Question

Use the formulas provided in class to construct a 95% confidence interval for the proportion of all Americans who have blue eyes.

- Interpret this confidence interval in the context of the research question.

- It is claimed that blue eyes are becoming less common, and it estimated that in 2006, one out of six Americans had blue eyes. Do a hypothesis test to see if our class provides evidence that blue eyes are becoming less common. What are the hypotheses?

- Compute the z-score and find the P-value

- Interpret your results in the context of the research question. Do the results from the confidence interval and hypothesis test agree or disagree? Explain.

Data: 4/13 have blue eyes in the class

PLEASE HELP!!

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=4
sample size(n)=13
success rate ( p )= x/n = 0.3077
I.
sample proportion = 0.3077
standard error = Sqrt ( (0.3077*0.6923) /13) )
= 0.128
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 * 0.128
= 0.2509
III.
CI = [ p ± margin of error ]
confidence interval = [0.3077 ± 0.2509]
= [ 0.0568 , 0.5586]
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DIRECT METHOD
given that,
possible chances (x)=4
sample size(n)=13
success rate ( p )= x/n = 0.3077
CI = confidence interval
confidence interval = [ 0.3077 ± 1.96 * Sqrt ( (0.3077*0.6923) /13) ) ]
= [0.3077 - 1.96 * Sqrt ( (0.3077*0.6923) /13) , 0.3077 + 1.96 * Sqrt ( (0.3077*0.6923) /13) ]
= [0.0568 , 0.5586]
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interpretations:
1. We are 95% sure that the interval [ 0.0568 , 0.5586] 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

b.
Given that,
possible chances (x)=4
sample size(n)=13
success rate ( p )= x/n = 0.3077
success probability,( po )=0.1667
failure probability,( qo) = 0.8333
null, Ho:p=0.1667
alternate, H1: p<0.1667
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.30769-0.1667/(sqrt(0.13891111)/13)
zo =1.364
| zo | =1.364
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =1.364 & | z α | =1.64
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: left tail - Ha : ( p < 1.36395 ) = 0.91371
hence value of p0.05 < 0.91371,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.1667
alternate, H1: p<0.1667
test statistic: 1.364
critical value: -1.64
decision: do not reject Ho
p-value: 0.91371
we do not have enough evidence to support the claim that blue eyes are becoming less common.