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AlekseyPX
2 years ago
5

For each of the following scenarios state whether H0 should be rejected or not. State any assumptions that you make beyond the i

nformation that is given.
(a) H0 : µ = 4, H1 : µ 6= 4, n = 15, X = 3.4, S = 1.5, α = .05.
(b) H0 : µ = 21, H1 : µ < 21, n = 75, X = 20.12, S = 2.1, α = .10.
(c) H0 : µ = 10, H1 : µ 6= 10, n = 36, p-value = 0.061.
Mathematics
1 answer:
scoundrel [369]2 years ago
7 0

Answer:

a)H_0 :\mu = 4\\ H_1 : \mu \neq 4 , n = 15 , X=3.4 , S=1.5 , α = .05

Formula : t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}

t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}

t =-1.549

p- value = 0.607(using calculator)

α = .05

p- value > α

So, we failed to reject null hypothesis

b)H_0 :\mu = 21\\ H_1 : \mu < 21 , n =75 , X=20.12 , S=2.1 , α = .10

Formula : t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}

t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}

t =-3.6290

p- value = 0.000412(using calculator)

α = .1

p- value< α

So, we reject null hypothesis

(c) H_0 :\mu = 10\\ H_1 : \mu \neq 10, n = 36, p-value = 0.061.

Assume α = .05

p-value = 0.061.

p- value > α

So, we failed to reject null hypothesis

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Answer:

\displaystyle   \begin{cases} \displaystyle  {x} _{1} =  - p \\   \displaystyle x _{2}   =  -  q \end{cases}

Step-by-step explanation:

we would like to solve the following equation for x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}  +  \frac{1}{x}  =  \frac{1}{p  + q + x}

to do so isolate \frac{1}{x} to right hand side and change its sign which yields:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{1}{p  + q + x}  -  \frac{1}{x}

simplify Substraction:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{x - (q + p +  x)}{x(p  + q + x)}

get rid of only x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{  - (q + p )}{x(p  + q + x)}

simplify addition of the left hand side:

\displaystyle  \frac{q + p}{pq}     =  \frac{  - (q + p )}{x(p  + q + x)}

divide both sides by q+p Which yields:

\displaystyle  \frac{1}{pq}     =  \frac{  -1}{x(p  + q + x)}

cross multiplication:

\displaystyle    x(p  + q + x)  =   - pq

distribute:

\displaystyle    xp  + xq +  {x}^{2} =   - pq

isolate -pq to the left hand side and change its sign:

\displaystyle    xp  + xq +  {x}^{2} + pq =  0

rearrange it to standard form:

\displaystyle   {x}^{2} +    xp  + xq  + pq =  0

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

\displaystyle  x( {x}^{} +   p ) + xq  + pq =  0

factor out q:

\displaystyle  x( {x}^{} +   p ) +q (x + p)=  0

group:

\displaystyle  ( {x}^{} +   p ) (x + q)=  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

\displaystyle   \begin{cases} \displaystyle  {x}^{} +   p  = 0 \\   \displaystyle x + q=  0 \end{cases}

cancel out p from the first equation and q from the second equation which yields:

\displaystyle   \begin{cases} \displaystyle  {x}^{}   =  - p \\   \displaystyle x  =  -  q \end{cases}

and we are done!

3 0
2 years ago
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