H0: Select an answer x̄₂ x̄₁ μ(red) p₂ p μ₁ μ₂ σ₁ s₁² σ(red) p₁ μ p̂₁ ? ≠ < > ≤ ≥ = Select an answer μ₁ μ₂ p₁ μ(blue) p σ₂ p₂ σ(blue) s₁² x̄₁ x̄₂ μ p̂₁ H1: Select an answer μ σ₁ x̄₁ μ(red) p̂₂ p₁ s₂² p₂ μ₂ μ₁ σ(red) x̄₂ p ? ≤ ≥ ≠ > = < Select an answer μ(blue) σ(blue) p μ μ₁ p₁ x̄₂ s₁² x̄₁ σ₂ p̂₁ p₂ μ₂ Original Claim = Select an answer H₁ H₀ Based on the hypotheses, find the following: Enter the critical values, along with the significance level and degrees of freedom χ2χ2(αα,df num, df den) below the graph. (Graph is for illustration only. No need to shade.) X2Χ2(,,) = Test Statistic = (Round to three decimal places.) p-value = (Round to four decimal places.) Decision: Select an answer Fail to reject the null hypothesis Accept the null hypothesis Reject the null hypothesis Accept the alternative hypothesis .
Does color enhance creativity?
Test the indicated claim about the standard deviations or variances of two populations. Subjects are given a creativity exercise on a computer with either a red background or a blue background. The scores are shown in the table. At the 0.05 significance level, test the claim that those tested with red background have creativity scores with a standard deviation equal to the standard deviation for those tested with a blue background.
| Red Background | Blue Background |
| n1 = 39 | n2 = 34 |
| xˉx̄1 = 19.6 | xˉx̄2 = 20.5 |
| s1 = 0.72 | s2 = 0.01 |
What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)
H0: Select an answer x̄₂ x̄₁ μ(red) p₂ p μ₁ μ₂ σ₁ s₁² σ(red) p₁ μ p̂₁ ? ≠ < > ≤ ≥ = Select an answer μ₁ μ₂ p₁ μ(blue) p σ₂ p₂ σ(blue) s₁² x̄₁ x̄₂ μ p̂₁
H1: Select an answer μ σ₁ x̄₁ μ(red) p̂₂ p₁ s₂² p₂ μ₂ μ₁ σ(red) x̄₂ p ? ≤ ≥ ≠ > = < Select an answer μ(blue) σ(blue) p μ μ₁ p₁ x̄₂ s₁² x̄₁ σ₂ p̂₁ p₂ μ₂
Based on the hypotheses, find the following:
Enter the critical values, along with the significance level and degrees of freedom χ2χ2(αα,df num, df den) below the graph. (Graph is for illustration only. No need to shade.)
Test Statistic = (Round to three decimal places.)
p-value = (Round to four decimal places.)
Decision: Select an answer Fail to reject the null hypothesis Accept the null hypothesis Reject the null hypothesis Accept the alternative hypothesis .
Conclusion: Select an answer There is not enough evidence to support There is not sufficient evidence to warrant rejection of There is sufficient evidence to warrant rejection of The sample data supports the claim that those tested with red background have creativity scores with a standard deviation equal to the standard deviation for those tested with a blue background.
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