Bicycle Deaths A researcher wanted to determine whether bicycle deaths were uniformly distributed over the days of the week. She randomly selected 200 deaths that involved a bicycle, recorded the day of the week on which the death occurred, and obtained the following results (the data are based on information obtained from the Insurance Institute for Highway Safety). Is there reason to believe that bicycle fatalities occur with equal frequency with respect to day of the week at the alpha = 0.05 level of significance?
Table of contents
- 1. Intro to Stats and Collecting Data1h 14m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
- 6. Normal Distribution and Continuous Random Variables2h 11m
- 7. Sampling Distributions & Confidence Intervals: Mean3h 23m
- Sampling Distribution of the Sample Mean and Central Limit Theorem19m
- Distribution of Sample Mean - Excel23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - Excel28m
- Confidence Intervals for Population Means - Excel25m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 25m
- 9. Hypothesis Testing for One Sample3h 29m
- 10. Hypothesis Testing for Two Samples4h 50m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - Excel28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - Excel12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - Excel9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - Excel21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - Excel12m
- 11. Correlation1h 24m
- 12. Regression1h 50m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA1h 57m
13. Chi-Square Tests & Goodness of Fit
Goodness of Fit Test
Problem 12.R.1
Textbook Question
Roulette Wheel A pit boss suspects that a roulette wheel is out of balance. A roulette wheel has 18 black slots, 18 red slots, and 2 green slots. The pit boss spins the wheel 500 times and records the following frequencies:

Is the wheel out of balance? Use the α = 0.05 level of significance.
Verified step by step guidance1
Step 1: Define the null and alternative hypotheses. The null hypothesis (H0) assumes the roulette wheel is balanced, meaning the probabilities of landing on Black, Red, and Green are proportional to the number of slots: Black = 18/38, Red = 18/38, Green = 2/38. The alternative hypothesis (H1) is that the wheel is not balanced, so the observed frequencies differ significantly from the expected frequencies.
Step 2: Calculate the expected frequencies for each outcome based on the total number of spins (500) and the theoretical probabilities. Use the formula: \(E_i = n \times p_i\), where \(n = 500\) and \(p_i\) is the probability for each color (Black, Red, Green).
Step 3: Use the Chi-square goodness-of-fit test to compare observed and expected frequencies. The test statistic is calculated as: \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\), where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each category.
Step 4: Determine the degrees of freedom for the test. Since there are 3 categories, degrees of freedom = number of categories - 1 = 2.
Step 5: Find the critical value from the Chi-square distribution table at \(\alpha = 0.05\) significance level and 2 degrees of freedom. Compare the calculated \(\chi^2\) statistic to this critical value to decide whether to reject or fail to reject the null hypothesis.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chi-Square Goodness-of-Fit Test
This test determines if observed categorical data differ significantly from expected frequencies. It compares the observed counts to expected counts under a specific hypothesis, such as a fair roulette wheel, to assess if deviations are due to chance or indicate imbalance.
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Goodness of Fit Test
Expected Frequencies in Probability
Expected frequencies are calculated by multiplying the total number of trials by the theoretical probabilities of each outcome. For a balanced roulette wheel, these probabilities are based on the proportion of black, red, and green slots, guiding the expected counts for comparison.
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Contingency Tables & Expected Frequencies
Significance Level and Hypothesis Testing
The significance level (α = 0.05) sets the threshold for rejecting the null hypothesis, which assumes the wheel is balanced. If the test statistic exceeds the critical value at this level, it suggests the wheel is out of balance with 95% confidence.
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