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? 32 +1.645,8908 32 + 1.9620 0.64 + 1.282,16.620.) © 264 265 CHAT 0.04 + 1.00, layanan

Answer 1

Answer:

0.64 ± 0.1117 or

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

Step-by-step explanation:

Sample size (n) = 50

Z-score for a 90% confidence interval (z) = 1.645

Proportion of students that read at least one book (p):

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

The confidence interval is given by:

$p\pm z*\sqrt{\frac{p*(1-p)}{n} }$

Applying the given data:

$0.64\pm 1.645*\sqrt{\frac{0.64*(1-0.64)}{50}}\\ 0.64\pm 0.1117$

The confidence interval is 0.64 ± 0.1117