Answer:
[tex]\mu_{\hat{p}}[/tex] = 0.42
[tex]\sigma_{\hat{p}}[/tex] = 0.0349
Step-by-step explanation:
Given:
Probability that students surveyed rarely or never use academic advising services = 32% = 0.32
In reality, students rarely or never use academic advising services at their college = 42% = 0.42
Sample size, n = 200
The sampling distribution of sample proportion will be approximately normal with mean
therefore,
Mean,[tex]\mu_{\hat{p}}[/tex] = p
or
[tex]\mu_{\hat{p}}[/tex] = 0.42
now,
the standard deviation is given using the formula
[tex]\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}[/tex]
on substituting the respective values, we get
[tex]\sigma_{\hat{p}}=\sqrt{\frac{0.42(1-0.42)}{200}}[/tex]
or
[tex]\sigma_{\hat{p}}[/tex]= √0.001218
or
[tex]\sigma_{\hat{p}}[/tex] = 0.0349