Problem 2 (20 points). You conduct a poll of 250 FIU like to see an Arby\'s rest
ID: 3317649 • Letter: P
Question
Problem 2 (20 points). You conduct a poll of 250 FIU like to see an Arby's restaurant on campus. You find that 180 out of 250 F Arby' students and inquire as to whether students would U students I would like to see an s restaurant on campus and 70 FIU students would not. a) Create a 99% confidence interval for the proportion p, the percentage of all FIU students who would like to see an Arby's restaurant on campus. conduct a two-sided Hypothesis test for Ho: p-0.65 (=0.01). Compute the resulting p-value State whether you would reject Ho and why.Explanation / Answer
PART A.
TRADITIONAL METHOD
given that,
possibile chances (x)=180
sample size(n)=250
success rate ( p )= x/n = 0.72
I.
sample proportion = 0.72
standard error = Sqrt ( (0.72*0.28) /250) )
= 0.0284
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
margin of error = 2.576 * 0.0284
= 0.0732
III.
CI = [ p ± margin of error ]
confidence interval = [0.72 ± 0.0732]
= [ 0.6468 , 0.7932]
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DIRECT METHOD
given that,
possibile chances (x)=180
sample size(n)=250
success rate ( p )= x/n = 0.72
CI = confidence interval
confidence interval = [ 0.72 ± 2.576 * Sqrt ( (0.72*0.28) /250) ) ]
= [0.72 - 2.576 * Sqrt ( (0.72*0.28) /250) , 0.72 + 2.576 * Sqrt ( (0.72*0.28) /250) ]
= [0.6468 , 0.7932]
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interpretations:
1. We are 99% sure that the interval [ 0.6468 , 0.7932] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
PART B.
Given that,
possibile chances (x)=180
sample size(n)=250
success rate ( p )= x/n = 0.72
success probability,( po )=0.65
failure probability,( qo) = 0.35
null, Ho:p=0.65
alternate, H1: p!=0.65
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.72-0.65/(sqrt(0.2275)/250)
zo =2.3205
| zo | =2.3205
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =2.32 & | z | =2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.32048 ) = 0.02032
hence value of p0.01 < 0.0203,here we do not reject Ho
ANSWERS
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null, Ho:p=0.65
alternate, H1: p!=0.65
test statistic: 2.3205
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.02032