Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Use a 0.05 significance level to test the claim that the samples are from populations with the same standard deviation. Assume that both samples are independent simple random samples from populations having normal distributions. Does the background color appear to have an effect on the variation of word recall scores? \begin{tabular}{lccc} & n & $\bar{x}$ & s \\ Red Background & 34 & 15.44 & 5.99 \\ Blue Background & 36 & 12.35 & 5.45 \end{tabular} What are the null and alternative hypotheses? A. $\mathrm{H}_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}$ \[ \mathrm{H}_{1}: \sigma_{1}^{2} \geq \sigma_{2}^{2} \] C. $\mathrm{H}_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}$ \[ H_{1}: \sigma_{1}^{2}<\sigma_{2}^{2} \] B. \[ \begin{array}{l} H_{0}: \sigma_{1}^{2} \neq \sigma_{2}^{2} \\ H_{1}: \sigma_{1}^{2}=\sigma_{2}^{2} \end{array} \] D. \[ \begin{array}{l} H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \\ H_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2} \end{array} \] Identify the test statistic. $F=1.21$ (Round to two decimal places as needed.) Use technology to identify the P-value. The $P$-value is $\square$. (Round to three decimalyplaces as needed.)
Solution
Step 1 ：The null hypothesis is a statement of no effect or no difference, and the alternative hypothesis is what we are trying to find evidence for. In this case, we are testing the claim that the samples are from populations with the same standard deviation.
Step 2 ：The null hypothesis would be that the two standard deviations are equal: \( H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \).
Step 3 ：The alternative hypothesis would be that the two standard deviations are not equal: \( H_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2} \).
Step 4 ：The correct hypotheses correspond to option D.
Step 5 ：The final answer for the first question is: \( \boxed{D. \begin{array}{l} H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \ H_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2} \end{array}} \).
Step 6 ：The test statistic provided is \( F=1.21 \).
Step 7 ：The final answer for the second question is: \( \boxed{1.21} \).
Step 8 ：In practice, one would use statistical software to find the P-value based on the F-statistic and the degrees of freedom for the two samples.
