Answer:
A sample size of at least 531 is required.
Explanation:
We are lacking the confidence level to solve this question, so i am going to use a 90% confidence level.
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
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
Now, find the margin of error M as such
In which
is the standard deviation of the population and n is the size of the sample.
Find the required sample size for the new study.
A sample size of at least n is required.
n is found when
![M = 0.05](https://img.qammunity.org/2021/formulas/mathematics/college/47yeiqzhq1ju91lm96d7vaqeh74q4mdkuw.png)
We have that
![\sigma = 0.7](https://img.qammunity.org/2021/formulas/mathematics/college/xph3id9lqkyakhrc974iyv9lp7qm6y994f.png)
So
![0.05√(n) = 1.645*0.7](https://img.qammunity.org/2021/formulas/mathematics/college/hawhjm7t96oe3reztnpdtnt1q8e15s6iuq.png)
![√(n) = (1.645*0.7)/(0.05)](https://img.qammunity.org/2021/formulas/mathematics/college/vct0n0hae8itl00bzcrgu5xlswrm2csqfn.png)
![(√(n))^(2) = ((1.645*0.7)/(0.05))^(2)](https://img.qammunity.org/2021/formulas/mathematics/college/7ruv7xojpvkcivj6637zcjneu8kt15yycn.png)
![n = 530.4](https://img.qammunity.org/2021/formulas/mathematics/college/3mh6m7dzijjcho018x4wgbgen58npyjw0i.png)
Rounding up
A sample size of at least 531 is required.