Mathematics, 18.03.2020 16:57 heids17043

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test millimeters versus millimeters, using the results of samples. Calculate the P-value if the observed statistic is Round your answer to four decimal places (e. g. 98.7654).

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Answer from: Janznznz4012

The answers to the questions are

(a) P = 0.0176

(b) P = 0.5

Step-by-step explanation:

The P value is the calculated probability of obtaining extreme reults which are as extreme as observed res ult when the study null hypothesis H₀ is true

The p value is given by

P ( $\overline{\rm x}$ = 185 when μ = 175)

z = $\frac{ \overline { \rm x} -\mu}{\sigma /\sqrt{n} }$

z( $\overline { \rm x}$ = 185) = $\frac{ 185 -175}{20/\sqrt{10} }$ = 1.5811

Therefore the P ( $\overline{\rm x}$ = 185 when μ = 175) = P(z = 1.5811)

0.94295

The probability density function is

$f(x) = \frac{1}{\sigma\sqrt{2\pi } } e^{-\frac{1}{2}(\frac{x-\mu}{\sigma} )^2 }$ Therefore the probability is

$f(x) = \frac{1}{20\sqrt{2\pi } } e^{-\frac{1}{2}(\frac{185-175}{20} )^2 }$ = 0.0176

b) The type II error is given by

β = P(type II error) = P(Fail to reject the null hypothesis when it is false)

Therefore

β = P ( $\overline { \rm x}$ < 185 when μ = 185)

P $(\frac{ \overline { \rm x} -\mu}{\sigma /\sqrt{n} }$ , which gives

P(z<0) and β = 0.5

β = P ( $\overline { \rm x}$ < 185 when μ = 195)

and P $(\frac{ \overline { \rm x} -\mu}{\sigma /\sqrt{n} }$

Which gives P(z < -1.58)

From z table the probability is β = 0.0571

Answer from: Quest

$y> \frac{1}{3}x$ and y≤ -x +4 .

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be = $\frac{1-0}{3-0} = \frac{1}{3}$ .

y-intercept of dotted line is y=0 only.

plugging slope m= $\frac{1}{3}$ and b=0 in slope-intercept form y=mx+b, we get

$y=\frac{1}{3}x$ .

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

$y> \frac{1}{3}x$ .

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be = $\frac{4-1}{0-3} = \frac{3}{-3}=-1$ .

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be

y≤ -x +4 .
Write a system of inequalities to represent the shaded portion of the graph.

Answer from: Quest

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

$\bf \textit{hyperbolas, vertical traverse axis } \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} [-0.35em] \rule{34em}{0.25pt} \begin{cases} h=0\\ k=0\\ a=\sqrt{5}\\ c=3 \end{cases}\implies c=\sqrt{a^2+b^2}\implies 3=\sqrt{(\sqrt{5})^2+b^2}\implies 3^2=5+b^2$

$\bf 3^2-5=b^2\implies 9-5=b^2\implies 4=b^2\implies \sqrt{4}=b\implies \boxed{2=b} [-0.35em] ~\dotfill \cfrac{(y-0)^2}{(\sqrt{5})^2}-\cfrac{(x-0)^2}{2^2}=1\implies \cfrac{y^2}{5}-\cfrac{x^2}{4}=1$

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A consumer products company is formulating a new shampoo and is interested in foam height (in millim...