Answer:
C.25
Explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = (1-0.95)/(2) = 0.025](https://img.qammunity.org/2021/formulas/mathematics/college/b2sgcgxued5x1354b5mv9i43o4qgtn8yk6.png)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so
![z = 1.96](https://img.qammunity.org/2021/formulas/mathematics/college/zv05k6fi2atwaveb38qmkwkmh0vcr5vhx2.png)
Now, find the margin of error M as such
![M = z*(\sigma)/(√(n))](https://img.qammunity.org/2021/formulas/mathematics/college/cvh8tdoppqkhyobio78yaazk1nqj1870w9.png)
In which
is the standard deviation of the population and n is the size of the sample.
Which of the following is the smallest sample the company can take to achieve the desired margin or error?
This is n when
![\sigma = 10, M = 4](https://img.qammunity.org/2021/formulas/mathematics/high-school/imaf3fz0w815b05tgsnnn892gy4hy2xfup.png)
![M = z*(\sigma)/(√(n))](https://img.qammunity.org/2021/formulas/mathematics/college/cvh8tdoppqkhyobio78yaazk1nqj1870w9.png)
![4 = 1.96*(10)/(√(n))](https://img.qammunity.org/2021/formulas/mathematics/high-school/2nzdqyzzl8gyyx5gd0oyotrlj47l3wdkdn.png)
![4√(n) = 1.96*10](https://img.qammunity.org/2021/formulas/mathematics/high-school/h1gkcm8mjjf47tz27cmwzdva4wdo5eevtq.png)
![√(n) = (1.96*10)/(4)](https://img.qammunity.org/2021/formulas/mathematics/high-school/5jfjnqfxtfjn8t08oamnqrxcdncdokxjlx.png)
![(√(n))^(2) = ((1.96*10)/(4))^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/aghb9vi2lduyhr93sdn07osp9trpij79mz.png)
![n = 24.01](https://img.qammunity.org/2021/formulas/mathematics/college/ur02prpz8v4g1u93t6836j7ldovq21kw2g.png)
We have to round up.
So the correct answer is:
C.25