Use the slider to change the value of m. As the value of m changes from 1 to 3, the graph . As the value of m changes from –6 to
–1, the graph . If the value of m is negative, the slope of the line is .
2 answers:
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Answer:
about 50 percent of its former width.
Step-by-step explanation:
Let's assume that our parameter of interest is given by and in order to construct a confidence interval we can use the following formula:
Where is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:
Where is a quantile from the normal standard distribution that accumulates
of the area on each tail of the distribution. And
represent the standard error for the parameter.
If our parameter of interest is the population proportion the standard of error is given by:
And if our parameter of interest is the sample mean the standard error is given by:
As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:
And since the margin of error is a multiple of the standard error we have that
Now if we find the width for a confidence interval we got this:
And we can express this as:
And we can define the function since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:
The subscript i is in order to say that is with the sample size n
If we increase the sample size from n to 4n now our width is:
The subscript f is in order to say that is the width for the sample size 4n.
So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:
about 50 percent of its former width.
Answer:
a) see your problem statement for the explanation
b) 2.54539334183
Step-by-step explanation:
(b) Many graphing calculators have a derivative function that lets you define the Newton's Method iterator as a function. That iterator is ...
x' = x - f(x)/f'(x)
where x' is the next "guess" and f'(x) is the derivative of f(x). In the attached, we use g(x) instead of x' for the iterated value.
Here, our f(x) is ...
f(x) = 3x^4 -8x^3 +6
An expression for f'(x) is
f'(x) = 12x^3 -24x^2
but we don't need to know that when we use the calculator's derivative function.
When we start with x=2.545 from the point displayed on the graph, the iteration function g(x) in the attached immediately shows the next decimal digits to be 393. Thus, after 1 iteration starting with 4 significant digits, we have a result good to the desired 6 significant digits: 2.545393. (The interactive nature of this calculator means we can copy additional digits from the iterated value to g(x) until the iterated value changes no more. We have shown that the iterator output is equal to the iterator input, but we get the same output for only 7 significant digits of input.)
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<em>Alternate iterator function</em>
If we were calculating the iterated value by hand, we might want to write the iterator as a rational function in Horner form.
g(x) = x - (3x^4 -8x^3 +6)/(12x^3 -24x^2) = (9x^4 -16x^3 -6)/(12x^3 -24x^2)
g(x) = ((9x -16)x^3 -6)/((12x -24)x^2) . . . . iterator suitable for hand calculation
