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mr Goodwill [35]
3 years ago
11

A manufacturer of flashlight batteries took a sample of 13 batteries from a day’s production and used them continuously until th

ey failed to work. The life lengths of the batteries, in hours, until they failed were: 342, 426, 317, 545, 264, 451, 1049, 631, 512, 266, 492, 562, and 298. At the .05 level of significance, is there evidence to suggest that the mean life length of the batteries produced by this manufacturer is more than 400 hours? A. Yes, because the test value 1.257 is less than the critical value 1.782 B. No, because the test value 1.257 is greater than the critical value 1.115 C. No, because the p-value for this test is equal to .1164 D. Yes, because the test value 1.257 is less than the critical value 2.179 Reset Selection
Mathematics
1 answer:
kogti [31]3 years ago
3 0

Answer:

A:

Step-by-step explanation:

***I'm pretty sure A should read "NO, because the test value 1.257 is less than the critical value 1.782.  Please check the wording of the problem***

H0 : μ ≤ 400

Ha : μ > 400 (claim)

Sample mean: 6,155/13

Sample standard deviation: √44422.4359

Critical test value:   t > 1.782

t =  (6,155/13 - 400)/[(√44422.4359)/√13]   =   1.257    

    1.257 < 1.782  ;  we fail to reject the null hypothesis

There is not enough evidence at the 5% level of significance to support the claim that the mean  battery life is at least 400 hours.  

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This problem asks for Taylor polynomials forf(x) = ln(1 +x) centered at= 0. Show Your work in an organized way.(a) Find the 4th,
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a) The 4th degree , 5th degree and sixth degree polynomials

f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}

f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}

f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}

b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.

log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\..  (-1)^{n-1}\frac{x^n}{n} +..

Step-by-step explanation:

Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0

      f(x) = log(1+x) ……(1)

using formula \frac{d}{dx} logx =\frac{1}{x}

Differentiating Equation(1) with respective to 'x' we get

f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)

f^{l} (x) = \frac{1}{1+x} (1)  ….(2)

At x= 0

f^{l} (0) = \frac{1}{1+0} (1)= 1

using formula \frac{d}{dx} x^{n-1}  =nx^{n-1}

Again Differentiating Equation(2) with respective to 'x' we get

f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))

f^{ll} (x) = \frac{-1}{(1+x)^2} (1)    ….(3)

At x=0

f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1

Again Differentiating Equation(3) with respective to 'x' we get

f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))

f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)=  \frac{(-1)^2 (2)!}{(1+x)^3} ….(4)

At x=0

f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)

f^{lll} (0) = 2

Again Differentiating Equation(4) with respective to 'x' we get

f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))

f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4} ....(5)

f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}

f^{lV} (0) = -6

Again Differentiating Equation(5) with respective to 'x' we get

f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))

f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5} .....(6)

At x=0

f^{V} (x) = 24

Again Differentiating Equation(6) with respective to 'x' we get

f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))

f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}

and so on...

The nth term is

f^{n} (x) =  = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}

Step :-2

Taylors theorem expansion of f(x) is

f(x) = f(a) + \frac{x}{1!} f^{l}(x) +\frac{(x-a)^2}{2!}f^{ll}(x)+\frac{(x-a)^3}{3!}f^{lll}(x)+\frac{(x-a)^4}{4!}f^{lV}(x)+\frac{(x-a)^5}{5!}f^{V}(x)+\frac{(x-a)^6}{6!}f^{VI}(x)+...….. \frac{(x-a)^n}{n!}f^{n}(x)

At x=a =0

f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)

Substitute  all values , we get

f(x) = f(0) + \frac{x}{1!} (1) +\frac{(x)^2}{2!}(-1)+\frac{(x)^3}{3!}(2)+\frac{(x)^4}{4!}(-6)+\frac{(x)^5}{5!}(24)+\frac{(x)^6}{6!}(-120)+...….. \frac{(x-0)^n}{n!}f^{n}(0)

On simplification we get

log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\..  (-1)^{n-1}\frac{x^n}{n} +..

4 0
3 years ago
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