Respuesta :
Answer:
D)
[tex](0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })[/tex]
90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month
(0.52846 , 0.75166)
Step-by-step explanation:
Step:-1
Given that the random sample size 'n' = 50
The sample proportion
[tex]p^{-} = \frac{32}{50} = 0.64[/tex]
90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month
[tex](p^{-} - Z_{0.90} \sqrt{\frac{p(1-p)}{n}, } , p^{-} +Z_{0.90} \sqrt{\frac{p(1-p)}{n} })[/tex]
Step(ii):-
Level of significance = 0.90
Z₀.₉₀ = 1.645
[tex](0.64 - 1.645 \sqrt{\frac{0.64(1-0.64)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(1-0.64)}{50} })[/tex]
[tex](0.64 - 1.645 \sqrt{\frac{0.64(0.36)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })[/tex]
(0.64 -1.645(0.06788) , (0.64 + 1.645(0.06788)
(0.52846 , 0.75166)
Final answer:-
90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month
(0.52846 , 0.75166)
