Teen Prayer In 1995, 40% of adolescents stated they prayed daily. A researcher wants to know whether this percentage has become higher since then. He surveys 40 adolescents and finds that 18 pray on a daily basis. Is there enough evidence to support the proportion of adolescents who pray daily has increased at the α = 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
9. Hypothesis Testing for One Sample
Performing Hypothesis Tests: Proportions
Problem 10.2.14b
Textbook Question
You Explain It! ESPSuppose an acquaintance claims to have the ability to determine the birth month of randomly selected individuals. To test such a claim, you randomly select 80 individuals and ask the acquaintance to state the birth month of the individual. If the individual has the ability to determine birth month, then the proportion of correct birth months should exceed 1/12 ≈ 0.083, the rate one would expect from simply guessing.
b. Suppose the individual was able to guess nine correct birth months. The P-value for such results is 0.1726. Explain what this P-value means and write a conclusion for the test.
Verified step by step guidance1
Understand the null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)): Here, \(H_0\) states that the individual is just guessing, so the true proportion of correct guesses is \(p = \frac{1}{12} \approx 0.083\). The alternative hypothesis \(H_a\) is that the individual has some ability, so \(p > 0.083\).
Interpret the P-value: The P-value of 0.1726 represents the probability of observing 9 or more correct guesses out of 80 purely by chance if the null hypothesis is true (i.e., if the person is just guessing).
Explain what a P-value means in hypothesis testing: A P-value measures how compatible the observed data is with the null hypothesis. A small P-value (usually less than a significance level like 0.05) suggests the data is unlikely under \(H_0\), leading us to reject \(H_0\). A large P-value suggests the data is consistent with \(H_0\).
Based on the P-value of 0.1726, which is greater than common significance levels (e.g., 0.05), we do not have strong evidence to reject the null hypothesis. This means the observed number of correct guesses could reasonably occur by random chance.
Write a conclusion: We conclude that there is insufficient evidence to support the claim that the individual can determine birth months better than random guessing. The data does not show a statistically significant ability to guess birth months.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
P-value
The P-value measures the probability of obtaining results at least as extreme as the observed data, assuming the null hypothesis is true. A small P-value indicates strong evidence against the null hypothesis, while a large P-value suggests the data is consistent with the null. It helps decide whether to reject or fail to reject the null hypothesis.
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Step 3: Get P-Value
Null Hypothesis and Alternative Hypothesis
The null hypothesis (H0) represents the default assumption, often stating no effect or no difference—in this case, that the individual guesses birth months correctly at the chance rate of 1/12. The alternative hypothesis (Ha) claims the individual’s accuracy is greater than chance. Hypothesis testing evaluates evidence to support or refute these claims.
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Step 1: Write Hypotheses
Statistical Significance and Decision Making
Statistical significance determines if observed results are unlikely under the null hypothesis, typically using a significance level (e.g., 0.05). If the P-value is less than this level, we reject H0; otherwise, we fail to reject it. Here, a P-value of 0.1726 is greater than 0.05, so we do not have enough evidence to conclude the individual can guess birth months better than chance.
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