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Accident data were analyzed to determine the numbers of fatal accidents for automo- biles of three sizes. The data for 346 accidents are as follows: Do the data indicate that the frequency of fatal accidents is dependent on the size of automobiles? Test using a \(5 \%\) significance level.

Short Answer

Expert verified
Explain your answer using the steps outlined. Answer: [Include your answer based on the comparison of the test statistic and critical value]

Step by step solution

01

Create a contingency table

We will create a contingency table with the given data. Assume that the table represents all the data collected from the 346 accidents.
02

Calculate the expected frequencies for each cell

Next, compute the expected frequencies for each cell by multiplying the row and column total and dividing by the overall total (in this case, 346). To do this for each cell, use the formula: $$Expected\ Frequency = \frac{Row\ Total \times Column\ Total}{Total\ Observations}$$
03

Compute the test statistic

Calculate the Chi-Square test statistic from the observed and expected frequencies using the formula: $$\chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} \frac{(Observed_{ij} - Expected_{ij})^2}{Expected_{ij}}$$ where \(Observed_{ij}\) is the observed frequency, \(Expected_{ij}\) is the expected frequency, \(r\) represents the number of rows, and \(c\) represents the number of columns.
04

Find the critical value and degrees of freedom

For a 5% significance level, find the critical value in a Chi-Square table based on the degrees of freedom. $$Degrees\ of\ Freedom = (number\ of\ rows\ -\ 1) \times (number\ of\ columns\ -\ 1)$$
05

Decide whether to reject or fail to reject the null hypothesis

Compare the test statistic calculated in step 3 with the critical value from step 4. If the test statistic exceeds the critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. In this case, after comparing the calculated test statistic and critical value, we can determine whether the frequency of fatal accidents is dependent or independent of the size of automobiles at a 5% significance level.

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Key Concepts

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

Contingency Table
When studying the relationship between categorical variables, the contingency table is a vital tool. Imagine a spreadsheet where data are organized into rows and columns, which makes it easy to visualize any connections. For the accident data in our exercise, we categorize the accidents by car sizes and outcomes (fatal or not fatal), with each accident falling into one specific cell.

This layout helps us to compare the observed frequencies—the actual data collected—with the expected frequencies—the numbers we'd anticipate if there was no relationship between car size and accident severity. Needless to say, creating a contingency table is always the first step in a Chi-Square Test of Independence and sets the stage for further calculations.
Expected Frequency
Once you've constructed your contingency table, the next step is to calculate the expected frequency for each cell in the table. This represents the frequency you would expect to observe if there was no association between the variables—in our case, if car size had no impact on the severity of accidents.

The formula for finding expected frequency is relatively straightforward; multiply the total for each row by each column's total, then divide by the grand total of all observations. However, it's critical to do this for every cell in the contingency table to compare the expected and observed values accurately.
Test Statistic
The test statistic is the number that tells you how much your observed data deviates from what was expected. In Chi-Square tests, the formula takes each cell's observed and expected frequencies and looks at the square of the difference—divided by the expected frequency.

This calculation is done for every cell, and then we sum up all these individual calculations. The larger this Chi-Square test statistic is, the more evidence we have that there could be an actual relationship between the variables—in this scenario, car size and accident fatality.
Degrees of Freedom
Degrees of freedom might sound like a complex term, but it essentially tells us how many values in a calculation are free to vary. It is crucial for determining the critical value against which we'll compare our test statistic. In a Chi-Square test, the degrees of freedom are calculated by subtracting one from the number of rows and the number of columns and then multiplying those two numbers together.

The degrees of freedom allow us to navigate the Chi-Square distribution table where we find the critical value. This value will guide us to either reject or keep the null hypothesis.
Null Hypothesis
The null hypothesis acts as our benchmark assumption. It's the 'status quo' that claims there is no relationship between the variables we are examining. For our car accident data, it says car size doesn't change the frequency of fatal accidents.

It's only after computing the Chi-Square test statistic and comparing it to the critical value linked with our degrees of freedom, that we decide if this null hypothesis can be rejected. If the test statistic is higher than the critical value, we reject it, pointing towards a significant association between the variables. Otherwise, we 'fail to reject' it, which means we stick with the null hypothesis as there's not enough evidence to suggest a relationship.

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Most popular questions from this chapter

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?

Suppose you are interested in following two independent traits in snap peas- seed texture \((\mathrm{S}=\mathrm{smooth}, \mathrm{s}=\) wrinkled \()\) and seed color \((\mathrm{Y}=\) yellow \(\mathrm{y}=\) green \()-\) in a second-generation cross of heterozygous parents. Mendelian theory states that the number of peas classified as smooth and yellow, wrinkled and yellow, smooth and green, and wrinkled and green should be in the ratio 9: 3: 3: 1 . Suppose that 100 randomly selected snap peas have \(56,19,17,\) and 8 in these respective categories. Do these data indicate that the 9: 3: 3: 1 model is correct? Test using \(\alpha=.01\).

The "sandwich generation" refers to middle-aged Americans who are either providing support for an aging parent while raising child under 18 or supporting a child over 18 . The information that follows summarizes the results of a poll \(^{8}\) regarding support for a parent 65 years or older by demographic group. $$ \begin{array}{lrrr} \hline \text { Provide Financial } & & & \\ \text { Support } & \text { Yes } & \text { No } & \text { Total } \\ \hline \text { Hispanics } & 134 & 66 & 200 \\ \text { Blacks } & 86 & 114 & 200 \\ \text { Whites } & 49 & 151 & 200 \\ \hline \end{array} $$ a. Find the proportion of individuals providing financial support for their parents for each demographic group. b. Are there significant differences among the proportions providing financial support for their parents for these demographic groups of Americans? Use \(\alpha=.01\). Are your results consistent with the observed proportions in part a?

Conduct the appropriate test of specified probabilities using the information given. Write the null and alternative hypotheses, give the rejection region with \(\alpha=.05\) and calculate the test statistic. Find the approximate \(p\) -value for the test. Conduct the test and state your conclusions. The five categories are equally likely to occur, and the category counts are shown in the table: $$ \begin{array}{l|ccccc} \text { Category } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Observed Count } & 47 & 63 & 74 & 51 & 65 \end{array} $$

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? $$ p_{1}=.25, p_{2}=.15, p_{3}=.10, p_{4}=? $$

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