Questions: Suppose you compared 2 independent samples (groups). The first sample contained 10 people, had a mean score of 74.20, and a standard deviation of 2.20. The second sample contained 10 people, had a mean score of 73.20, and a standard deviation of 2.10. In this study, what is the calculated value of t (see section 4 b of the lecture notes, the T-Test Example for Independent Data file, and/or pages 229-231 of the Handout file)? Send just the calculated value of t; send NO WORK.

Solution Steps

The Standard Error \( SE \) is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{2.20^2}{10} + \frac{2.10^2}{10}} = \sqrt{\frac{4.84}{10} + \frac{4.41}{10}} = \sqrt{0.484 + 0.441} = \sqrt{0.925} \approx 0.962 \]

The test statistic \( t \) is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{74.20 - 73.20}{0.962} \approx \frac{1.00}{0.962} \approx 1.039 \]

The degrees of freedom \( df \) are calculated using the formula:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{(0.484 + 0.441)^2}{\frac{(0.484)^2}{9} + \frac{(0.441)^2}{9}} \approx \frac{0.925^2}{\frac{0.234256}{9} + \frac{0.194481}{9}} \approx \frac{0.855625}{0.025} \approx 34.2 \]

The p-value \( P \) is calculated as follows:

\[ P = 2(1 - T(|t|)) \approx 2(1 - T(1.039)) \approx 0.30 \]

Final Answer