Question

ch 21 researchers compared the balance of a sample of elderly patients with that of a separate sample of younger patients. they exposed members of each age group to unpredictable noises and measured the amount of forward and backward sway. to compare the means, they should give a confidence interval for: μ(elderly) - μ(younger). μ(elderly) + μ(younger). μ(elderly) ÷ μ(younger). μ(elderly) × μ(younger). question 15 1 pts ch 21* the two - sample t statistic for the social media use study (\under 40\ mean minus \40 and over\ mean) is t = 3.18 with df = 420. the p - value for testing the hypotheses that younger adults use social media more than adults aged 40 or over. satisfies 0.001 < p < 0.002. 0.001 < p < 0.005. 0.0005 < p < 0.001.

Explanation:

Step1: Understand the confidence interval purpose

When comparing two means (elderly and younger patients' balance - related sway means here), we use the difference of means. The confidence interval for comparing two population means \(\mu_1\) and \(\mu_2\) (where \(\mu_1=\mu(\text{elderly})\) and \(\mu_2 = \mu(\text{younger})\)) is based on \(\mu_1-\mu_2\). Addition (\(\mu(\text{elderly})+\mu(\text{younger})\)), division (\(\mu(\text{elderly})\div\mu(\text{younger})\)) and multiplication (\(\mu(\text{elderly})\times\mu(\text{younger})\)) do not serve the purpose of comparing two means.

Step2: Analyze the \(P -\) value for the \(t -\) test

For a one - sided \(t -\) test with \(t = 3.18\) and \(df=420\). We know that for a \(t -\) distribution with large degrees of freedom (since \(df = 420\approx\infty\), and the \(t -\) distribution approaches the standard normal distribution). Looking at the standard normal (or large - \(df\) \(t -\)) critical values: \(P(Z>3.1)=0.001\) and \(P(Z > 3.29)=0.0005\). Since \(3.18\) is between \(3.1\) and \(3.29\), the \(P -\) value for the one - sided test \(P(t>3.18)\) (where \(t\) has \(df = 420\)) satisfies \(0.0005

Answer:

For the first question: \(\mu(\text{elderly})-\mu(\text{younger})\)
For the second question: \(0.0005 < P<0.001\)