Answer:
151 students must be selected.
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/2022/formulas/mathematics/college/k8m2vmetmk326pc3hdyvi0d7k37r14zn45.png)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error M as such
![M = z(\sigma)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/p19w5m3ctzqxc0b7ic9kz7y4ab19d7zpbv.png)
In which
is the standard deviation of the population and n is the size of the sample.
The population standard deviation is known to be $25.
This means that
![\sigma = 25](https://img.qammunity.org/2022/formulas/mathematics/college/musuwsiyss94shwufmpbx4v6npa2z1xcl0.png)
How many students must be randomly selected to estimate the mean weekly earnings of students at one college? Sample mean within $4 of the population mean
This is n for which M = 4. So
![M = z(\sigma)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/p19w5m3ctzqxc0b7ic9kz7y4ab19d7zpbv.png)
![4 = 1.96(25)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/x9x1zabcrtu81bldj8gcskygqwliwbtq3e.png)
![4√(n) = 1.96*25](https://img.qammunity.org/2022/formulas/mathematics/college/h02vv3alonogtr8lqpvkpwuexgi7607vfu.png)
![√(n) = (1.96*25)/(4)](https://img.qammunity.org/2022/formulas/mathematics/college/q8fl2doi2tatbzb223ngjd2y09yuqx4upm.png)
![(√(n))^2 = ((1.96*25)/(4))^2](https://img.qammunity.org/2022/formulas/mathematics/college/nnnsym535e50et5upz3rlu9bc0ej3hlu6t.png)
![n = 150.1](https://img.qammunity.org/2022/formulas/mathematics/college/kobdkpuobvce65q5mn1f46ti993w63kb6x.png)
Rounding up:
151 students must be selected.