Answer:
0.64 ± 0.1117 or
[tex]0.64\pm 1.645*\sqrt{\frac{0.64*(0.36)}{50}}[/tex]
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):
[tex]p=\frac{32}{50}=0.64[/tex]
The confidence interval is given by:
[tex]p\pm z*\sqrt{\frac{p*(1-p)}{n} }[/tex]
Applying the given data:
[tex]0.64\pm 1.645*\sqrt{\frac{0.64*(1-0.64)}{50}}\\ 0.64\pm 0.1117[/tex]
The confidence interval is 0.64 ± 0.1117
The 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month is [tex]0.64\pm 0.11178[/tex].
Given information:
From a random sample of 50 (n) students, Alma found that 32 students read at least 1 book last month.
Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book.
It is required to find the 90 percentage confidence interval for the proportion of all students in her district who read at least 1 book last month.
Now, from tha table, the z-score of 90 percent confidence level is,
[tex]z=1.645[/tex]
The probability p for the proportion of students who read atleast one book is,
[tex]p=\dfrac{32}{50}\\p=0.64[/tex]
So, the 90 percent confidence interval will be calculated as,
[tex]p\pm z\sqrt{\dfrac{p(1-p)}{n}}=0.64\pm 1.645\times \sqrt{\dfrac{0.64(1-0.64)}{50}}\\=0.64\pm 0.11178[/tex]
Therefore, the 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month is [tex]0.64\pm 0.11178[/tex].
For more details, refer to the link:
https://brainly.com/question/16029228
