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 56m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 17m
- 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 - ExcelBonus23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - ExcelBonus28m
- Confidence Intervals for Population Means - ExcelBonus25m
- 8. Sampling Distributions & Confidence Intervals: Proportion2h 10m
- 9. Hypothesis Testing for One Sample5h 8m
- Steps in Hypothesis Testing1h 6m
- Performing Hypothesis Tests: Means1h 4m
- Hypothesis Testing: Means - ExcelBonus42m
- Performing Hypothesis Tests: Proportions37m
- Hypothesis Testing: Proportions - ExcelBonus27m
- Performing Hypothesis Tests: Variance12m
- Critical Values and Rejection Regions28m
- Link Between Confidence Intervals and Hypothesis Testing12m
- Type I & Type II Errors16m
- 10. Hypothesis Testing for Two Samples5h 37m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - ExcelBonus28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - ExcelBonus12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - ExcelBonus9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - ExcelBonus21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - ExcelBonus12m
- Two Variances and F Distribution29m
- Two Variances - Graphing CalculatorBonus16m
- 11. Correlation1h 24m
- 12. Regression3h 33m
- Linear Regression & Least Squares Method26m
- Residuals12m
- Coefficient of Determination12m
- Regression Line Equation and Coefficient of Determination - ExcelBonus8m
- Finding Residuals and Creating Residual Plots - ExcelBonus11m
- Inferences for Slope31m
- Enabling Data Analysis ToolpakBonus1m
- Regression Readout of the Data Analysis Toolpak - ExcelBonus21m
- Prediction Intervals13m
- Prediction Intervals - ExcelBonus19m
- Multiple Regression - ExcelBonus29m
- Quadratic Regression15m
- Quadratic Regression - ExcelBonus10m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA2h 29m
13. Chi-Square Tests & Goodness of Fit
Goodness of Fit Test
Problem 12.T.2
Textbook Question
A researcher wanted to determine if the distribution of educational attainment of Americans today is different from the distribution in 2000. The distribution of educational attainment in 2000 was as follows:

Source: Statistical Abstract of the United States.
The researcher randomly selects 500 Americans, learns their levels of education, and obtains the data shown in the table. Do the data suggest that the distribution of educational attainment has changed since 2000? Use the α = 0.1 level of significance.

Verified step by step guidance1
Step 1: Identify the type of hypothesis test to use. Since we want to compare the observed frequencies of educational attainment in the sample to the expected frequencies based on the 2000 distribution, we will use a Chi-square goodness-of-fit test.
Step 2: State the null and alternative hypotheses. The null hypothesis \(H_0\) is that the distribution of educational attainment has not changed since 2000, meaning the observed frequencies follow the same distribution as in 2000. The alternative hypothesis \(H_a\) is that the distribution has changed.
Step 3: Calculate the expected frequencies for each education category using the relative frequencies from 2000 and the sample size of 500. For each category, the expected frequency \(E_i\) is calculated as \(E_i = 500 \times p_i\), where \(p_i\) is the relative frequency from the 2000 distribution.
Step 4: Compute the Chi-square test statistic using the formula:
\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]
where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each category.
Step 5: Determine the degrees of freedom, which is the number of categories minus 1 (i.e., \(df = 6 - 1 = 5\)). Then, compare the calculated Chi-square statistic to the critical value from the Chi-square distribution table at \(\alpha = 0.1\) and \(df = 5\). If the test statistic exceeds the critical value, reject the null hypothesis; otherwise, do not reject it.
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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 evaluates whether the observed frequency distribution of a categorical variable differs from a specified expected distribution. It compares observed counts to expected counts under the null hypothesis, using the chi-square statistic to measure discrepancies. A significant result suggests the distributions are different.
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Goodness of Fit Test
Null and Alternative Hypotheses
In hypothesis testing, the null hypothesis (H0) represents no change or difference—in this case, that the educational attainment distribution remains the same as in 2000. The alternative hypothesis (Ha) states that the distribution has changed. The test determines which hypothesis the data support.
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Step 1: Write Hypotheses
Level of Significance (α) and Decision Rule
The significance level α (here, 0.1) is the threshold for rejecting the null hypothesis. If the p-value from the test is less than α, we reject H0, indicating evidence of change. This controls the probability of a Type I error, or falsely detecting a difference when none exists.
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