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In the early 2000s, Deep Down Mining Company implemented new safety guidelines. Prior to these new guidelines, management expected there to be no accidents in 40 percent of the months, one accident in 30 percent of the months, two accidents in 20 percent of the months, and three accidents in 10 percent of the months. Over the last 10 years, or 120 months, there have been 46 months in which there were no accidents, 40 months in which there was one accident, 22 months in which there were two accidents, and 12 months in which there were 3 accidents. At the .05 significance level can the management at Deep Down conclude that there has been a change in the monthly accident distribution?

Short Answer

Expert verified
No, there's no evidence of a change in accident distribution at 0.05 significance.

Step by step solution

01

Define the Hypotheses

To start the hypothesis test, we define the null and alternative hypotheses. \( H_0 \): The observed distribution of accidents follows the expected distribution. \( H_a \): The observed distribution of accidents does not follow the expected distribution.
02

Set Up the Expected Frequencies

Given that there are 120 months of data, we calculate the expected frequencies for each accident category based on the management's expectations: - 0 accidents: 120 × 0.40 = 48 months - 1 accident: 120 × 0.30 = 36 months - 2 accidents: 120 × 0.20 = 24 months - 3 accidents: 120 × 0.10 = 12 months.
03

Compute the Chi-Square Test Statistic

The Chi-Square test statistic is calculated using the formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] Where \(O_i\) are the observed frequencies and \(E_i\) are the expected frequencies.Calculations:- For 0 accidents: \( \frac{(46-48)^2}{48} = \frac{4}{48} \approx 0.083 \)- For 1 accident: \( \frac{(40-36)^2}{36} = \frac{16}{36} \approx 0.444 \)- For 2 accidents: \( \frac{(22-24)^2}{24} = \frac{4}{24} \approx 0.167 \)- For 3 accidents: \( \frac{(12-12)^2}{12} = 0 \)Summing these values gives \( \chi^2 = 0.083 + 0.444 + 0.167 + 0 = 0.694 \).
04

Determine the Critical Value

For a Chi-Square test with 3 degrees of freedom (number of categories - 1) and a significance level of \(\alpha = 0.05\), we look up the critical value in a Chi-Square distribution table, which is approximately 7.815.
05

Make the Decision

Compare the calculated Chi-Square statistic \(0.694\) with the critical value \(7.815\). Since \(0.694 < 7.815\), we fail to reject the null hypothesis.
06

Conclude

At the 0.05 significance level, there is not enough evidence to conclude that there has been a change in the monthly accident distribution.

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Key Concepts

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

Chi-Square Test
The Chi-Square test is a statistical method used to examine if there is a significant difference between observed and expected frequencies in categorical data. It helps researchers determine whether any discrepancies between what is observed and what is expected are due to chance or indicate a real effect or change.

It's like a detective tool that tells you if some events happen purposely or just by random chance. In our particular example of accident rates at Deep Down Mining Company, the Chi-Square test checks whether the safety guidelines introduced really had a visible effect on reducing or changing accidents.

To calculate the Chi-Square statistic, we use the formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Where:
  • \( O_i \) is the observed frequency for a category.
  • \( E_i \) is the expected frequency for the same category.
After performing these calculations, we compare the computed test statistic to a critical value to make an informed decision about our hypotheses.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold used in hypothesis testing to determine whether the observed data is statistically significant. It's essentially a risk measure—the risk you're willing to take of making a wrong decision. In practical terms, it tells you how confident you can be about your test results.

In the Deep Down Mining Company example, the chosen significance level is 0.05. This means there is a 5% chance that any observed changes in the accident rate are due to random variation rather than a true alteration in distribution.

Deciding on a significance level beforehand helps in maintaining objectivity and consistency. If the probability (p-value) calculated from the data is lower than the significance level, it means there is enough evidence to reject the null hypothesis; otherwise, you fail to reject it.
Expected Frequencies
Expected frequencies are the number of occurrences you would anticipate for each category if everything happens perfectly according to the initial hypothesis. These expectations are calculated using prior knowledge or theory. In statistics, this acts as a benchmark to compare against actual data.

In our example, Deep Down Mining Company's management expected the distribution of accidents as 40% no accidents, 30% one accident, 20% two accidents, and 10% three accidents.

To calculate these expected frequencies for 120 months, you perform the following computations:
  • 0 accidents: \( 120 \times 0.40 = 48 \) months
  • 1 accident: \( 120 \times 0.30 = 36 \) months
  • 2 accidents: \( 120 \times 0.20 = 24 \) months
  • 3 accidents: \( 120 \times 0.10 = 12 \) months
By comparing these expected values with the observed ones, you can see if there has been a meaningful shift in the accident patterns.
Degrees of Freedom
Degrees of freedom (df) is a crucial concept in statistics that represents the number of values in a calculation that are free to vary. An easy way to think about it is to imagine degrees of freedom as the number of choices you have for changing data points without breaking any mathematical conditions.

In a Chi-Square test, the degrees of freedom is calculated as the number of categories minus one:
  • df = Number of categories - 1
In the case of Deep Down Mining, where there are four categories of accidents (0, 1, 2, and 3), the degrees of freedom are calculated as 4 - 1 = 3.

The degrees of freedom help determine the critical value from a Chi-Square distribution table. This value is crucial for deciding whether to reject the null hypothesis. It gives context to where our test statistic falls on the distribution and helps validate the test results.

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