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A fruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage. To test the claim, the grower sprays 200 trees with the new spray and 200 other trees with the standard spray. The following data were recorded. Standard Spray Mean yield per tree x (kg) Variance s² New Spray 109 3. Test statistic: z = 405 1-2. Null and alternative hypotheses: O Ho: (M₁M₂) = 0 versus H₂: (M₁ - M₂) # 0 O Ho: (M₁ M₂) = 0 versus H₂: (M₁M₂) = 0 O Ho: (M₁M₂) < 0 versus H₂: (μ₁ −μ₂) > O O Ho: (M₁-M₂) = 0 versus H₂: (μ₁ −μ₂) < 0 O Ho: (H₁-H₂) = 0 versus H₂: (M₁M₂) >0 Z< (a) Do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray, Myr exceeds that for trees treated with the standard spray, μ₂? Use α = 0.05. (Round your answers to two decimal places.) 105 375 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. Z > 5. Conclusion: O Ho is rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. (b) Construct a 95% confidence interval for the difference between the mean yields for the two sprays. (Round your answers to two decimal places.) kg to kg

A fruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage. To test the claim, the grower sprays 200 trees with the new spray and 200 other trees with the standard spray. The following data were recorded. Standard Spray Mean yield per tree x (kg) Variance s² New Spray 109 3. Test statistic: z = 405 1-2. Null and alternative hypotheses: O Ho: (M₁M₂) = 0 versus H₂: (M₁ - M₂) # 0 O Ho: (M₁ M₂) = 0 versus H₂: (M₁M₂) = 0 O Ho: (M₁M₂) < 0 versus H₂: (μ₁ −μ₂) > O O Ho: (M₁-M₂) = 0 versus H₂: (μ₁ −μ₂) < 0 O Ho: (H₁-H₂) = 0 versus H₂: (M₁M₂) >0 Z< (a) Do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray, Myr exceeds that for trees treated with the standard spray, μ₂? Use α = 0.05. (Round your answers to two decimal places.) 105 375 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. Z > 5. Conclusion: O Ho is rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. (b) Construct a 95% confidence interval for the difference between the mean yields for the two sprays. (Round your answers to two decimal places.) kg to kg

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A fruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage. To test the claim, the grower sprays 200 trees with the new spray and 200 other trees with the standard spray. The following data were recorded. Standard Spray Mean yield per tree x (kg) 105 Variance s² 375 (a) Do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray, μ₁, exceeds that for trees treated with the standard spray, μ₂? Use α = 0.05. (Round your answers to two decimal places.) New Spray 109 405 1-2. Null and alternative hypotheses: Ho: (M₁M₂) = 0 versus H₂: (μ₁ - M₂) = 0 O Ho: (M₁ M₂) = 0 versus H₂: (M₁M₂) = 0 O Ho: (M₁M₂) < 0 versus H₂: (M₁ M₂) > 0 O Ho: (M₁ M₂) = 0 versus H₂: (M₁M₂) <0 Ho: (M₁M₂) = 0 versus H₂: (M₁M₂) > 0 3. Test statistic: z = 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. Z > Z< 5. Conclusion: O Ho is rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. (b) Construct a 95% confidence interval for the difference between the mean yields for the two sprays. (Round your answers to two decimal places.) kg to kg
Transcribed Image Text:A fruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage. To test the claim, the grower sprays 200 trees with the new spray and 200 other trees with the standard spray. The following data were recorded. Standard Spray Mean yield per tree x (kg) 105 Variance s² 375 (a) Do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray, μ₁, exceeds that for trees treated with the standard spray, μ₂? Use α = 0.05. (Round your answers to two decimal places.) New Spray 109 405 1-2. Null and alternative hypotheses: Ho: (M₁M₂) = 0 versus H₂: (μ₁ - M₂) = 0 O Ho: (M₁ M₂) = 0 versus H₂: (M₁M₂) = 0 O Ho: (M₁M₂) < 0 versus H₂: (M₁ M₂) > 0 O Ho: (M₁ M₂) = 0 versus H₂: (M₁M₂) <0 Ho: (M₁M₂) = 0 versus H₂: (M₁M₂) > 0 3. Test statistic: z = 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. Z > Z< 5. Conclusion: O Ho is rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. O Ho is not rejected. There is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. (b) Construct a 95% confidence interval for the difference between the mean yields for the two sprays. (Round your answers to two decimal places.) kg to kg
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