Question

1. As beach temperatures rise, more female turtles are born. I want to show that more than 50% of the current turtle hatchlings are female, as this further threatens the turtle population. a. Define any relevant parameters and state the null and alternative hypotheses. b. If I want a margin of error of .01, what sample size should I get? c. I check 1000 eggs and find 676 of them are female. Use StatKey to find the SE, then use that SE to find a 99% confidence interval. d. Now use the formula discussed in chapter 6 for SE and create a 95% confidence interval. e. How do these intervals compare? f. Interpret the interval in the context of the problem. g. Find the z-score using the Statkey SE. What is the resulting p-value? h. Find the z-score using the formula SE. What is this resulting p-value? i. Compare these p-values. j. What is your formal decision and conclusion with context?

Answer 1

1.

a.
Given that,
possible chances (x)=676
sample size(n)=1000
success rate ( p )= x/n = 0.676
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.676-0.5/(sqrt(0.25)/1000)
zo =11.1312
| zo | =11.1312
critical value
the value of |z α| at los 0.01% is 2.33
we got |zo| =11.131 & | z α | =2.33
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 11.13122 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 11.1312
critical value: 2.33
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that more than 50% of the current turtle hatchlings are female,
as this further threatens the turtle population.
b.
margin of error =0.01
sample proportion=0.5
confidence level is 0.01
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.5
ME = 0.01
n = ( 2.576 / 0.01 )^2 * 0.5*0.5
= 16589.44 ~ 16590          
c.
TRADITIONAL METHOD
given that,
possible chances (x)=676
sample size(n)=1000
success rate ( p )= x/n = 0.676
I.
sample proportion = 0.676
standard error = Sqrt ( (0.676*0.324) /1000) )
= 0.0148
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.0148
= 0.0381
III.
CI = [ p ± margin of error ]
confidence interval = [0.676 ± 0.0381]
= [ 0.6379 , 0.7141]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=676
sample size(n)=1000
success rate ( p )= x/n = 0.676
CI = confidence interval
confidence interval = [ 0.676 ± 2.576 * Sqrt ( (0.676*0.324) /1000) ) ]
= [0.676 - 2.576 * Sqrt ( (0.676*0.324) /1000) , 0.676 + 2.576 * Sqrt ( (0.676*0.324) /1000) ]
= [0.6379 , 0.7141]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.6379 , 0.7141] 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
d.
TRADITIONAL METHOD
given that,
possible chances (x)=676
sample size(n)=1000
success rate ( p )= x/n = 0.676
I.
sample proportion = 0.676
standard error = Sqrt ( (0.676*0.324) /1000) )
= 0.0148
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 * 0.0148
= 0.029
III.
CI = [ p ± margin of error ]
confidence interval = [0.676 ± 0.029]
= [ 0.647 , 0.705]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=676
sample size(n)=1000
success rate ( p )= x/n = 0.676
CI = confidence interval
confidence interval = [ 0.676 ± 1.96 * Sqrt ( (0.676*0.324) /1000) ) ]
= [0.676 - 1.96 * Sqrt ( (0.676*0.324) /1000) , 0.676 + 1.96 * Sqrt ( (0.676*0.324) /1000) ]
= [0.647 , 0.705]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.647 , 0.705] 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
e.
99% sure that the interval [ 0.6379 , 0.7141]
95% sure that the interval [ 0.647 , 0.705]
f.
interpretations:
1. We are 95% sure that the interval [ 0.647 , 0.705] 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
g.
Given that,
possible chances (x)=676
sample size(n)=1000
success rate ( p )= x/n = 0.676
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,right tailed z α/2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.676-0.5/(sqrt(0.25)/1000)
zo =11.1312
| zo | =11.1312
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =11.131 & | z α | =1.64
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 11.13122 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
g.
test statistic: 11.1312
critical value: 1.64
decision: reject Ho
h.
p-value: 0
we have enough evidence to support the claim that more than 50% of the current turtle hatchlings are female,
i.
confidence level is 99% then p value =0
confidence level is 95%then p value = 0
j.
for both intervals 99%,95%
we have enough evidence to support the claim that more than 50% of the current turtle hatchlings are female