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Chapter 11: Problem 75

A chemical manufacturing company wants to locate a hazardous waste disposal site near a city of 50,000 residents and has offered substantial financial inducements to the city. Two hundred adults (110 women and 90 men) who are residents of this city are chosen at random. Sixty percent of these adults oppose the site, \(32 \%\) are in favor, and \(8 \%\) are undecided. Of those who oppose the site, \(65 \%\) are women; of those in favor, \(62.5 \%\) are men. Using the \(5 \%\) level of significance, can you conclude that opinions on the disposal site are dependent on gender?

Short Answer

Expert verified
The result of this statistical test will be whether or not to reject the null hypothesis. If the calculated chi-square value exceeds the critical value, we can conclude at the 5% significance level that opinion about the waste disposal site is dependent on gender.

Step by step solution

01

Degree of Freedom

The degrees of freedom (df) are given by (r-1) * (c-1), where r and c represent the number of rows and columns. There are two genders and three decisions hence df = (2-1) * (3-1) = 2.
02

Expected Values

The expected value of each cell in the table is calculated by (row_total * column_total) / overall_total. Firstly, calculate the total of males and females. Then calculate the total count for each opinion. Next, calculate the expected value for each cell in the table.
03

Chi-square Statistic

The Chi-square statistic is calculated as \(\sum_{i=1}^{n} \frac{(O_{i} - E_{i})^2}{E_{i}} \), where \(O_{i}\) is the observed count in each cell and \(E_{i}\) is the expected count. Calculate the chi-square statistic by subtracting expected from observed, squaring this value, dividing by expected values, and then summing all values.
04

Determine Significance

The critical value for Chi-square with 2 df at alpha = 0.05 (5% significance level) is 5.99. Compare the test statistic to the critical value. If the chi-square test statistic calculated in step 3 is greater than the critical value, reject the null hypothesis.

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Key Concepts

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

Degrees of Freedom
In statistics, degrees of freedom are important for understanding the variability in your data. In a chi-square test, they help to determine how much independence is available to make comparisons.
For this exercise, you calculate the degrees of freedom (often abbreviated as df) with the formula \((r-1) \times (c-1)\), where \(r\) is the number of different categories for one variable, and \(c\) is the number of categories for the second variable. In this case, there's two genders (men and women) and three opinions (oppose, favor, undecided). Thus, the degrees of freedom is 2: \((2-1) \times (3-1) = 2\).
Knowing the degrees of freedom helps in determining the critical value needed to validate your hypothesis against a statistical significance threshold.
Expected Values
Expected values in the chi-square test give a predicted count for each category, assuming there is no relationship between the variables. It is a crucial step, as we compare these with the observed values to assess independence or association.
To find the expected value for each category, use this formula: \( \text{Expected} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Overall Total}} \).
For the given problem, calculate the totals for each gender and each opinion category, then apply the formula. This prediction reflects how many individuals we expect in each category if gender and opinion were completely independent.
This calculation lays the groundwork for comparing observed data, to see if actual data deviates significantly from what was expected.
Observed Values
Observed values are the actual counts or quantities you record from your data. Unlike the expected values, these reflect real-world results.
In our example, these values are given as part of the survey results: the number of men and women in favor, opposed, or undecided about the hazardous waste disposal site. Observed values are the numbers tallied from people's responses during the survey.
The next step with these observed values is to compare them against the expected values, which gives insight into whether the variables鈥攊n this case, gender and opinion鈥攁ppear to be related or independent from one another.
Significance Level
The significance level, usually denoted as \(\alpha\), is the threshold of risk you're willing to take in rejecting a true null hypothesis. Commonly, this is set at 5% (or 0.05), indicating a 5% risk of concluding a relationship exists when it does not.
In this exercise, after calculating the chi-square test statistic, you compare it with a critical value from the chi-square distribution table with 2 degrees of freedom. If the test statistic exceeds this critical value, you reject the null hypothesis; meaning, you have significant evidence at the 5% level to suggest that opinions depend on gender.
Understanding the significance level helps establish the robustness and reliability of your statistical conclusions.

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Most popular questions from this chapter

A sample of 25 observations selected from a normally distributed population produced a sample variance of 35 . Construct a confidence interval for \(\sigma^{2}\) for each of the following confidence levels and comment on what happens to the confidence interval of \(\sigma^{2}\) when the confidence level decreases. a. \(1-\alpha=.99\) b. \(1-\alpha=.95\) c. \(1-\alpha=.90\)

A FOX News/Opinion Dynamics poll asked a question about gun control of random samples of 900 people each during May 2009 and March 2000 . The question asked was, "Which of the following do you think is more likely to decrease gun violence: better enforcement of existing gun laws or more laws and restrictions on obtaining guns?" The numbers in the following table are approximately the same as reported in the poll, which was reported to the nearest percent. \begin{tabular}{lcccc} \hline & Better Enforcement & More Laws and Restrictions & Both & Unsure \\ \hline May 2009 & 425 & 308 & 93 & 74 \\ March 2000 & 372 & 330 & 122 & 76 \\ \hline Source: http://www.pollingreport.com/guns.htm. \end{tabular} Test at the \(5 \%\) significance level whether the distributions of responses from May 2009 and March 2000 are significantly different.

A sample of 21 observations selected from a normally distributed population produced a sample variance of \(1.97\). a. Write the null and alternative hypotheses to test whether the population variance is greater than \(1.75\). b. Using \(\alpha=.025\), find the critical value of \(\chi^{2}\). Show the rejection and nonrejection regions on a chi-square distribution curve. c. Find the value of the test statistic \(x^{2}\). d. Using the \(2.5 \%\) significance level, will you reject the null hypothesis stated in part a?

Two random samples, one of 95 blue-collar workers and a second of 50 white- collar workers, were taken from a large company. These workers were asked about their views on a certain company issue. The following table gives the results of the survey. \begin{tabular}{lccc} \hline & \multicolumn{3}{c} { Opinion } \\ \cline { 2 - 4 } & Favor & Oppose & Uncertain \\ \hline Blue-collar workers & 44 & 39 & 12 \\ White-collar workers & 21 & 26 & 3 \\ \hline \end{tabular} Using the \(2.5 \%\) significance level, test the null hypothesis that the distributions of opinions are homogeneous for the two groups of workers.

Consider the following contingency table, which is based on a sample survey. \begin{tabular}{lrcr} \hline & Column 1 & Column 2 & Column 3 \\ \hline Row 1 & 137 & 64 & 105 \\ Row 2 & 98 & 71 & 65 \\ Row 3 & 115 & 81 & 115 \\ \hline \end{tabular} A. Write the null and alternative hypotheses for a test of independence for this table. b. Calculate the expected frequencies for all cells, assuming that the null hypothesis is true. c. For \(\alpha=.01\), find the critical value of \(\chi^{2}\). Show the rejection and nonrejection regions on the chi-square distribution curve. d. Find the value of the test statistic \(\chi^{2}\). e. Using \(\alpha=.01\), would you reject the null hypothesis?
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