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ankoles [38]
3 years ago
9

Some researchers have conjectured that stem-pitting disease in peach tree seedlings might be controlled with weed and soil treat

ment. An experiment was conducted to compare peach tree seedling growth with soil and weeds treated with one of the two herbicides. In a field containing 10 seedlings, five were randomly selected throughout the field and assigned to receive Herbicide A. The remainder was to receive Herbicide B. Soil and weeds for each seedling were treated with the appropriate herbicide, and, at the end of the study period, the height (cm) was recorded for each seedling. The following results were obtained: Herbicide A 87, 80, 80, 76, 73 Herbicide B 78, 77, 74, 68, 62 A 90% confidence interval for the difference in mean heights for the two herbicides is (0.2, 14.6). Which of the following statements is correct? The P-value for a test of the null hypothesis of equal means and the alternative of different means would be greater than 10% because the interval doesn't include 0. Neither of those statements is correct. A 95% confidence interval could not include 0, because we would be even more confident of a difference in the groups.
Mathematics
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

The first choice

Step-by-step explanation:

The choice of "The P-value for a test of the null hypothesis of equal means and the alternative of different means would be greater than 10% because the interval doesn't include 0" is the correct choice because the lower bound of the confidence interval is 0.2.  

Since we are looking at an interval based on the difference of heights, we are 95% sure the lowest difference is 0.2, or 20%.  Choosing  10% as a test value for a hypothesis test is a good value to choose since it's lower than the lowest bound of the confidence interval

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Step-by-step explanation:

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Step-by-step explanation:

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nordsb [41]

Answer:

18.66 ft/s

Step-by-step explanation:

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Applying the chain rule:

u=x^2+y^2\\\frac{dh}{dy}=\frac{d\sqrt u}{du} *\frac{du}{dy}\\\frac{dh}{dy}=\frac{1}{2\sqrt u} *2y\\\frac{dh}{dy}=\frac{y}{\sqrt {(x^2+y^2)}}

Therefore, at x=300 and y = 500, dy/dt is:

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