Test for Normality For the hypothesis test described in Exercise 2, the sample sizes are n1 = 2208 and n2 = 1986 When using the F test with these data, is it correct to reason that there is no need to check for normality because both samples have sizes that are greater than 30?
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- 1. Intro to Stats and Collecting Data1h 14m
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- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
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- 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
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- 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
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- Performing Hypothesis Tests: Variance12m
- Critical Values and Rejection Regions28m
- Link Between Confidence Intervals and Hypothesis Testing12m
- Type I & Type II Errors16m
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- Two Variances and F Distribution29m
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10. Hypothesis Testing for Two Samples
Two Variances and F Distribution
Problem 9.4.10a
Textbook Question
Second-Hand Smoke Samples from Data Set 15 “Passive and Active Smoke” include cotinine levels measured in a group of smokers ( n = 40, x_bar = 172.48 ng/mL, 119.50 ng/mL ) and a group of nonsmokers not exposed to tobacco smoke ( n = 40, x_bar = 16.35 ng/mL, 62.53 ng/mL ). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced.
a. Use a 0.05 significance level to test the claim that the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke.
Verified step by step guidance1
Step 1: Identify the hypotheses for the test. The null hypothesis (H₀) states that the variance of cotinine levels in smokers is less than or equal to the variance in nonsmokers not exposed to tobacco smoke (σ₁² ≤ σ₂²). The alternative hypothesis (H₁) states that the variance in smokers is greater than the variance in nonsmokers (σ₁² > σ₂²).
Step 2: Determine the test statistic to use. Since we are comparing variances, we use the F-test for equality of variances. The test statistic is calculated as F = (s₁² / s₂²), where s₁² is the sample variance of smokers and s₂² is the sample variance of nonsmokers.
Step 3: Calculate the sample variances. The sample variance is calculated using the formula s² = (standard deviation)². For smokers, s₁² = (119.50)², and for nonsmokers, s₂² = (62.53)².
Step 4: Determine the critical value for the F-distribution. Use the F-distribution table or software to find the critical value for a one-tailed test at a significance level of 0.05, with degrees of freedom df₁ = n₁ - 1 (for smokers) and df₂ = n₂ - 1 (for nonsmokers).
Step 5: Compare the calculated F-statistic to the critical value. If the calculated F-statistic is greater than the critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Interpret the result in the context of the problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (H0) and an alternative hypothesis (H1). In this context, the null hypothesis would state that the variation of cotinine levels in smokers is equal to or less than that in nonsmokers, while the alternative hypothesis would claim that the variation in smokers is greater.
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Step 1: Write Hypotheses
Variance and Standard Deviation
Variance measures the dispersion of a set of data points around their mean, indicating how much the values differ from the average. The standard deviation is the square root of the variance and provides a measure of spread in the same units as the data. In this question, comparing the variances of cotinine levels in smokers and nonsmokers is crucial to determine if there is a significant difference in their variability.
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Calculating Standard Deviation
Significance Level
The significance level, often denoted as alpha (α), is the threshold for determining whether to reject the null hypothesis. A common significance level is 0.05, which implies a 5% risk of concluding that a difference exists when there is none. In this scenario, using a 0.05 significance level means that if the p-value obtained from the test is less than 0.05, the null hypothesis can be rejected, indicating significant evidence that the variation in smokers is greater.
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Step 4: State Conclusion Example 4
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