Respuesta :
Answer:
a) [tex]ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423[/tex]
b) [tex]0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371[/tex]
Step-by-step explanation:
1) Data given and notation
n=900 represent the random sample taken
X=372 represent the students were pursuing liberal arts degrees
[tex]\hat p=\frac{372}{900}=0.413[/tex] estimated proportion of students were pursuing liberal arts degrees
[tex]\alpha=0.01[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
p= population proportion of students were pursuing liberal arts degrees
2) Solution to the problem
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
The margin of error is given by:
[tex]ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
If we replace we have:
[tex]ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423[/tex]
And replacing into the confidence interval formula we got:
[tex]0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371[/tex]
[tex]0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455[/tex]
And the 99% confidence interval would be given (0.371;0.455).
We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.
