Answer:
Answer:
0.25±0.095
Step-by-step explanation:
P^= 20/80=.25
sample size n= 80
z*=1.96
p^± sqr(p^(1-p^)/n)
=0.095
Step-by-step explanation:
The 95% confidence interval for the proportion of all students at the school that ate breakfast every day last week is 0.25 +0.080.
Fatimah wants to estimate the proportion of students at her school who ate breakfast every day last week.
She takes an SRS of 80 of the 1000 students at her school and finds that 20 of those sampled ate breakfast every day last week.
Based on this sample, which of the following is a 95% confidence interval for the proportion of all students at the school that ate breakfast every day last week?
She takes an SRS of 80 of the 1000 students at her school and finds that 20 of those sampled ate breakfast every day last week.
[tex]\rm P = \dfrac{Sample \ space}{Total \ SRS}\\ \\ P = \dfrac{20}{80}\\ \\ P = \dfrac{1}{4}\\ \\ P = 0.25[/tex]
Therefore,
The 95% confidence interval for the proportion of all students at the school that ate breakfast every day last week is
=[tex] = 1.96 \times \sqrt{\dfrac{0.25(1-0.25)}{80}}\\ \\ = 1.96 \times \sqrt{\dfrac{0.25\times 0.75}{80}}\\\\ = 1.96 \times 0.40\\\\ = 0.080[/tex]
Hence, The 95% confidence interval for the proportion of all students at the school that ate breakfast every day last week is 0.25 +0.080.
To know more about Confidence intervals click the link given below.
https://brainly.com/question/15682003
