Scientists are interested in the amount of time spent looking for predators by s

Question

Scientists are interested in the amount of time spent looking for predators by seals. They observe 12 seals at the beginning of haul-out. Furthermore, they observed them again at the end of haul-out, and their time looking for predators is recorded. The data looks like looks like

A) Construct a 99% confidence interval of the mean time the seal spent looking for predators at the beginning of the haul-out. B) Scientists are interested in whether there is a difference between the amounts of time spent looking for predators by seals at the beginning of a haul-out compared to the end of the haul-out. Formulate the hypothesis test using significance level 0.01. C) Can you test the hypothesis using confidence interval instead? If yes, how do you do it and make a conclusion? (just explain, no need to get the CI) If no, why?

DO NOT use excel, please go through all necessary steps

looks like 12 difference Time spent at the beginning of haul-out 1.83 1.12 Time spent at the End of haul-out 0.75 0.98 Seal #1 Seal #2 1.08 0.14 Seal #12 Mean Variance 1.43 1.21 0.20 1.18 0.87 0.08 3 0 2 0.25 0.34 0.15

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
sample mean, x =1.21
standard deviation, s =0.4472
sample size, n =12
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.4472/ sqrt ( 12) )
= 0.129
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 3.106
margin of error = 3.106 * 0.129
= 0.401
III.
CI = x ± margin of error
confidence interval = [ 1.21 ± 0.401 ]
= [ 0.809 , 1.611 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1.21
standard deviation, s =0.4472
sample size, n =12
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 3.106
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 1.21 ± t a/2 ( 0.4472/ Sqrt ( 12) ]
= [ 1.21-(3.106 * 0.129) , 1.21+(3.106 * 0.129) ]
= [ 0.809 , 1.611 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 0.809 , 1.611 ] contains the true population mean
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 mean

b.
Given that,
mean(x)=1.21
standard deviation , s.d1=0.4472
number(n1)=12
y(mean)=0.87
standard deviation, s.d2 =0.2828
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.106
since our test is two-tailed
reject Ho, if to < -3.106 OR if to > 3.106
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.21-0.87/sqrt((0.19999/12)+(0.07998/12))
to =2.226
| to | =2.226
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 3.106
we got |to| = 2.22597 & | t | = 3.106
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.226 ) = 0.048
hence value of p0.01 < 0.048,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.226
critical value: -3.106 , 3.106
decision: do not reject Ho
p-value: 0.048

c.
yes,
confidence interval using instead of hypothesis test

TRADITIONAL METHOD
given that,
mean(x)=1.21
standard deviation , s.d1=0.7742
number(n1)=12
y(mean)=0.87
standard deviation, s.d2 =0.2828
number(n2)=12
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0.599/12)+(0.08/12))
= 0.238
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 11 d.f is 3.106
margin of error = 3.106 * 0.238
= 0.739
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1.21-0.87) ± 0.739 ]
= [-0.399 , 1.079]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=1.21
standard deviation , s.d1=0.7742
sample size, n1=12
y(mean)=0.87
standard deviation, s.d2 =0.2828
sample size,n2 =12
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 1.21-0.87) ± t a/2 * sqrt((0.599/12)+(0.08/12)]
= [ (0.34) ± t a/2 * 0.238]
= [-0.399 , 1.079]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-0.399 , 1.079] 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

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