Question

Out of a survey of 20 people 3 people said they do not like the taste of cinnamon and 17 people said they liked the taste.

For this sample, complete the following:

a. Calculate the proportion who answered “yes” and the proportion who answered “no.”

b. Calculate the 95% Confidence Interval for the proportion who answered “yes.”

c. Using your survey results, test your hypothesis.

Show work for all steps including the formulas. Calculate the p-value for the test. Use α = 0.05 to find your conclusion.

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=17
sample size(n)=20
success rate ( p )= x/n = 0.85
I.
sample proportion = 0.85
standard error = Sqrt ( (0.85*0.15) /20) )
= 0.08
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.08
= 0.156
III.
CI = [ p ± margin of error ]
confidence interval = [0.85 ± 0.156]
= [ 0.694 , 1.006]
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DIRECT METHOD
given that,
possible chances (x)=17
sample size(n)=20
success rate ( p )= x/n = 0.85
CI = confidence interval
confidence interval = [ 0.85 ± 1.96 * Sqrt ( (0.85*0.15) /20) ) ]
= [0.85 - 1.96 * Sqrt ( (0.85*0.15) /20) , 0.85 + 1.96 * Sqrt ( (0.85*0.15) /20) ]
= [0.694 , 1.006]
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interpretations:
1. We are 95% sure that the interval [ 0.694 , 1.006] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
a.
proportion who answered yes is 0.85
the proportion who answered no is 0.15
b.
95% sure that the interval [ 0.694 , 1.006]
c.
Given that,
possibile chances (x)=17
sample size(n)=20
success rate ( p )= x/n = 0.85
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.85-0.5/(sqrt(0.25)/20)
zo =3.1305
| zo | =3.1305
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =3.13 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.1305 ) = 0.00175
hence value of p0.05 > 0.0017,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: 3.1305
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00175
we have enough evidence to support the claim that the proportion who answered yes.