Construction About 63% of the residents in a town are in favor of building a new high school. One hundred five residents are randomly selected. What is the probability that the sample proportion in favor of building a new school is less than 55%? Interpret your result.
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
8. Sampling Distributions & Confidence Intervals: Proportion
Sampling Distribution of Sample Proportion
Problem 8.T.7
Textbook Question
Net worth is defined as total assets (value of house, cars, money, etc.) minus total liabilities (mortgage balance, credit card debt, etc.). According to a recent study by TNS Financial Services, 7% of American households had a net worth in excess of \$1 million (excluding their primary residence). A random sample of 1000 American households results in 82 having a net worth in excess of \$1 million. Explain why the results of this survey do not necessarily imply that the proportion of households with a net worth in excess of \$1 million has increased.
Verified step by step guidance1
Identify the population proportion given by the study, which is \(p = 0.07\) (7% of households have net worth over \$1 million).
Recognize that the sample proportion from the survey is \(\hat{p} = \frac{82}{1000} = 0.082\) (8.2%).
Understand that sample proportions vary due to random sampling variability, so the observed sample proportion may differ from the true population proportion by chance.
Calculate the standard error (SE) of the sample proportion using the formula:
\(SE = \sqrt{\frac{p(1-p)}{n}}\)
where \(n=1000\) is the sample size.
Use the standard error to assess whether the observed sample proportion is significantly different from the population proportion by considering how many standard errors away 0.082 is from 0.07; if it is within a reasonable range, the difference can be attributed to sampling variability rather than a true increase.
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.
Sampling Variability
Sampling variability refers to the natural differences that occur between samples drawn from the same population. Even if the true population proportion remains constant, different random samples can yield different results due to chance. This means the observed 82 out of 1000 households might not reflect a real change but just random fluctuation.
Recommended video:
Sampling Distribution of Sample Proportion
Hypothesis Testing and Statistical Significance
Hypothesis testing helps determine if an observed difference is likely due to chance or represents a true change in the population. Without conducting a formal test, such as a z-test for proportions, we cannot conclude that the increase from 7% to 8.2% is statistically significant or meaningful.
Recommended video:
Guided course
Step 2: Calculate Test Statistic
Confidence Intervals
A confidence interval provides a range of plausible values for the population proportion based on the sample data. If the interval for the sample proportion includes the original 7%, it suggests the observed increase might not be a real change but within expected sampling error.
Recommended video:
Introduction to Confidence Intervals
Watch next
Master Using the Normal Distribution to Approximate Binomial Probabilities with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
Textbook Question
44
views
