & 9.641 & 9.625 & 9.626 & 9.579 \\ \hline \end{array} \end{aligned}

In Exercises 9-12, find the critical \(F\)-value for a right-tailed test using the level of significance \(\alpha\) and degrees of freedom d.f. \(_{\cdot N}\) and \(d . f \cdot D\). $$ \alpha=0.01, \text { d.f. }_{\mathrm{N}}=12, \text { d.f. }_{\mathrm{D}}=10 $$

The contingency table shows the distribution of a random sample of fatal pedestrian motor vehicle collisions by time of day and gender in a recent year. At \(\alpha=0.10\), can you conclude that time of day and gender are related? (Adapted from National Highway Traffic Safety Administration) $$ \begin{array}{|l|c|c|c|c|} \cline { 3 - 5 } \multicolumn{1}{c|}{} & \multicolumn{4}{c|}{\text { Time of day }} \\ \hline & \mathbf{1 2}_{\text {A.M. }-} & \mathbf{6}_{\text {A.M. }-} & \mathbf{1 2}_{\text {P.M. }-} & \mathbf{6}_{\text {P.M. }-} \\ \hline \text { Gender } & \mathbf{5 : 5 9}_{\text {A.M. }} & \mathbf{1 1 : 5 9}_{\text {A.M. }} & \mathbf{5 : 5 9} \text { P.M. } & \mathbf{1 1 : 5 9} \text { P.M. } \\ \hline \text { Male } & 611 & 595 & 884 & 911 \\ \text { Female } & 260 & 354 & 563 & 552 \\ \hline \end{array} $$

In Exercises 13-16, find the critical \(F\)-value for a two-tailed test using the level of significance \(\alpha\) and degrees of freedom d.f. \(_{\cdot N}\) and \(d . f_{\cdot D}\). $$ \alpha=0.05, \text { d.f. }_{\mathrm{N}}=9, \text { d.f. }_{\mathrm{D}}=8 $$

An agricultural analyst is comparing the wheat production in Oklahoma counties. The analyst claims that the variation in wheat production is greater in Garfield County than in Kay County. A sample of 21 Garfield County farms has a standard deviation of \(0.76\) bushel per acre. A sample of 16 Kay County farms has a standard deviation of \(0.58\) bushel per acre. At \(\alpha=0.10\), can you support the analyst's claim?
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