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Tcecarenko [31]
3 years ago
12

A market research analyst claims that 32% of the people who visit the mall actually make a purchase. You think that less than 32

% buy something and decide to test the claim. You stand by the exit door of the mall starting at noon and ask 82 people as they are leaving whether they bought anything. You find that only 20 people made a purchase.
Required:
a. State the null and alternative hypotheses to test that less than 32% of man visitors make a purchase.
b. Find the value of the test statistics z.
c. The p-value of the test is about:______
Mathematics
1 answer:
Ksju [112]3 years ago
4 0

Answer:

p=32 p<32

Step-by-step explanation:

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1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given
neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}

2)  Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: x^2-2x-4=0

x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}

As for

x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\

3) Rewriting and calculating its derivative. Remember to do it, in radians.

5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1

x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}

For the second root, let's try -1.5

x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\

For x=-3.9, last root.

x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}

f(x)=x^6-x^4+3x^3-2x

\mathbf{f'(x)=6x^5-4x^3+9x^2-2}

\mathbf{f''(x)=30x^4-12x^2+18x}

For -1.2

x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx  -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx  \mathbf{-1.29322}\\

For x=0.4

x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx  \mathbf{0.50785}\\

and for x=-0.4

x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\

These roots (in bold) are the critical numbers

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20 POINTS! WILL MARK BRAINIEST FOR FIRST CORRECT ANSWER!
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Answer:

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Step-by-step explanation:

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