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The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Given:
f(x) = ln(x)
n = 4
c = 3
nth Taylor polynomial for the function, centered at c
The Taylor series for f(x) = ln x centered at 5 is:
Since, c = 5 so,
Now
f(5) = ln 5
f'(x) = 1/x ⇒ f'(5) = 1/5
f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25
f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125
f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625
So Taylor polynomial for n = 4 is:
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Hence,
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
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Answer:
Reject <em>H</em>₀. There is a significant difference in drug resistance between the two states.
Step-by-step explanation:
In this case we need to determine whether the data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states.
The significance level of the test is, <em>α</em> = 0.02.
(1)
The hypothesis can be defined as follows:
<em>H</em>₀: There is no difference between the proportions of drug-resistant cases in the two states, i.e. .
<em>Hₐ</em>: There is a statistically significant difference between the proportions of drug-resistant cases in the two states, i.e. .
(2)
Compute the sample proportions and total proportion as follows:
Compute the test statistic value as follows:
The test statistic value is 2.86.
(3)
The decision rule is:
The null hypothesis will be rejected if the p-value of the test is less than the significance level.
Compute the p-value as follows:
<em>p</em>-value = 0.00424 < <em>α</em> = 0.02.
The null hypothesis will be rejected at 0.02 significance level.
Reject <em>H</em>₀. There is a significant difference in drug resistance between the two states.