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Healthy men aged 21-35 were randomly assigned to one of two groups: half received \(0.82\) gram of alcohol per kilogram of body weight; half received a placebo. Participants were then given 30 minutes to read up to 34 pages of Tolstoy's War and Peace (beginning at Chapter 1 , with each page containing approximately 22 lines of text). Every two to four minutes, participants were prompted to indicate whether they were "zoning out." The proportion of times participants indicated they were zoning out was recorded for each subject. The following table summarizes data on the proportion of episodes of zoning out. \({ }^{13}\) (The study report gave the standard error of the mean \(s / n \sqrt{n}\), abbreviated as SEM, rather than the standard deviation s.) $$ \begin{array}{llll} \hline \text { Group } & n & x^{-} \bar{x} & \text { SEM } \\ \hline \text { Alcohol } & 25 & 0.25 & 0.05 \\ \hline \text { Placebo } & 25 & 0.12 & 0.03 \\ \hline \end{array} $$ (a) What are the two sample standard deviations? (b) What degrees of freedom does the conservative Option 2 use for twosample \(t\) procedures for these samples? (c) Using Option 2, give a \(90 \%\) confidence interval for the mean difference between the two groups.

Step by step solution

01

Calculate the Sample Standard Deviations

To find the sample standard deviations for each group, use the relationship between the SEM, standard deviation \(s\), and the sample size \(n\): \( SEM = \frac{s}{\sqrt{n}} \). Rearrange this to solve for \(s\): \( s = SEM \times \sqrt{n} \). For the Alcohol group: \( s_1 = 0.05 \times \sqrt{25} = 0.05 \times 5 = 0.25 \).For the Placebo group: \( s_2 = 0.03 \times \sqrt{25} = 0.03 \times 5 = 0.15 \).

02

Determine Degrees of Freedom Using Option 2

Option 2 for a two-sample \(t\) procedure uses the smaller of \(n_1 - 1\) and \(n_2 - 1\) as the degrees of freedom.For both groups, \(n_1 = 25\) and \(n_2 = 25\), thus:Degrees of freedom \(df = \min(25-1, 25-1) = 24\).

03

Compute the 90% Confidence Interval

The formula for the confidence interval for the difference between two means is:\[ (\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Where \(t^*\) is the critical value from the \(t\)-distribution with 24 degrees of freedom for a 90% confidence interval. Using a \(t\)-distribution table or calculator, \(t^* \approx 1.711\).The difference in means is \(\bar{x}_1 - \bar{x}_2 = 0.25 - 0.12 = 0.13\).Standard error of the difference is:\[ \sqrt{\frac{0.25^2}{25} + \frac{0.15^2}{25}} = \sqrt{0.01 + 0.009} = \sqrt{0.019} \approx 0.138 \]The confidence interval is:\[ 0.13 \pm 1.711 \times 0.138 = 0.13 \pm 0.236 \]This gives the interval: \([-0.106, 0.366]\), suggesting there is no statistically significant difference between the means at a 90% confidence level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error of the Mean

Understanding the Standard Error of the Mean (SEM) is crucial when working with data samples. The SEM measures how much the sample mean of your data is expected to differ from the true population mean. It is a way to express the precision of the sample mean estimate.
To calculate the SEM, you can use the formula:\[SEM = \frac{s}{\sqrt{n}}\]where:

• \(s\) is the sample standard deviation.
• \(n\) is the sample size.

The SEM decreases as the sample size increases, which implies more reliable estimation of the population mean with larger samples. In the original exercise, SEM was used to summarize how often participants were zoning out.
For the alcoholic group, SEM was given as 0.05. This small SEM value suggests that the sample mean of the zoning out proportion is close to what we might expect from the overall population. Similarly, for the placebo group, the SEM was 0.03, indicating a similar understanding of the group's tendency to zone out.

Confidence Interval

A confidence interval provides a range of values within which we can be fairly certain the true population parameter lies. In statistical terms, it's a way to capture uncertainty around an estimate. If we compute a 90% confidence interval, for example, it implies that if we were to take 100 different samples and build a CI from each sample, we would expect the true parameter to be contained in approximately 90 of those intervals.
This is computed using the formula:\[(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]where:

• \(\bar{x}_1 - \bar{x}_2\) is the difference in sample means.
• \(t^*\) is the critical value from the \(t\)-distribution.
• \(s_1\) and \(s_2\) are sample standard deviations.
• \(n_1\) and \(n_2\) are sample sizes.

In the exercise, the 90% CI was calculated for the difference in means between the alcohol and placebo groups. The computed interval was \([-0.106, 0.366]\). Such intervals help in determining if differences are statistically significant. In this case, as the interval includes zero, it suggests that the difference in means is not statistically significant at the 90% level.

Degrees of Freedom

Degrees of freedom are a crucial concept in statistics that essentially tell us how many values in a calculation are free to vary. It's a foundational piece in the calculation of many statistical tests, including the \(t\)-test used in the exercise.
For independent samples \(t\)-tests, degrees of freedom can be calculated easily for equal sample sizes using the formula:\[df = \min(n_1 - 1, n_2 - 1)\]where \(n_1\) and \(n_2\) are the sizes of the two samples involved.
In the case of the exercise, both groups consisted of 25 participants. Therefore, the degrees of freedom used for the \(t\)-test was 24, calculated from \(25 - 1 = 24\).
Having the correct degrees of freedom allows for the determination of the appropriate critical value \(t^*\) when constructing confidence intervals or conducting hypothesis tests. This parameter is pivotal in ensuring the accuracy and reliability of inferential statistics.