Question

a polling agency reported that the proportion of 12th grade students who are engaged is 0.34. engagement is defined as students involvement and enthusiasm with school. the superintendent of one high school district believes the engagement of the districts students is higher than the proportion reported. the district randomly surveys 90 of the 3600 students in the district and finds that 38 reported being engaged in school. complete parts (a) through (e).
the variable is qualitative.
(b) what is the sample proportion of students in this high school district who are engaged?
\\( \hat { p } = 0.4222 \\)
(round to four decimal places as needed.)
(c) verify that the distribution of the sample proportion is approximately normal.
the sample size, \\( n = 90 \\), is less than 5% of the population size, or \\( 0.05 n = 180 \\)
(type integers or decimals. do not round.)
for this sample, \\( n \hat { p } ( 1 - \hat { p } ) = 22.0 \geq 10 \\)
(round to one decimal place as needed.)
(d) assuming that the proportion of students in this high school district is the same as what was reported by the polling agency, determine the mean and standard deviation of the distribution of the sample proportion.
\\( mu _ { hat { p } } = \\)
\\( sigma _ { hat { p } } = \\)
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the mean of the sample proportion

The mean of the sample proportion \(\mu_{\hat{p}}\) is equal to the population proportion \(p\). Given \(p = 0.34\), so \(\mu_{\hat{p}}=p\).
\(\mu_{\hat{p}} = 0.34\)

Step2: Calculate the standard deviation of the sample proportion

The formula for the standard deviation of the sample proportion is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\).
Substitute \(p = 0.34\) and \(n = 90\) into the formula:

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Answer:

\(\mu_{\hat{p}} = 0.34\), \(\sigma_{\hat{p}}\approx0.0499\)