Question

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random sample of 50 students, she found that 32 students read at least 1 book last month. Assuming all conditions for Inference are met, which of the following defines a 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month?
A. 32 +1.645,√(32)(18/50)
B. 32+ 1.9620√ √(32)(18/50)
C. 0.64 + 1.282√(0.64 )(0.36)/ 50
D. 0.64 + 1.645√(0.64 )(0.36)/ 50
E. 0.64 + 1.96√(0.64 )(0.36)/ 50

Answer 1

Answer:

D)

$(0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })$

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

                       $p^{-} = \frac{32}{50} = 0.64$

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

$(p^{-} - Z_{0.90} \sqrt{\frac{p(1-p)}{n}, } , p^{-} +Z_{0.90} \sqrt{\frac{p(1-p)}{n} })$

Step(ii):-

Level of significance = 0.90

Z₀.₉₀ = 1.645

$(0.64 - 1.645 \sqrt{\frac{0.64(1-0.64)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(1-0.64)}{50} })$

$(0.64 - 1.645 \sqrt{\frac{0.64(0.36)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })$

(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)