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).
(a) what type of variable is \engaged\?
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 } ) = \square \\) 10.
(round to one decimal place as needed.)

Explanation:

Step1: Substitute the values into the formula

We know that \(n = 90\) and \(\hat{p}=0.4222\). The formula is \(n\hat{p}(1 - \hat{p})\).

$$n\hat{p}(1 - \hat{p})=90\times0.4222\times(1 - 0.4222)$$

Step2: Calculate \((1-\hat{p})\)

First, calculate \(1 - 0.4222=0.5778\)

$$90\times0.4222\times0.5778$$

Step3: Calculate \(90\times0.4222\)

\(90\times0.4222 = 37.998\)

$$37.998\times0.5778$$

Step4: Calculate the final result

\(37.998\times0.5778\approx21.97\approx22.0\) (rounded to one decimal place)

Answer:

\(22.0\)