[DATA] Pedestrian Deaths A researcher wanted to determine whether pedestrian deaths were uniformly distributed over the days of the week. She randomly selected 300 pedestrian deaths, 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). Test the belief that the day of the week on which a fatality happens involving a pedestrian occurs with equal frequency 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 12.2.7a
Textbook Question
[NOW WORK] Job Satisfaction Is there an association between one’s level of education and satisfaction with work? A random sample of 5244 employed individuals were asked to disclose their highest level of education and satisfaction with their work/job. The results are shown in the table below. The data are from the General Social Survey.
[Image]
a. Compute the expected values of each cell under the assumption of independence.
Verified step by step guidance1
Identify the variables involved: one is the level of education (categorical variable with several categories), and the other is satisfaction with work (also categorical with several categories). The goal is to test for independence between these two categorical variables.
Recall that under the assumption of independence, the expected frequency for each cell in a contingency table is calculated using the formula:
\[ E_{ij} = \frac{(\text{Row total for row } i) \times (\text{Column total for column } j)}{\text{Grand total}} \]
Obtain the row totals (sum of counts for each education level), the column totals (sum of counts for each satisfaction level), and the grand total (total number of individuals, which is 5244 in this case) from the given data table.
For each cell in the table, multiply the corresponding row total by the column total, then divide by the grand total to find the expected count for that cell under the assumption of independence.
Repeat this calculation for every cell in the table to complete the expected frequency table, which will be used later for hypothesis testing (such as a chi-square test of independence).
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chi-Square Test of Independence
This test evaluates whether two categorical variables are related or independent. It compares observed frequencies in a contingency table to expected frequencies calculated under the assumption of independence. A significant difference suggests an association between the variables.
Recommended video:
Guided course
Independence Test
Expected Cell Frequencies
Expected frequencies represent the counts we would expect in each cell of a contingency table if the two variables were independent. They are calculated by multiplying the row total by the column total and dividing by the overall sample size.
Recommended video:
Guided course
Contingency Tables & Expected Frequencies
Contingency Table
A contingency table displays the frequency distribution of variables and helps summarize the relationship between categorical variables. It organizes data into rows and columns, showing counts for each combination of categories.
Recommended video:
Guided course
Contingency Tables & Expected Frequencies
Watch next
Master Goodness of Fit Test with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
Textbook Question
15
views
