\(\boxed{\begin{array}{l}
\Delta \\
\text{Single Sample T-Test: Alcohol and Reaction Time} \\
\circ \\
\text{Hypotheses} \\
\odot \\
\text{Null Hypothesis (\(H_0\)): Alcohol has no effect on reaction time. The mean reaction time after alcohol consumption is equal to the population mean under normal circumstances. \(H_0: \mu = 195\)} \\
\text{Alternative Hypothesis (\(H_1\)): Alcohol affects reaction time. The mean reaction time after alcohol consumption is different from the population mean under normal circumstances. \(H_1: \mu \neq 195\)} \\
\circ \\
\text{Steps to calculate t} \\
\odot \\
\text{Calculate the sample mean (\(\bar{x}\)) of the reaction times after alcohol consumption.} \\
\text{Calculate the sample standard deviation (\(s\)) of the reaction times.} \\
\text{Calculate the standard error of the mean (\(SE\)): \(SE = \frac{s}{\sqrt{n}}\), where \(n\) is the sample size.} \\
\text{Calculate the t-statistic: \(t = \frac{\bar{x} - \mu}{SE}\), where \(\mu\) is the population mean (195).} \\
\circ \\
\text{Choose the t-cutoff score} \\
\odot \\
\text{Degrees of freedom (\(df\)) = \(n - 1 = 15 - 1 = 14\)} \\
\text{Alpha level (\(\alpha\)) = 0.05, two-tailed test} \\
\text{Using a t-table or calculator, find the critical t-value for \(df = 14\) and \(\alpha = 0.05\) (two-tailed). The t-cutoff score is approximately \(\pm 2.145\).} \\
\circ \\
\text{Calculate the t-statistic using SPSS} \\
\odot \\
\text{Assuming SPSS output provides:} \\
\text{Sample Mean (\(\bar{x}\)) = 222} \\
\text{Sample Standard Deviation (\(s\)) = 7.937} \\
\text{t-statistic = 13.15} \\
\text{Degrees of freedom (\(df\)) = 14} \\
\text{p-value = 0.000} \\
\text{Effect Size (Cohen's d) = 6.78} \\
\circ \\
\text{Write the t-statistic using proper APA format} \\
\odot \\
\text{\(t(14) = 13.15, p < .001, d = 6.78\)} \\
\circ \\
\text{Decide whether or not you would reject the null hypothesis} \\
\odot \\
\text{Since the calculated t-statistic (13.15) is greater than the critical t-value (2.145), we reject the null hypothesis. Also, the p-value (0.000) is less than the alpha level (0.05), which also leads to rejecting the null hypothesis.} \\
\circ \\
\text{Interpret this result} \\
\odot \\
\text{The results suggest that alcohol consumption significantly affects reaction time. Specifically, alcohol consumption appears to increase reaction time compared to normal circumstances. The effect size is very large, indicating a substantial practical significance.} \\
\star \\
\text{The results indicate a significant effect of alcohol on reaction time, with a large effect size.} \\
\Delta \\
\text{Independent Sample T-Test: Anxiety Levels and Counseling Location} \\
\circ \\
\text{Hypotheses} \\
\odot \\
\text{Null Hypothesis (\(H_0\)): There is no difference in anxiety levels between individuals receiving in-home counseling and those receiving out-of-home counseling. \(H_0: \mu_{in-home} = \mu_{out-of-home}\)} \\
\text{Alternative Hypothesis (\(H_1\)): There is a difference in anxiety levels between individuals receiving in-home counseling and those receiving out-of-home counseling. \(H_1: \mu_{in-home} \neq \mu_{out-of-home}\)} \\
\circ \\
\text{Steps to calculate t} \\
\odot \\
\text{Calculate the sample mean (\(\bar{x}_1\)) and sample standard deviation (\(s_1\)) for the in-home counseling group.} \\
\text{Calculate the sample mean (\(\bar{x}_2\)) and sample standard deviation (\(s_2\)) for the out-of-home counseling group.} \\
\text{Calculate the pooled variance (\(s_p^2\)):} \\
\text{\(s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\), where \(n_1\) and \(n_2\) are the sample sizes of the two groups.} \\
\text{Calculate the standard error of the difference between the means:} \\
\text{\(SE = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}\)} \\
\text{Calculate the t-statistic:} \\
\text{\(t = \frac{(\bar{x}_1 - \bar{x}_2)}{SE}\)} \\
\circ \\
\text{Choose the t-cutoff score} \\
\odot \\
\text{Degrees of freedom (\(df\)) = \(n_1 + n_2 - 2 = 20 + 20 - 2 = 38\)} \\
\text{Alpha level (\(\alpha\)) = 0.05, two-tailed test} \\
\text{Using a t-table or calculator, find the critical t-value for \(df = 38\) and \(\alpha = 0.05\) (two-tailed). The t-cutoff score is approximately \(\pm 2.024\).} \\
\circ \\
\text{Calculate the t-statistic using SPSS} \\
\odot \\
\text{Assuming SPSS output provides:} \\
\text{Mean (In-Home) = 3.7} \\
\text{Standard Deviation (In-Home) = 2.16} \\
\text{Mean (Out-of-Home) = 5.6} \\
\text{Standard Deviation (Out-of-Home) = 1.46} \\
\text{t-statistic = -3.34} \\
\text{Degrees of freedom (\(df\)) = 38} \\
\text{p-value = 0.002} \\
\text{Effect Size (Cohen's d) = -1.07} \\
\circ \\
\text{Write the t-statistic using proper APA format} \\
\odot \\
\text{\(t(38) = -3.34, p = .002, d = -1.07\)} \\
\circ \\
\text{Decide whether or not you would reject the null hypothesis} \\
\odot \\
\text{Since the absolute value of the calculated t-statistic (3.34) is greater than the critical t-value (2.024), we reject the null hypothesis. Also, the p-value (0.002) is less than the alpha level (0.05), which also leads to rejecting the null hypothesis.} \\
\circ \\
\text{Interpret this result} \\
\odot \\
\text{The results suggest that there is a significant difference in anxiety levels between individuals receiving in-home counseling and those receiving out-of-home counseling. Specifically, individuals receiving in-home counseling appear to have significantly lower anxiety scores than those receiving out-of-home counseling. The effect size is large, indicating a substantial practical significance.} \\
\star \\
\text{The results indicate a significant difference in anxiety levels based on counseling location, with a large effect size.} \\
\end{array}}\)
\(\boxed{\begin{array}{l}
\smiley \\
\text{The results indicate a significant effect of alcohol on reaction time, with a large effect size.} \\
\text{The results indicate a significant difference in anxiety levels based on counseling location, with a large effect size.} \\
\end{array}}\)
Answered
