Answer:
Step-by-step explanation:
Given that the proportion for the random sample of 75 residents from the town who would like to see more trees in the park is
[tex]p=\frac{27}{75} =\frac{9}{25} \\
q=1-p=\frac{16}{25} \
\Mean of the sample proportion P = \frac{9}{25}\\
Std error of proportion =\sqrt{\frac{pq}{n} } \\
=\sqrt{\frac{9(16)}{25(25)(75)} }\\
=0.003072\\
Margin of error = 1.95(std error) = \\
=0.006\\[/tex]
We assumed 95% confidence level
The random sample gives an idea about the whole town preferences for trees planted
We have sample proportion as 0.36 and
confidence interval 95% =(0,36-0.006, 0.+0.006)
=(0.354,0.366)
We are 95% confident that when different samples for large sizes are taken, people who prefer trees to be planted would be within 0.354 and 0.366
or between 35.4% and 36.6%