Please i need help FAST

The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval ab

Dmitry_Shevchenko [17]

Answer:

The minimum sample size needed is 125.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$\pi = 0.25$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when $M = 0.1$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.1\sqrt{n} = 2.575{0.25*0.75}$

$\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}$

$(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}$

$n = 124.32$

Rounding up

The minimum sample size needed is 125.

5 0

3 years ago

The volume

Sedaia [141]

$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&& y={{ k }}x
\end{array}\\ \quad \\
$

and also

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{ k}}}{x}
\end{array}
$

now, we know that V varies directly to T and inversely to P simultaneously
thus $\bf V=T\cdot \cfrac{k}{P}$

so $\bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k
\\\\\\
V=\cfrac{4T}{P}\qquad now\quad 
\begin{cases}
V=74\\
P=10
\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T$