Suppose that the New York City apartments in a certain complex that have dogs is

Question

Suppose that the New York City apartments in a certain complex that have dogs is of interest to the city. A sample of 120 housing units in the complex is randomly done and they find that 32 of them contain those small foofy dogs you find in small apartments. They want to make sure that less than 25% of them contain a dog due to some regulation some politician put in at some point in time.

A) Test if the true proportion of dogs in the city apartments is less than 25%. Assume alpha is 10%. Do all the steps and interpret your results specifically.

B) Calculate a 90% confidence interval for the true proportion of apartments in the complex that contain dogs.

Explanation / Answer

PART A.
Given that,
possibile chances (x)=32
sample size(n)=120
success rate ( p )= x/n = 0.2667
success probability,( po )=0.25
failure probability,( qo) = 0.75
null, Ho:p<0.25
alternate, H1: p>0.25
level of significance, = 0.1
from standard normal table,right tailed z /2 =1.28
since our test is right-tailed
reject Ho, if zo > 1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.26667-0.25/(sqrt(0.1875)/120)
zo =0.4216
| zo | =0.4216
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =0.422 & | z | =1.28
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 0.42164 ) = 0.33664
hence value of p0.1 < 0.33664,here we do not reject Ho
ANSWERS
---------------
null, Ho:p<=0.25
alternate, H1: p>0.25
test statistic: 0.4216
critical value: 1.28
decision: do not reject Ho
p-value: 0.33664
evidence to support true proportion of dogs in the city apartments is less than 25%
PART B.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=32
Sample Size(n)=120
Sample proportion = x/n =0.267
Confidence Interval = [ 0.267 ±Z a/2 ( Sqrt ( 0.267*0.733) /120)]
= [ 0.267 - 1.645* Sqrt(0.002) , 0.267 + 1.65* Sqrt(0.002) ]
= [ 0.201,0.333]
Interpretations:
1) We are 90% sure that the interval [0.201 , 0.333 ] contains the true population proportion
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion  

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