Problem
Researchers conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given in the accompanying table. Higher scores correspond to more creativity. The researchers make the claim that "blue enhances performance on a creative task." Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b). Click the icon to view the summary statistics. a. Use a 0.05 significance level to test the claim that blue enhances performance on a creative task. What are the null and alternative hypotheses? YA. \[ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1}>\mu_{2} \end{array} \] B. $\mathrm{H}_{0}: \mu_{1} \neq \mu_{2}$ \[ H_{1}: \mu_{1}<\mu_{2} \] C. \[ \begin{array}{l} H_{0}: \mu_{1} \geq \mu_{2} \\ H_{1}: \mu_{1}<\mu_{2} \end{array} \] D. \[ \begin{array}{l} \mathrm{H}_{0}: \mu_{1}=\mu_{2} \\ \mathrm{H}_{1}: \mu_{1} \neq \mu_{2} \end{array} \] The test statistic, $\mathrm{t}$, is (Round to two decimal places as needed.) Response Summary Statistics \begin{tabular}{|l|c|c|c|c|} \hline Background & $\boldsymbol{\mu}$ & $\mathbf{n}$ & $\overline{\mathbf{x}}$ & $\mathbf{s}$ \\ \hline Blue & $\boldsymbol{\mu}_{1}$ & 36 & 4.42 & 0.54 \\ \hline Red & $\boldsymbol{\mu}_{2}$ & 31 & 3.72 & 0.98 \\ \hline \end{tabular} Print Done View an Check answer
Solution
Step 1 ：The null and alternative hypotheses are: \[H_{0}: \mu_{1}=\mu_{2}\] and \[H_{1}: \mu_{1}>\mu_{2}\]
Step 2 ：The test statistic, t, is calculated using the sample means, sample sizes, and sample standard deviations for the two groups. The formula for the t-statistic is: \[t = \frac{\overline{x}_{1} - \overline{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}\]
Step 3 ：Substitute the given values into the formula: \[t = \frac{4.42 - 3.72}{\sqrt{\frac{0.54^{2}}{36} + \frac{0.98^{2}}{31}}}\]
Step 4 ：Calculate the t-statistic to get \(t = 3.54\)
Step 5 ：The final answer is: \[\boxed{3.54}\]
