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{~A}\( forestry official is comparing the causes of forest fires in two regions, \)\mathrm{A}\( and \)\mathrm{B}\(. The following table shows the causes of fire for 76 randomly selected recent fires in these two regions. \begin{tabular}{lcccc} \hline & Arson & Accident & Lightning & Unknown \\ \hline Region A & 6 & 9 & 6 & 10 \\ Region B & 7 & 14 & 15 & 9 \\ \hline \end{tabular} Test at the \)5 \%$ significance level whether causes of fire and regions of fires are related.

Short Answer

Expert verified
The solution involves a chi-square test. After calculating expected frequencies, a chi-square statistic value is found. The null hypothesis (no association between causes of fire and regions of fires) is rejected if the calculated chi-square statistic is greater than the critical value from chi-square distribution table at 5% significance level. The exact decision (reject or fail to reject null hypothesis) would depend on the calculated chi-square value.

Step by step solution

01

State the Hypotheses

The null hypothesis is that the causes of fire are independent of the region, and the alternative hypothesis is that they are not independent. Symbolically, these can be written as: \[ H_0 : Causes of fire \ and \ region \ are \ independent \] \[ H_a : Causes of fire \ and \ region \ are \ not \ independent \]
02

Calculate Observed and Expected Frequencies

First calculate the row and column totals, which are required to find the expected frequencies. Then calculate expected frequencies using the formula: \[ E_{i,j} = (row \ total_i * column \ total_j) / grand \ total \] For example, for the 'Arson' cause in 'Region A', the expected frequency would be calculated as: \[ E_{A,Arson} = (25*13)/76 \] Repeat this process for all cells in the table.
03

Calculate the Chi-Square Statistic

The Chi-square statistic is calculated as: \[ \chi^2 = \sum((O_{i,j} - E_{i,j})^2/E_{i,j}) \] where \(O_{i,j}\) refers to the observed frequency and \(E_{i,j}\) refers to the expected frequency. Calculate the chi-square statistic using the observed and expected frequencies.
04

Determine the Critical Value and Make a Decision

The critical value corresponds to the 5% significance level and is found in the chi-square distribution table. The degrees of freedom for this test would be (number of rows - 1) * (number of columns - 1) = (2-1)*(4-1) = 3. If the calculated chi-square statistic is greater than the critical value, then reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical technique used to decide whether there is enough evidence in a set of data to infer that a certain condition is true for the entire population. In our example of forest fires, we want to determine if the cause of fires depends on the region they occur in. This exercise begins with formulating two hypotheses:
  • The null hypothesis (\(H_0\) ): Assumes there is no dependency between the cause of fire and the region, meaning they are independent.
  • The alternative hypothesis (\(H_a\) ): Suggests that the cause of fire and the region are not independent, implying a relationship exists.
We test these hypotheses using a statistical method called the Chi-Square test. The goal is to determine whether any observed differences between data sets are due to chance or indicate a real variance.
Significance Level
The significance level, often denoted by alpha (\( \alpha \)), is a threshold set by the researcher before conducting hypothesis testing. It helps to decide whether to reject the null hypothesis. In this case, a 5% significance level (\( \alpha = 0.05 \)) is used.
A 5% significance level means that there is a 5% risk of concluding that a difference exists when there is none. This is called a Type I error. When the p-value obtained from our test is less than or equal to 0.05, the null hypothesis is rejected, indicating that the observed pattern is statistically significant.
Setting the significance level helps in making informed and reliable decisions based on the data, ensuring the results of the test are meaningful and not due to random fluctuations.
Degrees of Freedom
Degrees of freedom (df) are a crucial part of statistical tests, representing the number of values in the final calculation that are free to vary. They are determined based on the sample size and the nature of the data.
For the Chi-Square test, the degrees of freedom are calculated using the formula:
\[(df) = (number \ of \ rows - 1) \times (number \ of \ columns - 1)\]
In the forest fire example, there are 2 regions (rows) and 4 causes (columns). Therefore, the degrees of freedom are calculated as \((2-1) \times (4-1) = 3\).Knowing the degrees of freedom helps us determine the critical value from the Chi-Square distribution table. This critical value is used to compare against the calculated Chi-Square statistic to decide if we reject or fail to reject the null hypothesis.
Observed and Expected Frequencies
Observed frequencies are the actual counts obtained from the data under investigation. In the context of the forest fire example, these frequencies show how often each cause occurred in both regions. On the other hand, expected frequencies represent what would be expected if the null hypothesis were true.
The expected frequency for any cell in the table is calculated by:
\[E_{i,j} = \frac{(\text{row total}_i \times \text{column total}_j)}{\text{grand total}}\]
For example, to find the expected frequency of arson-caused fires in Region A, the formula accounts for the proportion of total fires and applies it to specific categories.
The combination of comparing observed and expected frequencies is crucial in the Chi-Square test. It helps determine if deviations between observed and expected counts are significant or just due to random variations, ultimately guiding the hypothesis outcome.

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Most popular questions from this chapter

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.

Two drugs were administered to two groups of randomly assigned 60 and 40 patients, respectively, to cure the same disease. The following table gives information about the number of patients who were cured and not cured by each of the two drugs. \begin{tabular}{lcc} \hline & Cured & Not Cured \\ \hline Drug I & 44 & 16 \\ Drug II & 18 & 22 \\ \hline \end{tabular} Test at the \(1 \%\) significance level whether or not the two drugs are similar in curing and not curing the patients.

The makers of Flippin' Out Pancake Mix claim that one cup of their mix contains 11 grams of sugar. However, the mix is not uniform, so the amount of sugar varies from cup to cup. One cup of mix was taken from each of 24 randomly selected boxes. The sample variance of the sugar measurements from these 24 cups was \(1.47\) grams. Assume that the distribution of sugar content is approximately normal. a. Construct the \(98 \%\) confidence intervals for the population variance and standard deviation. b. Test at the \(1 \%\) significance level whether the variance of the sugar content per cup is greater than \(1.0\) gram.

The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at the \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.

Find the value of \(\chi^{2}\) for 12 degrees of freedom and an area of \(.025\) in the right tail of the chi-square distribution curve.

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