contestada

A recent poll of 80 randomly selected Californians showed that 38% ( = 0.38) believe they are doing all that they can to conserve water.

The state government would like to know, within a 99% confidence level, the margin of error for this poll. (99% confidence level z*-score of 2.58)

Remember, the margin of error, E, can be determined using the formula E = z*.

To the nearest whole percent, the margin of error for the poll is %.

Respuesta :

merlynthewhizz
The margin of error given the proportion can be found using the formula

[tex]z* \sqrt{ \frac{p(1-p)}{n} } [/tex]

Where
[tex]z*[/tex] is the z-score of the confidence level
[tex]p[/tex] is the sample proportion
[tex]n[/tex] is the sample size

We have
[tex]z*=2.58[/tex]
[tex]p=0.38[/tex]
[tex]n=80[/tex]

Plugging these values into the formula, we have:
[tex]2.58 \sqrt{ \frac{0.38(1-0.38)}{80} } =0.14[/tex]

The result 0.14 as percentage is 14%

Margin error is 38% ⁺/₋ 14%
JeanaShupp

Answer:  14%


Step-by-step explanation:

We know that Margin error [tex]M.E.=z\sqrt{\frac{p(1-p)}{n}}[/tex]

Given: Sample size n=80

sample proportion p=0.38

Confidence level z=2.58

Substitute the values in the formula, we get

Margin error [tex]M.E.=2.58\sqrt{\frac{0.38(1-0.38)}{80}}[/tex]

[tex]\Rightarrow\ M.E.=2.58\sqrt{\frac{0.62\ \cdot\ 0.38}{80}}\\\Rightarrow\ M.E.=2.58 \sqrt{0.002945}\\\Rightarrow\ M.E.=2.58\times0.054\\\Rightarrow\ M.E.=0.13932\\\Rightarrow\ M.E.=13.932\%\approx14\%[/tex]


Thus, to the nearest whole percent, the margin of error for the poll is 14%




Otras preguntas