No Variation in a Sample An experiment was conducted to test the effects of alcohol. Researchers measured the breath alcohol levels for a treatment group of people who drank ethanol and another group given a placebo. The results are given below (based on data from “Effects of Alcohol Intoxication on Risk Taking, Strategy, and Error Rate in Visuomotor Performance,” by Streufert et al., Journal of Applied Psychology, Vol. 77, No. 4). Use a 0.05 significance level to test the claim that the two sample groups come from populations with the same mean.
Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 53m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample2h 19m
- 10. Hypothesis Testing for Two Samples3h 22m
- 11. Correlation1h 6m
- 12. Regression1h 4m
- 13. Chi-Square Tests & Goodness of Fit1h 20m
- 14. ANOVA1h 0m
9. Hypothesis Testing for One Sample
Performing Hypothesis Tests: Means
Problem 9.3.7c
Textbook Question
In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.
The Freshman 15 The “Freshman 15” refers to the belief that college students gain 15 lb (or 6.8 kg) during their freshman year. Listed below are weights (kg) of randomly selected male college freshmen (from Data Set 13 “Freshman 15” in Appendix B). The weights were measured in September and later in April.
c. What do you conclude about the Freshman 15 belief?


1
Step 1: Calculate the differences between the paired weights for September and April. For each individual, subtract the September weight from the April weight to find the change in weight.
Step 2: Compute the mean of the differences. Add all the differences together and divide by the number of paired samples to find the average change in weight.
Step 3: Calculate the standard deviation of the differences. Use the formula for standard deviation: \( \sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \), where \( x_i \) are the differences, \( \bar{x} \) is the mean of the differences, and \( n \) is the number of paired samples.
Step 4: Perform a hypothesis test to determine if the mean difference is significantly different from 15 lb (or 6.8 kg). Use a t-test for paired samples, with the null hypothesis \( H_0: \mu = 6.8 \) and the alternative hypothesis \( H_a: \mu \neq 6.8 \). Calculate the t-statistic using \( t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \), where \( \bar{x} \) is the mean difference, \( \mu \) is the hypothesized mean, \( s \) is the standard deviation, and \( n \) is the sample size.
Step 5: Compare the calculated t-statistic to the critical t-value from the t-distribution table at the chosen significance level (e.g., \( \alpha = 0.05 \)). If the t-statistic falls outside the critical range, reject the null hypothesis. Based on the results, draw a conclusion about the validity of the Freshman 15 belief.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Paired Sample Data
Paired sample data involves two related groups where measurements are taken from the same subjects at two different times or under two different conditions. In this context, the weights of male college freshmen are measured in September and again in April, allowing for a direct comparison of weight changes over time.
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Normal Distribution
Normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. The assumption that the differences in weights have an approximately normal distribution is crucial for applying statistical tests that rely on this property, such as the paired t-test.
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Statistical Inference
Statistical inference involves drawing conclusions about a population based on sample data. In this case, the analysis of the weight differences will help determine whether the belief in the 'Freshman 15'—that students gain 15 pounds during their freshman year—is supported by the data collected from the sample of students.
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