Question

\(\bullet CI = \bar{x} \pm Z_{\alpha/2} \times \sigma_{\bar{x}} \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \\ \bullet CI = \bar{x} \pm (t_{\alpha/2, n-1} \times se) \qquad se = \frac{s}{\sqrt{n}} \\ \bullet CI = \hat{p} \pm Z_{\alpha/2} \times se \qquad se = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

          \(\bullet CI = \bar{x} \pm Z_{\alpha/2} \times \sigma_{\bar{x}} \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \\ \bullet CI = \bar{x} \pm (t_{\alpha/2, n-1} \times se) \qquad se = \frac{s}{\sqrt{n}} \\ \bullet CI = \hat{p} \pm Z_{\alpha/2} \times se \qquad se = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

∙ CI = x̅± Zα/2×σx̅ σx̅ = (σ)/(√(n))
∙ CI = x̅± (tα/2, n-1× se) se = (s)/(√(n))
∙ CI = p̂± Zα/2× se se = √((p̂(1-p̂))/(n))

Added by Marc V.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Dominador Tan

08/11/2023

Step 1

For the standard program (control group): s = √(variance/n) = √(23.2/30) = 1.52 For the new program (treatment group): s = √(variance/n) = √(11.5/30) = 0.76 Show more…

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Texts: Thirty alcoholics enrolled to participate in a new program aimed at reducing alcohol use. Half of the participants were randomly selected to receive the new treatment program, while the other half were provided with the current standard program. At the end of the 60 days, the mean number of drinks consumed per week for each subject was as follows: Standard Program (Control): Mean = 6.2 drinks, variance = 23.2 drinks New Program (Treatment): Mean = 4.8 drinks, variance = 11.5 drinks Construct a 95% confidence interval for the mean difference of drinks. (Only use one of the formulas below) CI = x ± (tα/2 * s/√n) s = √(variance/n) CI = x ± (tα/2 * s/√(n-1)) s = √(variance/n) CI = p̂ ± (tα/2 * √((p̂ * (1-p̂))/n)) s = √((p̂ * (1-p̂))/n)