I need help with 14 and 15 ASAP
1 answer:
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Answer:
A. The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation:
Hello!
When all other things remain the same, which of the following statements about the width of the interval is correct?
A. The interval will be narrower if 15 men are used in the sample.
B. The interval will be wider if 15 men are used in the sample.
C. The interval will be narrower if 5 men are used in the sample.
D. The interval will be narrower if the level is increased to 99% confidence.
E. The interval will be wider if the level is decreased to 90% confidence.
Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.
The general structure for the CI is "point estimate"±" margin fo error"
Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)
The margin of error of the interval is:
d= * (Sd/n)
1) The sample size changes, all other terms of the interval stay the same.
As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.
↓d= * (Sd/↑n)
↑d= * (Sd/↓n)
Correct option: A. The interval will be narrower if 15 men are used in the sample.
2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:
↓d= ↓ * (Sd/n)
↑d= ↑ * (Sd/n)
I hope it helps!
Answer:
(a) -3/4
(b) -0.75
(c) -0.75
Step-by-step explanation:
It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...
- evaluating the function at the midpoint of the bracketing interval
- choosing a smaller bracketing interval
- identifying the x-value known to be closest to the solution
Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.
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(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.
For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.
For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.
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(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.
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(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.
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See the attached table for function values.
_____
<em>Comment on bisection iteration</em>
Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.
Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.
